Chapter 9

Sources of Magnetic Fields

9.1 Biot-Savart Law....................................................................................................... 2

Interactive Simulation 9.1: Magnetic Field of a Current Element.......................... 3 Example 9.1: Magnetic Field due to a Finite Straight Wire ...................................... 3 Example 9.2: Magnetic Field due to a Circular Current Loop .................................. 6 9.1.1 Magnetic Field of a Moving Point Charge ....................................................... 9 Animation 9.1: Magnetic Field of a Moving Charge ............................................. 10 Animation 9.2: Magnetic Field of Several Charges Moving in a Circle................ 11 Interactive Simulation 9.2: Magnetic Field of a Ring of Moving Charges .......... 11

9.2 Force Between Two Parallel Wires ....................................................................... 12

Animation 9.3: Forces Between Current-Carrying Parallel Wires......................... 13

9.3 Ampere’s Law........................................................................................................ 13

Example 9.3: Field Inside and Outside a Current-Carrying Wire............................ 16 Example 9.4: Magnetic Field Due to an Infinite Current Sheet .............................. 17

9.4 Solenoid ................................................................................................................. 19

Examaple 9.5: Toroid............................................................................................... 22

9.5 Magnetic Field of a Dipole .................................................................................... 23

9.5.1 Earth’s Magnetic Field at MIT ....................................................................... 24 Animation 9.4: A Bar Magnet in the Earth’s Magnetic Field ................................ 26

9.6 Magnetic Materials ................................................................................................ 27

9.6.1 Magnetization ................................................................................................. 27 9.6.2 Paramagnetism................................................................................................ 30 9.6.3 Diamagnetism ................................................................................................. 31 9.6.4 Ferromagnetism .............................................................................................. 31

9.7 Summary................................................................................................................ 32

9.8 Appendix 1: Magnetic Field off the Symmetry Axis of a Current Loop............... 34

9.9 Appendix 2: Helmholtz Coils ................................................................................ 38

Animation 9.5: Magnetic Field of the Helmholtz Coils ......................................... 40 Animation 9.6: Magnetic Field of Two Coils Carrying Opposite Currents ........... 42 Animation 9.7: Forces Between Coaxial Current-Carrying Wires......................... 43

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Animation 9.8: Magnet Oscillating Between Two Coils ....................................... 43 Animation 9.9: Magnet Suspended Between Two Coils........................................ 44

9.10 Problem-Solving Strategies ................................................................................. 45

9.10.1 Biot-Savart Law:........................................................................................... 45 9.10.2 Ampere’s law: ............................................................................................... 47

9.11 Solved Problems .................................................................................................. 48

9.11.1 Magnetic Field of a Straight Wire ................................................................ 48 9.11.2 Current-Carrying Arc.................................................................................... 50 9.11.3 Rectangular Current Loop............................................................................. 51 9.11.4 Hairpin-Shaped Current-Carrying Wire........................................................ 53 9.11.5 Two Infinitely Long Wires ........................................................................... 54 9.11.6 Non-Uniform Current Density ...................................................................... 56 9.11.7 Thin Strip of Metal........................................................................................ 58 9.11.8 Two Semi-Infinite Wires .............................................................................. 60

9.12 Conceptual Questions .......................................................................................... 61

9.13 Additional Problems ............................................................................................ 62

9.13.1 Application of Ampere's Law ....................................................................... 62 9.13.2 Magnetic Field of a Current Distribution from Ampere's Law..................... 62 9.13.3 Cylinder with a Hole..................................................................................... 63 9.13.4 The Magnetic Field Through a Solenoid ...................................................... 64 9.13.5 Rotating Disk ................................................................................................ 64 9.13.6 Four Long Conducting Wires ....................................................................... 64 9.13.7 Magnetic Force on a Current Loop............................................................... 65 9.13.8 Magnetic Moment of an Orbital Electron..................................................... 65 9.13.9 Ferromagnetism and Permanent Magnets..................................................... 66 9.13.10 Charge in a Magnetic Field......................................................................... 67 9.13.11 Permanent Magnets..................................................................................... 67 9.13.12 Magnetic Field of a Solenoid...................................................................... 67 9.13.13 Effect of Paramagnetism............................................................................. 68

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Sources of Magnetic Fields 9.1 Biot-Savart Law Currents which arise due to the motion of charges are the source of magnetic fields. When charges move in a conducting wire and produce a current I, the magnetic field at any point P due to the current can be calculated by adding up the magnetic field contributions, , from small segments of the wire ddB s , (Figure 9.1.1).

Figure 9.1.1 Magnetic field dB at point P due to a current-carrying element I d s .

These segments can be thought of as a vector quantity having a magnitude of the length of the segment and pointing in the direction of the current flow. The infinitesimal current source can then be written as I d s .

Let r denote as the distance form the current source to the field point P, and the corresponding unit vector. The Biot-Savart law gives an expression for the magnetic field contribution, , from the current source,

r

dB Id s ,

02

ˆ4

I ddr

µπ

×=

s rB (9.1.1)

where 0µ is a constant called the permeability of free space: (9.1.2) 7

0 4 10 T m/Aµ π −= × ⋅ Notice that the expression is remarkably similar to the Coulomb’s law for the electric field due to a charge element dq:

20

1 ˆ4

dqdrπε

=E r (9.1.3)

Adding up these contributions to find the magnetic field at the point P requires integrating over the current source,

2

02

wire wire

ˆ4

I ddr

µπ

×= =∫ ∫

s rB B (9.1.4)

The integral is a vector integral, which means that the expression for B is really three integrals, one for each component of B . The vector nature of this integral appears in the cross product . Understanding how to evaluate this cross product and then perform the integral will be the key to learning how to use the Biot-Savart law.

ˆI d ×s r

Interactive Simulation 9.1: Magnetic Field of a Current Element Figure 9.1.2 is an interactive ShockWave display that shows the magnetic field of a current element from Eq. (9.1.1). This interactive display allows you to move the position of the observer about the source current element to see how moving that position changes the value of the magnetic field at the position of the observer.

Figure 9.1.2 Magnetic field of a current element. Example 9.1: Magnetic Field due to a Finite Straight Wire A thin, straight wire carrying a current I is placed along the x-axis, as shown in Figure 9.1.3. Evaluate the magnetic field at point P. Note that we have assumed that the leads to the ends of the wire make canceling contributions to the net magnetic field at the point . P

Figure 9.1.3 A thin straight wire carrying a current I.

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Solution: This is a typical example involving the use of the Biot-Savart law. We solve the problem using the methodology summarized in Section 9.10. (1) Source point (coordinates denoted with a prime) Consider a differential element ˆ'd dx=s i carrying current I in the x-direction. The location of this source is represented by ˆ' 'x=r i . (2) Field point (coordinates denoted with a subscript “P”) Since the field point P is located at ( , ) (0, )x y a= , the position vector describing P is

. ˆP a=r j

(3) Relative position vector The vector is a “relative” position vector which points from the source point

to the field point. In this case,

'P= −r r rˆ 'a x= −r j i , and the magnitude 2| | 'r a= = +r 2x is the

distance from between the source and P. The corresponding unit vector is given by

2 2

ˆ ˆ' ˆ ˆˆ sin cos'

a xr a x

θ θ−= = = −

+

r j ir j i

(4) The cross product ˆd ×s r The cross product is given by ˆ ˆ ˆ ˆˆ ( ' ) ( cos sin ) ( 'sin )d dx dxθ θ θ× = × − + =s r i i j k (5) Write down the contribution to the magnetic field due to Id s The expression is

0 02 2

ˆ sin ˆ4 4

I Id dxdr r

µ µ θπ π

×= =

s rB k

which shows that the magnetic field at P will point in the ˆ+k direction, or out of the page. (6) Simplify and carry out the integration

4

The variables θ, x and r are not independent of each other. In order to complete the integration, let us rewrite the variables x and r in terms of θ. From Figure 9.1.3, we have

2

/ sin csc

cot csc

r a a

x a dx a

θ θ

dθ θ θ

= =⎧⎪⎨

= ⇒ = −⎪⎩

Upon substituting the above expressions, the differential contribution to the magnetic field is obtained as

2

0 02

( csc )sin sin4 ( csc ) 4

I Ia ddB da a

µ µθ θ θ θ θπ θ π

−= = −

Integrating over all angles subtended from 1θ− to 2θ (a negative sign is needed for 1θ in order to take into consideration the portion of the length extended in the negative x axis from the origin), we obtain

2

1

0 02sin (cos cos )

4 4I IB da a

θ

θ 1µ µθ θ θπ π−

= − = +∫ θ (9.1.5)

The first term involving 2θ accounts for the contribution from the portion along the +x axis, while the second term involving 1θ contains the contribution from the portion along the axis. The two terms add! x− Let’s examine the following cases: (i) In the symmetric case where 2 1θ θ= − , the field point P is located along the

perpendicular bisector. If the length of the rod is 2L , then 21cos / 2L L aθ = + and the

magnetic field is

0 01 2 2

cos2 2

I I LBa a L a

µ µθπ π

= =+

(9.1.6)

(ii) The infinite length limit L → ∞ This limit is obtained by choosing ( ,1 2 ) (0,0)θ θ = . The magnetic field at a distance a away becomes

0

2IBa

µπ

= (9.1.7)

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Note that in this limit, the system possesses cylindrical symmetry, and the magnetic field lines are circular, as shown in Figure 9.1.4.

Figure 9.1.4 Magnetic field lines due to an infinite wire carrying current I. In fact, the direction of the magnetic field due to a long straight wire can be determined by the right-hand rule (Figure 9.1.5).

Figure 9.1.5 Direction of the magnetic field due to an infinite straight wire If you direct your right thumb along the direction of the current in the wire, then the fingers of your right hand curl in the direction of the magnetic field. In cylindrical coordinates ( , , )r zϕ where the unit vectors are related by ˆ ˆ ˆ× =r φ z , if the current flows in the +z-direction, then, using the Biot-Savart law, the magnetic field must point in the ϕ -direction. Example 9.2: Magnetic Field due to a Circular Current Loop A circular loop of radius R in the xy plane carries a steady current I, as shown in Figure 9.1.6. (a) What is the magnetic field at a point P on the axis of the loop, at a distance z from the center? (b) If we place a magnetic dipole ˆ

zµ=µ k at P, find the magnetic force experienced by the dipole. Is the force attractive or repulsive? What happens if the direction of the dipole is reversed, i.e., ˆ

zµ= −µ k

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Figure 9.1.6 Magnetic field due to a circular loop carrying a steady current. Solution: (a) This is another example that involves the application of the Biot-Savart law. Again let’s find the magnetic field by applying the same methodology used in Example 9.1. (1) Source point In Cartesian coordinates, the differential current element located at

ˆ' (cos ' sin 'R ˆ)φ φ= +r i j can be written as ˆ ˆ( '/ ') ' '( sin ' cos ' )Id I d d d IRdφ φ φ φ φ= = − +s r i j . (2) Field point Since the field point P is on the axis of the loop at a distance z from the center, its position vector is given by ˆ

P z=r k . (3) Relative position vector 'P= −r r r The relative position vector is given by ˆ ˆ ˆ' cos ' sin 'P R Rφ φ− = − − +r = r r i j kz and its magnitude

( )22 2( cos ') sin 'r R R z Rφ φ= = − + − + = +r 2 2z (9.1.9) is the distance between the differential current element and P. Thus, the corresponding unit vector from Id s to P can be written as

'ˆ| '

P

Pr |−

= =−

r rrrr r

7

(9.1.8)

(4) Simplifying the cross product The cross product can be simplified as ( ')Pd × −s r r

( )ˆ ˆ ˆ ˆ ˆ( ') ' sin ' cos ' [ cos ' sin ' ]

ˆ ˆ ˆ'[ cos ' sin ' ]

Pd R d R R z

R d z z R

φ φ φ φ φ

φ φ φ

× − = − + × − − +

= + +

s r r i j i j k

i j k (9.1.10)

(5) Writing down dB Using the Biot-Savart law, the contribution of the current element to the magnetic field at P is

0 0 02 3

02 2 3/ 2

ˆ ( '4 4 4 |

ˆ ˆ ˆcos ' sin ' '4 ( )

P

P

I I I dd ddr r

IR z z R dR z

3

)' |

µ µ µπ π π

µ φ φ φπ

× −× ×= = =

−

+ +=

+

s r rs r s rBr r

i j k (9.1.11)

(6) Carrying out the integration Using the result obtained above, the magnetic field at P is

20

2 2 3/ 20

ˆ ˆ ˆcos ' sin ' '4 ( )

IR z z R dR z

πµ φ φ φπ

+ +=

+∫i j kB (9.1.12)

The x and the y components of B can be readily shown to be zero:

20 0

2 2 3/ 2 2 2 3/ 20

2cos ' ' sin ' 0

04 ( ) 4 ( )xIRz IRzB d

R z R zπ πµ µφ φ φ

π π= =

+ +∫ = (9.1.13)

20 0

2 2 3/ 2 2 2 3/ 20

2sin ' ' cos ' 0

04 ( ) 4 ( )yIRz IRzB d

R z R zπ πµ µφ φ φ

π π= = −

+ +∫ = (9.1.14)

On the other hand, the z component is

22 220 0

2 2 3/ 2 2 2 3/ 2 2 2 3/ 20

2'4 ( ) 4 ( ) 2( )z

IRIR IRB dR z R z R z

πµ µ πφπ π

= = =+ +∫ 0µ

+ (9.1.15)

Thus, we see that along the symmetric axis, zB is the only non-vanishing component of the magnetic field. The conclusion can also be reached by using the symmetry arguments.

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The behavior of 0/zB B where 0 0 / 2B I Rµ= is the magnetic field strength at , as a function of

0z =/z R is shown in Figure 9.1.7:

Figure 9.1.7 The ratio of the magnetic field, 0/zB B , as a function of /z R (b) If we place a magnetic dipole ˆ

zµ=µ k at the point P, as discussed in Chapter 8, due to the non-uniformity of the magnetic field, the dipole will experience a force given by

ˆ( ) ( ) zB z z z

dBBdz

µ µ ⎛ ⎞= ∇ ⋅ = ∇ = ⎜ ⎟⎝ ⎠

F µ B k (9.1.16)

Upon differentiating Eq. (9.1.15) and substituting into Eq. (9.1.16), we obtain

2

02 2 5/ 2

3 ˆ2( )

zB

IR zR zµ µ

= −+

F k (9.1.17)

Thus, the dipole is attracted toward the current-carrying ring. On the other hand, if the direction of the dipole is reversed, ˆ

zµ= −µ k , the resulting force will be repulsive. 9.1.1 Magnetic Field of a Moving Point Charge Suppose we have an infinitesimal current element in the form of a cylinder of cross-sectional area A and length ds consisting of n charge carriers per unit volume, all moving at a common velocity v along the axis of the cylinder. Let I be the current in the element, which we define as the amount of charge passing through any cross-section of the cylinder per unit time. From Chapter 6, we see that the current I can be written as n Aq I=v (9.1.18)

The total number of charge carriers in the current element is simply , so that using Eq. (9.1.1), the magnetic field d

dN n A ds=

B due to the dN charge carriers is given by

9

0 0 02 2

ˆ ˆ( | |) ( ) ( )4 4 4

nAq d n A ds q dN qdr r 2

ˆr

µ µ µπ π π

× ×= = =

v s r v r v rB × (9.1.19)

where r is the distance between the charge and the field point P at which the field is being measured, the unit vector ˆ points from the source of the field (the charge) to P. The differential length vector is defined to be parallel to v

/ r=r rd s . In case of a single charge,

, the above equation becomes 1dN =

02

ˆ4

qr

µπ

×=

v rB (9.1.20)

Note, however, that since a point charge does not constitute a steady current, the above equation strictly speaking only holds in the non-relativistic limit where v , the speed of light, so that the effect of “retardation” can be ignored.

c

The result may be readily extended to a collection of N point charges, each moving with a different velocity. Let the ith charge be located at (iq , , )i i ix y z and moving with velocity

. Using the superposition principle, the magnetic field at P can be obtained as: iv

03/ 22 2 21

ˆ ˆ ˆ( ) ( ) ( )4 ( ) ( ) ( )

Ni i i

i ii i i i

x x y y z zqx x y y z z

µπ=

⎡ ⎤− + − + −⎢ ⎥= ×⎢ ⎥⎡ ⎤− + − + −⎣ ⎦⎣ ⎦

∑ i j kB v (9.1.21)

Animation 9.1: Magnetic Field of a Moving Charge Figure 9.1.8 shows one frame of the animations of the magnetic field of a moving positive and negative point charge, assuming the speed of the charge is small compared to the speed of light.

Figure 9.1.8 The magnetic field of (a) a moving positive charge, and (b) a moving negative charge, when the speed of the charge is small compared to the speed of light.

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Animation 9.2: Magnetic Field of Several Charges Moving in a Circle Suppose we want to calculate the magnetic fields of a number of charges moving on the circumference of a circle with equal spacing between the charges. To calculate this field we have to add up vectorially the magnetic fields of each of charges using Eq. (9.1.19).

Figure 9.1.9 The magnetic field of four charges moving in a circle. We show the magnetic field vector directions in only one plane. The bullet-like icons indicate the direction of the magnetic field at that point in the array spanning the plane. Figure 9.1.9 shows one frame of the animation when the number of moving charges is four. Other animations show the same situation for N =1, 2, and 8. When we get to eight charges, a characteristic pattern emerges--the magnetic dipole pattern. Far from the ring, the shape of the field lines is the same as the shape of the field lines for an electric dipole. Interactive Simulation 9.2: Magnetic Field of a Ring of Moving Charges

Figure 9.1.10 shows a ShockWave display of the vectoral addition process for the case where we have 30 charges moving on a circle. The display in Figure 9.1.10 shows an observation point fixed on the axis of the ring. As the addition proceeds, we also show the resultant up to that point (large arrow in the display).

Figure 9.1.10 A ShockWave simulation of the use of the principle of superposition to find the magnetic field due to 30 moving charges moving in a circle at an observation point on the axis of the circle.

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Figure 9.1.11 The magnetic field due to 30 charges moving in a circle at a given observation point. The position of the observation point can be varied to see how the magnetic field of the individual charges adds up to give the total field.

In Figure 9.1.11, we show an interactive ShockWave display that is similar to that in Figure 9.1.10, but now we can interact with the display to move the position of the observer about in space. To get a feel for the total magnetic field, we also show a “iron filings” representation of the magnetic field due to these charges. We can move the observation point about in space to see how the total field at various points arises from the individual contributions of the magnetic field of to each moving charge. 9.2 Force Between Two Parallel Wires We have already seen that a current-carrying wire produces a magnetic field. In addition, when placed in a magnetic field, a wire carrying a current will experience a net force. Thus, we expect two current-carrying wires to exert force on each other. Consider two parallel wires separated by a distance a and carrying currents I1 and I2 in the +x-direction, as shown in Figure 9.2.1.

Figure 9.2.1 Force between two parallel wires

The magnetic force, , exerted on wire 1 by wire 2 may be computed as follows: Using the result from the previous example, the magnetic field lines due to I

12F2 going in the +x-

direction are circles concentric with wire 2, with the field 2B pointing in the tangential

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direction. Thus, at an arbitrary point P on wire 1, we have 2 0 2ˆ( / 2 )I aµ π= −B j , which

points in the direction perpendicular to wire 1, as depicted in Figure 9.2.1. Therefore,

( ) 0 2 0 1 212 1 2 1

ˆ ˆ ˆ2 2

I I I lI I la a

µ µπ π

⎛ ⎞= × = × − = −⎜ ⎟⎝ ⎠

F B i jl k (9.2.1)

Clearly points toward wire 2. The conclusion we can draw from this simple calculation is that two parallel wires carrying currents in the same direction will attract each other. On the other hand, if the currents flow in opposite directions, the resultant force will be repulsive.

12F

Animation 9.3: Forces Between Current-Carrying Parallel Wires Figures 9.2.2 shows parallel wires carrying current in the same and in opposite directions. In the first case, the magnetic field configuration is such as to produce an attraction between the wires. In the second case the magnetic field configuration is such as to produce a repulsion between the wires.

(a) (b) Figure 9.2.2 (a) The attraction between two wires carrying current in the same direction. The direction of current flow is represented by the motion of the orange spheres in the visualization. (b) The repulsion of two wires carrying current in opposite directions. 9.3 Ampere’s Law We have seen that moving charges or currents are the source of magnetism. This can be readily demonstrated by placing compass needles near a wire. As shown in Figure 9.3.1a, all compass needles point in the same direction in the absence of current. However, when

, the needles will be deflected along the tangential direction of the circular path (Figure 9.3.1b).

0I ≠

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Figure 9.3.1 Deflection of compass needles near a current-carrying wire Let us now divide a circular path of radius r into a large number of small length vectors

, that point along the tangential direction with magnitude ˆs∆ ∆s = φ s∆ (Figure 9.3.2).

Figure 9.3.2 Amperian loop In the limit , we obtain 0∆ →s

( )002

2Id B ds rr

µ Iπ µπ

⎛ ⎞⋅ = = =⎜ ⎟⎝ ⎠∫ ∫B s (9.3.1)

The result above is obtained by choosing a closed path, or an “Amperian loop” that follows one particular magnetic field line. Let’s consider a slightly more complicated Amperian loop, as that shown in Figure 9.3.3

Figure 9.3.3 An Amperian loop involving two field lines

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The line integral of the magnetic field around the contour abcda is

(9.3.2) 2 2 1 10 ( ) 0 [ (2 )]

abcda ab bc cd cd

d d d d

B r B rθ π θ

⋅ = ⋅ + ⋅ + ⋅ + ⋅

= + + + −

∫ ∫ ∫ ∫ ∫B s B s B s B s B sd

where the length of arc bc is 2r θ , and 1(2 )r π θ− for arc da. The first and the third integrals vanish since the magnetic field is perpendicular to the paths of integration. With

1 0 / 2 1B I rµ π= and 2 0 / 2 2B I rµ π= , the above expression becomes

0 0 0 02 1

2 1

( ) [ (2 )] (2 )2 2 2 2abcda

I I I Id r rr r 0Iµ µ µ µθ π θ θ π θ µ

π π π π⋅ = + − = + − =∫ B s (9.3.3)

We see that the same result is obtained whether the closed path involves one or two magnetic field lines. As shown in Example 9.1, in cylindrical coordinates ( , , )r zϕ with current flowing in the +z-axis, the magnetic field is given by 0 ˆ( / 2 )I rµ π=B φ . An arbitrary length element in the cylindrical coordinates can be written as ˆ ˆ ˆd dr r d dzϕ= + +s r φ z (9.3.4) which implies

0 0 00

closed path closed path closed path

(2 )2 2 2

I I Id r d dr

µ µ µ Iϕ ϕ ππ π π

⎛ ⎞⋅ = = = =⎜ ⎟⎝ ⎠∫ ∫ ∫B s µ (9.3.5)

In other words, the line integral of d⋅∫ B s around any closed Amperian loop is

proportional to encI , the current encircled by the loop.

Figure 9.3.4 An Amperian loop of arbitrary shape.

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The generalization to any closed loop of arbitrary shape (see for example, Figure 9.3.4) that involves many magnetic field lines is known as Ampere’s law: 0 encd Iµ⋅∫ B s = (9.3.6) Ampere’s law in magnetism is analogous to Gauss’s law in electrostatics. In order to apply them, the system must possess certain symmetry. In the case of an infinite wire, the system possesses cylindrical symmetry and Ampere’s law can be readily applied. However, when the length of the wire is finite, Biot-Savart law must be used instead.

Biot-Savart Law 02

ˆ4

I dr

µπ

×= ∫

s rB general current source ex: finite wire

Ampere’s law 0 encd Iµ⋅∫ B s = current source has certain symmetry ex: infinite wire (cylindrical)

Ampere’s law is applicable to the following current configurations: 1. Infinitely long straight wires carrying a steady current I (Example 9.3) 2. Infinitely large sheet of thickness b with a current density J (Example 9.4). 3. Infinite solenoid (Section 9.4). 4. Toroid (Example 9.5). We shall examine all four configurations in detail. Example 9.3: Field Inside and Outside a Current-Carrying Wire Consider a long straight wire of radius R carrying a current I of uniform current density, as shown in Figure 9.3.5. Find the magnetic field everywhere.

Figure 9.3.5 Amperian loops for calculating the B field of a conducting wire of radius R.

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Solution: (i) Outside the wire where r , the Amperian loop (circle 1) completely encircles the current, i.e.,

R≥encI I= . Applying Ampere’s law yields

( ) 02d B ds B r Iπ µ⋅ = = =∫ ∫B s which implies

0

2IBr

µπ

=

(ii) Inside the wire where r , the amount of current encircled by the Amperian loop (circle 2) is proportional to the area enclosed, i.e.,

R<

2

enc 2

rI IR

ππ

⎛ ⎞= ⎜ ⎟

⎝ ⎠

Thus, we have

( )2

00 2 22

2Irrd B r I B

R Rµππ µ

π π⎛ ⎞

⋅ = = ⇒ =⎜ ⎟⎝ ⎠

∫ B s

We see that the magnetic field is zero at the center of the wire and increases linearly with r until r=R. Outside the wire, the field falls off as 1/r. The qualitative behavior of the field is depicted in Figure 9.3.6 below:

Figure 9.3.6 Magnetic field of a conducting wire of radius R carrying a steady current I . Example 9.4: Magnetic Field Due to an Infinite Current Sheet Consider an infinitely large sheet of thickness b lying in the xy plane with a uniform current density 0

ˆJ=J i . Find the magnetic field everywhere.

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Figure 9.3.7 An infinite sheet with current density 0ˆJ=J i .

Solution: We may think of the current sheet as a set of parallel wires carrying currents in the +x-direction. From Figure 9.3.8, we see that magnetic field at a point P above the plane points in the −y-direction. The z-component vanishes after adding up the contributions from all wires. Similarly, we may show that the magnetic field at a point below the plane points in the +y-direction.

Figure 9.3.8 Magnetic field of a current sheet We may now apply Ampere’s law to find the magnetic field due to the current sheet. The Amperian loops are shown in Figure 9.3.9.

Figure 9.3.9 Amperian loops for the current sheets For the field outside, we integrate along path . The amount of current enclosed by is

1C 1C

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enc 0 ( )I d J b= ⋅ =∫∫ J A (9.3.7) Applying Ampere’s law leads to 0 enc 0 0(2 ) ( )d B I J bµ µ⋅ = = =∫ B s or 0 0 / 2B J bµ= . Note that the magnetic field outside the sheet is constant, independent of the distance from the sheet. Next we find the magnetic field inside the sheet. The amount of current enclosed by path is 2C enc 0 (2 | | )I d J z= ⋅ =∫∫ J A (9.3.9) Applying Ampere’s law, we obtain 0 enc 0 0(2 ) (2 | | )d B I J zµ µ⋅ = = =∫ B s or 0 0 | |B J zµ= . At , the magnetic field vanishes, as required by symmetry. The results can be summarized using the unit-vector notation as

0z =

0 0

0 0

0 0

ˆ, / 22

ˆ, / 2 / 2

ˆ , / 22

J b z b

J z b z bJ b z b

µ

µµ

⎧− >⎪⎪⎪= − − < <⎨⎪⎪ < −⎪⎩

j

B j

j

(9.3.11)

Let’s now consider the limit where the sheet is infinitesimally thin, with . In this case, instead of current density , we have surface current

0b →

0ˆJ=J i ˆK=K i , where 0K J b= .

Note that the dimension of K is current/length. In this limit, the magnetic field becomes

0

0

ˆ, 02

ˆ , 02

K z

K z

µ

µ

⎧− >⎪⎪= ⎨⎪ <⎪⎩

jB

j (9.3.12)

9.4 Solenoid A solenoid is a long coil of wire tightly wound in the helical form. Figure 9.4.1 shows the magnetic field lines of a solenoid carrying a steady current I. We see that if the turns are closely spaced, the resulting magnetic field inside the solenoid becomes fairly uniform,

19

(9.3.8)

(9.3.10)

provided that the length of the solenoid is much greater than its diameter. For an “ideal” solenoid, which is infinitely long with turns tightly packed, the magnetic field inside the solenoid is uniform and parallel to the axis, and vanishes outside the solenoid.

Figure 9.4.1 Magnetic field lines of a solenoid

We can use Ampere’s law to calculate the magnetic field strength inside an ideal solenoid. The cross-sectional view of an ideal solenoid is shown in Figure 9.4.2. To compute B , we consider a rectangular path of length l and width w and traverse the path in a counterclockwise manner. The line integral of B along this loop is

(9.4.1) 0 0 0

d d d d

Bl

⋅ ⋅ + ⋅ + ⋅ + ⋅

= + + +

∫ ∫ ∫ ∫ ∫1 2 3 4

B s = B s B s B s B d s

Figure 9.4.2 Amperian loop for calculating the magnetic field of an ideal solenoid. In the above, the contributions along sides 2 and 4 are zero because B is perpendicular to

. In addition, along side 1 because the magnetic field is non-zero only inside the solenoid. On the other hand, the total current enclosed by the Amperian loop is d s =B 0

encI NI= , where N is the total number of turns. Applying Ampere’s law yields 0d Bl Nµ⋅ = = I∫ B s (9.4.2) or

20

00

NIB nIl

µ µ= = (9.4.3)

where represents the number of turns per unit length., In terms of the surface current, or current per unit length

/n N l=K nI= , the magnetic field can also be written as,

0B Kµ= (9.4.4) What happens if the length of the solenoid is finite? To find the magnetic field due to a finite solenoid, we shall approximate the solenoid as consisting of a large number of circular loops stacking together. Using the result obtained in Example 9.2, the magnetic field at a point P on the z axis may be calculated as follows: Take a cross section of tightly packed loops located at z’ with a thickness ' , as shown in Figure 9.4.3 dz The amount of current flowing through is proportional to the thickness of the cross section and is given by , where ( ') ( / )dI I ndz I N l dz= = ' /n N l= is the number of turns per unit length.

Figure 9.4.3 Finite Solenoid The contribution to the magnetic field at P due to this subset of loops is

2 2

0 02 2 3/ 2 2 2 3/ 2 (

2[( ') ] 2[( ') ]zR RdB dI nIdz

z z R z z Rµ µ

= =− + − +

') (9.4.5)

Integrating over the entire length of the solenoid, we obtain

2 2/ 20 02 2 3/ 2 2 2 2/ 2

02 2 2 2

/ 2

/ 2

' '2 [( ') ] 2 ( ')

( / 2) ( / 2)2 ( / 2) ( / 2)

l

z l

l

l

nIR nIRdz z zBz z R R z z R

nI l z l zz l R z l R

µ µ

µ

−

−

−= =

− + − +

⎡ ⎤− += +⎢ ⎥

⎢ − + + + ⎥⎣ ⎦

∫ (9.4.6)

21

A plot of 0/zB B , where 0 0B nIµ= is the magnetic field of an infinite solenoid, as a function of /z R is shown in Figure 9.4.4 for 10l R= and 20l R= .

Figure 9.4.4 Magnetic field of a finite solenoid for (a) 10l R= , and (b) . 20l R= Notice that the value of the magnetic field in the region| | / 2z l< is nearly uniform and approximately equal to 0B . Examaple 9.5: Toroid Consider a toroid which consists of N turns, as shown in Figure 9.4.5. Find the magnetic field everywhere.

Figure 9.4.5 A toroid with N turns Solutions: One can think of a toroid as a solenoid wrapped around with its ends connected. Thus, the magnetic field is completely confined inside the toroid and the field points in the azimuthal direction (clockwise due to the way the current flows, as shown in Figure 9.4.5.) Applying Ampere’s law, we obtain

22

0(2 )d Bds B ds B r Nπ µ⋅ = = = =∫ ∫ ∫B s I or

0

2NIBr

µπ

= (9.4.8)

where r is the distance measured from the center of the toroid.. Unlike the magnetic field of a solenoid, the magnetic field inside the toroid is non-uniform and decreases as1/ . r 9.5 Magnetic Field of a Dipole Let a magnetic dipole moment vector ˆµ= −µ k be placed at the origin (e.g., center of the Earth) in the plane. What is the magnetic field at a point (e.g., MIT) a distance r away from the origin?

yz

Figure 9.5.1 Earth’s magnetic field components In Figure 9.5.1 we show the magnetic field at MIT due to the dipole. The y- and z- components of the magnetic field are given by

20 03 3

3 sin cos , (3cos 1)4 4y zB B

r rµ µµ µθ θ θπ π

= − = − − (9.5.1)

Readers are referred to Section 9.8 for the detail of the derivation. In spherical coordinates (r,θ,φ) , the radial and the polar components of the magnetic field can be written as

03

2sin cos cos4r y zB B B

rµ µθ θ θπ

= + = − (9.5.2)

23

(9.4.7)

and

03cos sin sin

4y zB B Brθ

µ µθ θ θπ

= − = − (9.5.3)

respectively. Thus, the magnetic field at MIT due to the dipole becomes

03

ˆ ˆˆ (sin 2cos )4rB B

rθ ˆµ µ θ θπ

= + = − +B θ r θ r (9.5.4)

Notice the similarity between the above expression and the electric field due to an electric dipole p (see Solved Problem 2.13.6):

30

1 ˆ ˆ(sin 2cos )4

pr

θ θπε

= +E θ r

The negative sign in Eq. (9.5.4) is due to the fact that the magnetic dipole points in the −z-direction. In general, the magnetic field due to a dipole moment µ can be written as

03

ˆ ˆ3( )4 rµπ

⋅ −=

µ r r µB (9.5.5)

The ratio of the radial and the polar components is given by

0

3

03

2 cos4 2cot

sin4

rB rB

rθ

µ µ θπ θµ µ θπ

−= =

− (9.5.6)

9.5.1 Earth’s Magnetic Field at MIT The Earth’s field behaves as if there were a bar magnet in it. In Figure 9.5.2 an imaginary magnet is drawn inside the Earth oriented to produce a magnetic field like that of the Earth’s magnetic field. Note the South pole of such a magnet in the northern hemisphere in order to attract the North pole of a compass. It is most natural to represent the location of a point P on the surface of the Earth using the spherical coordinates ( , , )r θ φ , where r is the distance from the center of the Earth, θ is the polar angle from the z-axis, with 0 θ π≤ ≤ , and φ is the azimuthal angle in the xy plane, measured from the x-axis, with 0 2φ π≤ ≤ (See Figure 9.5.3.) With the distance fixed at , the radius of the Earth, the point P is parameterized by the two anglesEr r= θ and φ .

24

Figure 9.5.2 Magnetic field of the Earth In practice, a location on Earth is described by two numbers – latitude and longitude. How are they related to θ and φ ? The latitude of a point, denoted as δ , is a measure of the elevation from the plane of the equator. Thus, it is related to θ (commonly referred to as the colatitude) by 90δ θ= ° − . Using this definition, the equator has latitude 0 , and the north and the south poles have latitude

°90± ° , respectively.

The longitude of a location is simply represented by the azimuthal angle φ in the spherical coordinates. Lines of constant longitude are generally referred to as meridians. The value of longitude depends on where the counting begins. For historical reasons, the meridian passing through the Royal Astronomical Observatory in Greenwich, UK, is chosen as the “prime meridian” with zero longitude.

Figure 9.5.3 Locating a point P on the surface of the Earth using spherical coordinates. Let the z-axis be the Earth’s rotation axis, and the x-axis passes through the prime meridian. The corresponding magnetic dipole moment of the Earth can be written as

0 0 0 0 0ˆ ˆ ˆ(sin cos sin sin cos )

ˆ ˆ ˆ( 0.062 0.18 0.98 )E E

E

µ θ φ θ φ θ

µ

= + +

= − + −

µ i j

i j k

k

25

(9.5.7)

where , and we have used 22 27.79 10 A mEµ = × ⋅ 0 0( , ) (169 ,109 )θ φ = ° ° . The expression shows that Eµ has non-vanishing components in all three directions in the Cartesian coordinates. On the other hand, the location of MIT is for the latitude and 71 for the longitude ( north of the equator, and 71 west of the prime meridian), which means that

42 N° W°42° °

90 42 48mθ = ° − ° = ° , and 360 71 289mφ = ° − ° = ° . Thus, the position of MIT can be described by the vector

(9.5.8) MITˆ ˆ ˆ(sin cos sin sin cos )

ˆ ˆ ˆ(0.24 0.70 0.67 )E m m m m m

E

r

r

θ φ θ φ θ= + +

= − +

r i j

i j k

k

The angle between E−µ and is given by MITr

1 1MIT

MIT

cos cos (0.80) 37| || |

EME

E

θ − −⎛ ⎞− ⋅= =⎜ ⎟−⎝ ⎠

r µr µ

= ° (9.5.9)

Note that the polar angle θ is defined as 1 ˆˆcos ( )θ −= ⋅r k , the inverse of cosine of the dot product between a unit vector r for the position, and a unit vector ˆ+k in the positive z-direction, as indicated in Figure 9.6.1. Thus, if we measure the ratio of the radial to the polar component of the Earth’s magnetic field at MIT, the result would be

2cot 37 2.65rBBθ

= ° ≈ (9.5.10)

Note that the positive radial (vertical) direction is chosen to point outward and the positive polar (horizontal) direction points towards the equator. Animation 9.4: Bar Magnet in the Earth’s Magnetic Field Figure 9.5.4 shows a bar magnet and compass placed on a table. The interaction between the magnetic field of the bar magnet and the magnetic field of the earth is illustrated by the field lines that extend out from the bar magnet. Field lines that emerge towards the edges of the magnet generally reconnect to the magnet near the opposite pole. However, field lines that emerge near the poles tend to wander off and reconnect to the magnetic field of the earth, which, in this case, is approximately a constant field coming at 60 degrees from the horizontal. Looking at the compass, one can see that a compass needle will always align itself in the direction of the local field. In this case, the local field is dominated by the bar magnet. Click and drag the mouse to rotate the scene. Control-click and drag to zoom in and out.

26

Figure 9.5.4 A bar magnet in Earth’s magnetic field 9.6 Magnetic Materials The introduction of material media into the study of magnetism has very different consequences as compared to the introduction of material media into the study of electrostatics. When we dealt with dielectric materials in electrostatics, their effect was always to reduce E below what it would otherwise be, for a given amount of “free” electric charge. In contrast, when we deal with magnetic materials, their effect can be one of the following: (i) reduce B below what it would otherwise be, for the same amount of "free" electric current (diamagnetic materials); (ii) increase B a little above what it would otherwise be (paramagnetic materials); (iii) increase B a lot above what it would otherwise be (ferromagnetic materials). Below we discuss how these effects arise. 9.6.1 Magnetization Magnetic materials consist of many permanent or induced magnetic dipoles. One of the concepts crucial to the understanding of magnetic materials is the average magnetic field produced by many magnetic dipoles which are all aligned. Suppose we have a piece of material in the form of a long cylinder with area A and height L, and that it consists of N magnetic dipoles, each with magnetic dipole moment µ , spread uniformly throughout the volume of the cylinder, as shown in Figure 9.6.1.

27

Figure 9.6.1 A cylinder with N magnetic dipole moments

We also assume that all of the magnetic dipole moments µ are aligned with the axis of the cylinder. In the absence of any external magnetic field, what is the average magnetic field due to these dipoles alone? To answer this question, we note that each magnetic dipole has its own magnetic field associated with it. Let’s define the magnetization vector M to be the net magnetic dipole moment vector per unit volume:

1i

iV= ∑M µ (9.6.1)

where V is the volume. In the case of our cylinder, where all the dipoles are aligned, the magnitude of is simply M /M N ALµ= . Now, what is the average magnetic field produced by all the dipoles in the cylinder?

Figure 9.6.2 (a) Top view of the cylinder containing magnetic dipole moments. (b) The equivalent current. Figure 9.6.2(a) depicts the small current loops associated with the dipole moments and the direction of the currents, as seen from above. We see that in the interior, currents flow in a given direction will be cancelled out by currents flowing in the opposite direction in neighboring loops. The only place where cancellation does not take place is near the edge of the cylinder where there are no adjacent loops further out. Thus, the average current in the interior of the cylinder vanishes, whereas the sides of the cylinder appear to carry a net current. The equivalent situation is shown in Figure 9.6.2(b), where there is an equivalent current eqI on the sides.

28

The functional form of eqI may be deduced by requiring that the magnetic dipole moment produced by eqI be the same as total magnetic dipole moment of the system. The condition gives eqI A Nµ= (9.6.2) or

eqNIAµ

= (9.6.3)

Next, let’s calculate the magnetic field produced by eqI . With eqI running on the sides, the equivalent configuration is identical to a solenoid carrying a surface current (or current per unit length) K . The two quantities are related by

eqI NKL AL

µ M= = = (9.6.4)

Thus, we see that the surface current K is equal to the magnetization M , which is the average magnetic dipole moment per unit volume. The average magnetic field produced by the equivalent current system is given by (see Section 9.4) 0 0MB K Mµ µ= = (9.6.5)

Since the direction of this magnetic field is in the same direction as M , the above expression may be written in vector notation as 0M µ=B M (9.6.6) This is exactly opposite from the situation with electric dipoles, in which the average electric field is anti-parallel to the direction of the electric dipoles themselves. The reason is that in the region interior to the current loop of the dipole, the magnetic field is in the same direction as the magnetic dipole vector. Therefore, it is not surprising that after a large-scale averaging, the average magnetic field also turns out to be parallel to the average magnetic dipole moment per unit volume. Notice that the magnetic field in Eq. (9.6.6) is the average field due to all the dipoles. A very different field is observed if we go close to any one of these little dipoles. Let’s now examine the properties of different magnetic materials

29

9.6.2 Paramagnetism The atoms or molecules comprising paramagnetic materials have a permanent magnetic dipole moment. Left to themselves, the permanent magnetic dipoles in a paramagnetic material never line up spontaneously. In the absence of any applied external magnetic field, they are randomly aligned. Thus, =M 0 and the average magnetic field MB is also

zero. However, when we place a paramagnetic material in an external field , the

dipoles experience a torque 0B

0= ×τ µ B that tends to align µ with 0B , thereby producing a

net magnetization parallel toM 0B . Since MB is parallel to 0B , it will tend to enhance

. The total magnetic field is the sum of these two fields: 0B B 0 0M µ= + = +B B B B M0 (9.6.7) Note how different this is than in the case of dielectric materials. In both cases, the torque on the dipoles causes alignment of the dipole vector parallel to the external field. However, in the paramagnetic case, that alignment enhances the external magnetic field, whereas in the dielectric case it reduces the external electric field. In most paramagnetic substances, the magnetization M is not only in the same direction as B , but also

linearly proportional to . This is plausible because without the external field there would be no alignment of dipoles and hence no magnetization M

0

0B 0B. The linear relation

between M and is expressed as 0B

0

0mχ

µ=

BM (9.6.8)

where mχ is a dimensionless quantity called the magnetic susceptibility. Eq. (10.7.7) can then be written as 0(1 )m mχ κ= + =B B 0B (9.6.9) where 1m mκ χ= + (9.6.10) is called the relative permeability of the material. For paramagnetic substances, , or equivalently,

1mκ >

0mχ > , although mχ is usually on the order of 10 to . The magnetic permeability

6− 310−

mµ of a material may also be defined as 0(1 )m m 0mµ χ µ κ µ= + = (9.6.11)

30

Paramagnetic materials have 0mµ µ> . 9.6.3 Diamagnetism In the case of magnetic materials where there are no permanent magnetic dipoles, the presence of an external field will induce magnetic dipole moments in the atoms or

molecules. However, these induced magnetic dipoles are anti-parallel to , leading to a

magnetization M and average field

0B

0B

MB anti-parallel to 0B , and therefore a reduction in the total magnetic field strength. For diamagnetic materials, we can still define the magnetic permeability, as in equation (8-5), although now 1mκ < , or 0mχ < , although

mχ is usually on the order of 510−− to 910−− . Diamagnetic materials have 0mµ µ< . 9.6.4 Ferromagnetism In ferromagnetic materials, there is a strong interaction between neighboring atomic dipole moments. Ferromagnetic materials are made up of small patches called domains, as illustrated in Figure 9.6.3(a). An externally applied field 0B will tend to line up those magnetic dipoles parallel to the external field, as shown in Figure 9.6.3(b). The strong interaction between neighboring atomic dipole moments causes a much stronger alignment of the magnetic dipoles than in paramagnetic materials.

Figure 9.6.3 (a) Ferromagnetic domains. (b) Alignment of magnetic moments in the direction of the external field . 0B The enhancement of the applied external field can be considerable, with the total magnetic field inside a ferromagnet 10 or times greater than the applied field. The permeability of a ferromagnetic material is not a constant, since neither the total field

or the magnetization M increases linearly with

3 410mκ

B 0B . In fact the relationship between

and is not unique, but dependent on the previous history of the material. The M 0B

31

phenomenon is known as hysteresis. The variation of M as a function of the externally applied field is shown in Figure 9.6.4. The loop abcdef is a hysteresis curve. 0B

Figure 9.6.4 A hysteresis curve. Moreover, in ferromagnets, the strong interaction between neighboring atomic dipole moments can keep those dipole moments aligned, even when the external magnet field is reduced to zero. And these aligned dipoles can thus produce a strong magnetic field, all by themselves, without the necessity of an external magnetic field. This is the origin of permanent magnets. To see how strong such magnets can be, consider the fact that magnetic dipole moments of atoms typically have magnitudes of the order of 23 210 A m− ⋅ . Typical atomic densities are atoms/m3. If all these dipole moments are aligned, then we would get a magnetization of order

2910

23 2 29 3 6(10 A m )(10 atoms/m ) 10 A/mM − ⋅∼ ∼

M The magnetization corresponds to values of 0M µ=B of order 1 tesla, or 10,000 Gauss, just due to the atomic currents alone. This is how we get permanent magnets with fields of order 2200 Gauss. 9.7 Summary

• Biot-Savart law states that the magnetic field dB at a point due to a length element d carrying a steady current I and located at rs away is given by

02

ˆ4

I ddr

µπ

×=

s rB

where r = r and is the permeability of free space. 7

0 4 10 T m/Aµ π −= × ⋅

• The magnitude of the magnetic field at a distance r away from an infinitely long straight wire carrying a current I is

32

(9.6.12)

0

2IBr

µπ

=

• The magnitude of the magnetic force between two straight wires of length

carrying steady current of BF

1 and 2I I and separated by a distance r is

0 1 2

2BI IF

rµ

π=

• Ampere’s law states that the line integral of d⋅B s around any closed loop is

proportional to the total steady current passing through any surface that is bounded by the close loop:

0 encd Iµ⋅ =∫ B s

• The magnetic field inside a toroid which has N closely spaced of wire carrying a current I is given by

0

2NIBr

µπ

=

where r is the distance from the center of the toroid.

• The magnetic field inside a solenoid which has N closely spaced of wire carrying current I in a length of l is given by

0 0NB I nl

µ µ= = I

where n is the number of number of turns per unit length.

• The properties of magnetic materials are as follows:

Materials Magnetic susceptibility

mχ Relative permeability

1m mκ χ= + Magnetic permeability

0m mµ κ µ=

Diamagnetic 5 910 10− −− −∼ 1mκ < 0mµ µ<

Paramagnetic 5 310 10− −∼ 1mκ > 0mµ µ>

Ferromagnetic 1mχ 1mκ 0mµ µ

33

9.8 Appendix 1: Magnetic Field off the Symmetry Axis of a Current Loop In Example 9.2 we calculated the magnetic field due to a circular loop of radius R lying in the xy plane and carrying a steady current I, at a point P along the axis of symmetry. Let’s see how the same technique can be extended to calculating the field at a point off the axis of symmetry in the yz plane.

Figure 9.8.1 Calculating the magnetic field off the symmetry axis of a current loop.

Again, as shown in Example 9.1, the differential current element is

ˆ ˆ'( sin ' cos ' )Id R dφ φ φ= − +s i j

ˆ)

and its position is described by ˆ' (cos ' sin 'R φ φ= +r i j

k. On the other hand, the field point

P now lies in the yz plane with r j ˆP y z= + , as shown in Figure 9.8.1. The

corresponding relative position vector is ( )ˆ ˆ' cos ' sin 'P

ˆR y R zφ φ− = − + − +r = r r i j k (9.8.1) with a magnitude

( )22 2 2 2 2( cos ') sin ' 2 sinr R y R z R y z yRφ φ φ= = − + − + = + + −r (9.8.2) and the unit vector

'ˆ| '

P

Pr |−

= =−

r rrrr r

pointing from Id s to P. The cross product ˆd ×s r can be simplified as

(9.8.3) ( )

( )

ˆ ˆ ˆ ˆ ˆˆ ' sin ' cos ' [ cos ' ( sin ') ]

ˆ ˆ ˆ'[ cos ' sin ' sin ' ]

d R d R y R z

R d z z R y

φ φ φ φ φ

φ φ φ φ

× = − + × − + − +

= + + −

s r i j i j k

i j k

34

Using the Biot-Savart law, the contribution of the current element to the magnetic field at P is

( )

( )0 0 0

3/ 22 3 2 2 2

ˆ ˆ ˆcos ' sin ' sin 'ˆ'

4 4 4 2 sin '

z z R yI I IRd dd dr r R y z yR

φ φ φµ µ µ φπ π π φ

+ + −× ×= = =

+ + −

i j ks r s rB (9.8.4)

Thus, magnetic field at P is

( ) ( )( )

203/ 20 2 2 2

ˆ ˆ ˆcos ' sin ' sin '0, , '

4 2 sin '

z z R yIRy z dR y z yR

π φ φ φµ φπ φ

+ + −=

+ + −∫

i j kB (9.8.5)

The x-component of B can be readily shown to be zero

( )

203/ 20 2 2 2

cos ' ' 04 2 sin '

xIRz dB

R y z yR

πµ φ φπ φ

= =+ + −

∫ (9.8.6)

by making a change of variable 2 2 2 2 sinw R y z yR 'φ= + + − , followed by a straightforward integration. One may also invoke symmetry arguments to verify that

xB must vanish; namely, the contribution at 'φ is cancelled by the contribution at 'π φ− . On the other hand, the y and the z components of B ,

( )

203/ 20 2 2 2

sin ' '4 2 sin '

yIRz dB

R y z yR

πµ φ φπ φ

=+ + −

∫ (9.8.7)

and

( )( )

203/ 20 2 2 2

sin ' '4 2 sin '

z

R y dIRBR y z yR

π φ φµπ φ

−=

+ + −∫ (9.8.8)

involve elliptic integrals which can be evaluated numerically. In the limit , the field point P is located along the z-axis, and we recover the results obtained in Example 9.2:

0y =

20 0

2 2 3/ 2 2 2 3/ 20

2sin ' ' cos ' 0

04 ( ) 4 ( )yIRz IRzB d

R z R zπ πµ µφ φ φ

π π= = −

+ +∫ = (9.8.9)

and

35

22 220 0

2 2 3/ 2 2 2 3/ 2 2 2 3/ 20

2'4 ( ) 4 ( ) 2( )z

IRIR IRB dR z R z R z

πµ µ πφπ π

= = =+ +∫ 0µ

+ (9.8.10)

Now, let’s consider the “point-dipole” limit where 2 2 1/ 2( )R y z r+ = , i.e., the characteristic dimension of the current source is much smaller compared to the distance where the magnetic field is to be measured. In this limit, the denominator in the integrand can be expanded as

( )

3/ 223/ 22 2 23 2

2

3 2

1 2 sin '2 sin ' 1

1 3 2 sin '1 2

R yRR y z yRr r

R yRr r

φφ

φ

−− ⎡ ⎤−

+ + − = +⎢ ⎥⎣ ⎦⎡ ⎤⎛ ⎞−

= − +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

…

(9.8.11)

This leads to

2203 20

2 22 20 05 50

3 2 sin '1 s4 2

3 3sin ' '4 4

yI Rz R yR in ' 'B d

r r

I IR yz R yzdr r

π

π

µ φ φ φπ

µ µ πφ φπ π

⎡ ⎤⎛ ⎞−≈ −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

= =

∫

∫ (9.8.12)

and

220

3 20

3 2 22 203 2 2 20

3 20

3 2 2

2 20

3 2

3 2 sin '1 ( sin ')4 2

3 9 31 sin ' sin ' '4 2 2

3 324 2

32 higher order ter4

zI R R yRB R

r r

I R R R Ry

'y d

R dr r r r

I R R RyRr r r

I R yr r

π

π

µ φ φ φπ

µ φ φ φπ

µ πππ

µ ππ

⎡ ⎤⎛ ⎞−≈ − −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦⎡ ⎤⎛ ⎞ ⎛ ⎞

= − − − −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞= − −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

= − +

∫

∫

ms⎡ ⎤⎢ ⎥⎣ ⎦

(9.8.13)

The quantity 2( )I Rπ may be identified as the magnetic dipole moment IAµ = , where

2A Rπ= is the area of the loop. Using spherical coordinates where siny r θ= and cosz r θ= , the above expressions may be rewritten as

2

0 05

( ) 3( sin )( cos ) 3 sin cos4 4yI R r rB

r rµ π µ

3

θ θ µ θπ π

= =θ (9.8.14)

36

and

2 2 2

20 0 03 2 3 3

( ) 3 sin2 (2 3sin ) (34 4 4z

I R rBr r r r

µ µ µπ θ µ µθπ π π

⎛ ⎞= − = − =⎜ ⎟

⎝ ⎠2cos 1)θ − (9.8.15)

Thus, we see that the magnetic field at a point r due to a current ring of radius R may be approximated by a small magnetic dipole moment placed at the origin (Figure 9.8.2).

R

Figure 9.8.2 Magnetic dipole moment ˆµ=µ k

The magnetic field lines due to a current loop and a dipole moment (small bar magnet) are depicted in Figure 9.8.3.

Figure 9.8.3 Magnetic field lines due to (a) a current loop, and (b) a small bar magnet.

The magnetic field at P can also be written in spherical coordinates ˆˆrB Bθ= +B r θ (9.8.16) The spherical components rB and Bθ are related to the Cartesian components yB and zB by sin cos , cos sinr y z y zB B B B B Bθθ θ θ= + = − θ

ˆ

(9.8.17) In addition, we have, for the unit vectors, ˆ ˆ ˆˆˆ sin cos , cos sinθ θ θ= + = −r j k θ j θ k (9.8.18) Using the above relations, the spherical components may be written as

37

( )

2 203/ 20 2 2

cos '4 2 sin sin '

rIR dB

R r rR

πµ θ φπ θ φ

=+ −

∫ (9.8.19)

and

( ) ( )( )

203/ 20 2 2

sin ' sin ',

4 2 sin sin '

r R dIRB rR r rR

π

θ

φ θ φµθπ θ φ

−=

+ −∫ (9.8.20)

In the limit where R r , we obtain

2 220 0 0

3 30

cos 2 cos 2 cos'4 4 4r

IR IRB dr r

πµ θ µ µ3r

π θ µφπ π π

≈ = =∫θ (9.8.21)

and

( )( )

( )

203/ 20 2 2

2 2 2 22 203 20

20 0

3 3

03

sin ' sin '4 2 sin sin '

3 3 3 sinsin 1 sin ' 3 sin sin ' '4 2 2 2

( )sin2 sin 3 sin4 4

sin4

r R dIRBR r rR

IR R R RR r Rr r r r

IR I RR Rr r

r

π

θ

π

φ θ φµπ θ φ

µ θ dθ φ θ φπ

µ µ π θπ θ π θπ π

µ µ θ

φ

π

−=

+ −

⎡ ⎤⎛ ⎞ ⎛ ⎞≈ − − + − − +⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦

≈ − + =

=

∫

∫

(9.8.22) 9.9 Appendix 2: Helmholtz Coils Consider two N-turn circular coils of radius R, each perpendicular to the axis of symmetry, with their centers located at / 2z l= ± . There is a steady current I flowing in the same direction around each coil, as shown in Figure 9.9.1. Let’s find the magnetic field B on the axis at a distance z from the center of one coil.

Figure 9.9.1 Helmholtz coils

38

Using the result shown in Example 9.2 for a single coil and applying the superposition principle, the magnetic field at (a point at a distance ( ,0)P z / 2z l− away from one center and / 2z l+ from the other) due to the two coils can be obtained as:

2

0top bottom 2 2 3/ 2 2 2 3/ 2

1 12 [( / 2) ] [( / 2) ]z

NIRB B Bz l R z l R

µ ⎡ ⎤= + = +⎢ ⎥− + + +⎣ ⎦

(9.9.1)

A plot of 0/zB B with 00 3/ 2(5 / 4)

NIBR

µ= being the field strength at and 0z = l R= is

depicted in Figure 9.9.2.

Figure 9.9.2 Magnetic field as a function of /z R . Let’s analyze the properties of zB in more detail. Differentiating zB with respect to z, we obtain

2

02 2 5/ 2 2 2 5/ 2

3( / 2) 3( / 2)( )2 [( / 2) ] [( / 2) ]

zz

NIRdB z l z lB zdz z l R z l R

µ ⎧ ⎫− +′ = = − −⎨ ⎬− + + +⎩ ⎭ (9.9.2)

One may readily show that at the midpoint, 0z = , the derivative vanishes:

0

0z

dBdz =

= (9.9.3)

Straightforward differentiation yields

22 20

2 2 2 5/ 2

2

2 2 5/ 2 2 2 7 / 2

3 15( / 2)( )2 [( / 2) ] [( / 2) ]

3 15( / 2) [( / 2) ] [( / 2) ]

zN IRd B z lB z

dz z l R z l R

z lz l R z l R

µ ⎧ −′′ = = − +⎨ − + − +⎩⎫+

− + ⎬+ + + + ⎭

2 2 7 / 2

(9.9.4)

39

At the midpoint , the above expression simplifies to 0z =

22 20

2 2 2 5/ 2 20

2 2 20

2 2 7 / 2

6 15(0)2 [( / 2) ] 2[( / 2) ]

6( )2 [( / 2) ]

zz

NId B lBdz l R l R

NI R ll R

µ

µ=

2 7 / 2

⎧ ⎫′′ = = − +⎨ ⎬+ +⎩ ⎭

−= −

+

(9.9.5)

Thus, the condition that the second derivative of zB vanishes at 0z = is . That is, the distance of separation between the two coils is equal to the radius of the coil. A configuration with l is known as Helmholtz coils.

l R=

R= For small z, we may make a Taylor-series expansion of ( )zB z about 0z = :

21( ) (0) (0) (0) ...2!z z z zB z B B z B z′ ′′= + + + (9.9.6)

The fact that the first two derivatives vanish at 0z = indicates that the magnetic field is fairly uniform in the small z region. One may even show that the third derivative

vanishes at as well. (0)zB′′′ 0z = Recall that the force experienced by a dipole in a magnetic field is (B = ∇ ⋅F µ )B . If we

place a magnetic dipole ˆzµ=µ k at 0z = , the magnetic force acting on the dipole is

ˆ( ) zB z z z

dBBdz

µ µ ⎛ ⎞= ∇ = ⎜ ⎟⎝ ⎠

F k (9.9.7)

which is expected to be very small since the magnetic field is nearly uniform there. Animation 9.5: Magnetic Field of the Helmholtz Coils The animation in Figure 9.9.3(a) shows the magnetic field of the Helmholtz coils. In this configuration the currents in the top and bottom coils flow in the same direction, with their dipole moments aligned. The magnetic fields from the two coils add up to create a net field that is nearly uniform at the center of the coils. Since the distance between the coils is equal to the radius of the coils and remains unchanged, the force of attraction between them creates a tension, and is illustrated by field lines stretching out to enclose both coils. When the distance between the coils is not fixed, as in the animation depicted in Figure 9.9.3(b), the two coils move toward each other due to their force of attraction. In this animation, the top loop has only half the current as the bottom loop. The field configuration is shown using the “iron filings” representation.

40

(a) (b) Figure 9.9.3 (a) Magnetic field of the Helmholtz coils where the distance between the coils is equal to the radius of the coil. (b) Two co-axial wire loops carrying current in the same sense are attracted to each other. Next, let’s consider the case where the currents in the loop flow in the opposite directions, as shown in Figure 9.9.4.

Figure 9.9.4 Two circular loops carrying currents in the opposite directions. Again, by superposition principle, the magnetic field at a point with is (0,0, )P z 0z >

2

01 2 2 2 3/ 2 2 2 3/ 2

1 12 [( / 2) ] [( / 2) ]z z z

NIRB B Bz l R z l R

µ ⎡ ⎤= + = −⎢ ⎥− + + +⎣ ⎦

(9.9.8)

A plot of 0/zB B with 0 0 / 2B NI Rµ= and l R= is depicted in Figure 9.9.5.

41Figure 9.9.5 Magnetic field as a function of /z R .

Differentiating zB with respect to z, we obtain

2

02 2 5/ 2 2 2 5/ 2

3( / 2) 3( / 2)( )2 [( / 2) ] [( / 2) ]

zz

NIRdB z l z lB zdz z l R z l R

µ ⎧ ⎫− +′ = = − +⎨ ⎬− + + +⎩ ⎭ (9.9.9)

At the midpoint, , we have 0z =

2

02 2 5/ 2

3(0) 00 2 [( / 2) ]

zz

NIRdB lBzdz l R

µ′ = == +

≠ (9.9.10)

Thus, a magnetic dipole ˆ

zµ=µ k placed at 0z = will experience a net force:

2

02 2 5/ 2

(0) 3ˆ ˆ( ) ( ) 2 [( / 2) ]

zzB z z z

NIRdB lBdz l R

µ µµ µ ⎛ ⎞= ∇ ⋅ = ∇ = =⎜ ⎟ +⎝ ⎠F µ B k k (9.9.11)

For , the above expression simplifies to l R=

05/ 2 2

3 ˆ 2(5 / 4)

zB

NIR

µ µ=F k (9.9.12)

Animation 9.6: Magnetic Field of Two Coils Carrying Opposite Currents The animation depicted in Figure 9.9.6 shows the magnetic field of two coils like the Helmholtz coils but with currents in the top and bottom coils flowing in the opposite directions. In this configuration, the magnetic dipole moments associated with each coil are anti-parallel.

(a) (b) Figure 9.9.6 (a) Magnetic field due to coils carrying currents in the opposite directions. (b) Two co-axial wire loops carrying current in the opposite sense repel each other. The

42

field configurations here are shown using the “iron filings” representation. The bottomwire loop carries twice the amount of current as the top wire loop.

At the center of the coils along the axis of symmetry, the magnetic field is zero. With the distance between the two coils fixed, the repulsive force results in a pressure between them. This is illustrated by field lines that are compressed along the central horizontal axis between the coils. Animation 9.7: Forces Between Coaxial Current-Carrying Wires

Figure 9.9.7 A magnet in the TeachSpin ™ Magnetic Force apparatus when the current in the top coil is counterclockwise as seen from the top. Figure 9.9.7 shows the force of repulsion between the magnetic field of a permanent magnet and the field of a current-carrying ring in the TeachSpin ™ Magnetic Force apparatus. The magnet is forced to have its North magnetic pole pointing downward, and the current in the top coil of the Magnetic Force apparatus is moving clockwise as seen from above. The net result is a repulsion of the magnet when the current in this direction is increased. The visualization shows the stresses transmitted by the fields to the magnet when the current in the upper coil is increased. Animation 9.8: Magnet Oscillating Between Two Coils Figure 9.9.8 illustrates an animation in which the magnetic field of a permanent magnet suspended by a spring in the TeachSpinTM apparatus (see TeachSpin visualization), plus the magnetic field due to current in the two coils (here we see a "cutaway" cross-section of the apparatus).

43Figure 9.9.8 Magnet oscillating between two coils

The magnet is fixed so that its north pole points upward, and the current in the two coils is sinusoidal and 180 degrees out of phase. When the effective dipole moment of the top coil points upwards, the dipole moment of the bottom coil points downwards. Thus, the magnet is attracted to the upper coil and repelled by the lower coil, causing it to move upwards. When the conditions are reversed during the second half of the cycle, the magnet moves downwards. This process can also be described in terms of tension along, and pressure perpendicular to, the field lines of the resulting field. When the dipole moment of one of the coils is aligned with that of the magnet, there is a tension along the field lines as they attempt to "connect" the coil and magnet. Conversely, when their moments are anti-aligned, there is a pressure perpendicular to the field lines as they try to keep the coil and magnet apart. Animation 9.9: Magnet Suspended Between Two Coils Figure 9.9.9 illustrates an animation in which the magnetic field of a permanent magnet suspended by a spring in the TeachSpinTM apparatus (see TeachSpin visualization), plus the magnetic field due to current in the two coils (here we see a "cutaway" cross-section of the apparatus). The magnet is fixed so that its north pole points upward, and the current in the two coils is sinusoidal and in phase. When the effective dipole moment of the top coil points upwards, the dipole moment of the bottom coil points upwards as well. Thus, the magnet the magnet is attracted to both coils, and as a result feels no net force (although it does feel a torque, not shown here since the direction of the magnet is fixed to point upwards). When the dipole moments are reversed during the second half of the cycle, the magnet is repelled by both coils, again resulting in no net force. This process can also be described in terms of tension along, and pressure perpendicular to, the field lines of the resulting field. When the dipole moment of the coils is aligned with that of the magnet, there is a tension along the field lines as they are "pulled" from both sides. Conversely, when their moments are anti-aligned, there is a pressure perpendicular to the field lines as they are "squeezed" from both sides.

44Figure 9.9.9 Magnet suspended between two coils

9.10 Problem-Solving Strategies In this Chapter, we have seen how Biot-Savart and Ampere’s laws can be used to calculate magnetic field due to a current source. 9.10.1 Biot-Savart Law: The law states that the magnetic field at a point P due to a length element carrying a steady current I located at away is given by

dsr

0 0

2 3

ˆ4 4

I Id ddr r

µ µπ π

× ×= =

s r s rB

The calculation of the magnetic field may be carried out as follows: (1) Source point: Choose an appropriate coordinate system and write down an expression for the differential current element I ds , and the vector 'r describing the position of I ds . The magnitude is the distance between ' | ' |r = r I ds and the origin. Variables with a “prime” are used for the source point. (2) Field point: The field point P is the point in space where the magnetic field due to the current distribution is to be calculated. Using the same coordinate system, write down the position vector Pr for the field point P. The quantity | |P Pr = r is the distance between the origin and P. (3) Relative position vector: The relative position between the source point and the field point is characterized by the relative position vector 'P= −r r r . The corresponding unit vector is

'ˆ| '

P

Pr |−

= =−

r rrrr r

where is the distance between the source and the field point P. | | | ' |Pr = = −r r r (4) Calculate the cross product ˆd ×s r or d ×s r . The resultant vector gives the direction of the magnetic field , according to the Biot-Savart law. B (5) Substitute the expressions obtained to dB and simplify as much as possible.

45

(6) Complete the integration to obtain Bif possible. The size or the geometry of thesystem is reflected in the integration limits. Change of variables sometimes may help tocomplete the integration.

Below we illustrate how these steps are executed for a current-carrying wire of length L and a loop of radius R.

Current distribution Finite wire of length L Circular loop of radius R

Figure

(1) Source point ˆ ' '

ˆ( '/ ') ' '

x

d d dx dx dx

=

= =

r i

s r i

ˆ ˆ' (cos ' sin ' )ˆ ˆ( '/ ') ' '( sin ' cos ' )

R

d d d d Rd

φ φ

φ φ φ φ φ

= +

= = − +

r i j

s r i j

(2) Field point P ˆP y=r j ˆ

P z=r k

(3) Relative position vector 'P= −r r r

2 2

2 2

ˆ ˆ'

| | 'ˆ ˆ'ˆ'

y x

r x

y xx y

= −

= = +

−=

+

r j i

r

j ir

y 2 2

2 2

ˆ ˆ ˆcos ' sin '

| |ˆ ˆ ˆcos ' sin 'ˆ

R R

r R z

R RR z

φ φ

φ φ

= − − +

= = +

− − +=

+

r i j

r

i jr

z

z

k

k

(4) The cross product ˆd ×s r 2 2

ˆˆ y dxd

y x

′× =

′+

ks r 2 2

ˆ ˆ ˆ'( cos ' sin ' )ˆ R d z z RdR z

φ φ φ+ +× =

+

i js r k

(5) Rewrite dB 02 2 3/

ˆ

4 ( )I y dxd

y xµ

π′

=′+kB 2

02 2 3/ 2

ˆ ˆ ˆ'( cos ' sin ' )4 ( )

I R d z z RdR z

µ φ φ φπ

+ +=

+i jB k

(6) Integrate to get B / 20

2 2 3// 2

02 2

00

'4 ( ' )

4 ( / 2)

x

y

L

z L

BB

Iy dxBy x

I Ly y L

µπ

µπ

−

=

=

=+

=+

∫ 2

202 2 3/ 2 0

202 2 3/ 2 0

2 220 02 2 3/ 2 2 2 3/ 20

cos ' ' 0 4 ( )

sin ' ' 04 ( )

'4 ( ) 2( )

x

y

z

IRzB dR z

IRzB dR z

IR IRB dR z R z

π

π

π

µ φ φπ

µ φ φπ

µ µφπ

= =+

= =+

= =+ +

∫

∫

∫

46

9.10.2 Ampere’s law: Ampere’s law states that the line integral of d⋅B s around any closed loop is proportional to the total current passing through any surface that is bounded by the closed loop:

0 encd Iµ⋅ =∫ B s

To apply Ampere’s law to calculate the magnetic field, we use the following procedure: (1) Draw an Amperian loop using symmetry arguments. (2) Find the current enclosed by the Amperian loop. (3) Calculate the line integral d⋅∫ B s around the closed loop.

(4) Equate with d⋅∫ B s 0 encIµ and solve for B . Below we summarize how the methodology can be applied to calculate the magnetic field for an infinite wire, an ideal solenoid and a toroid.

System Infinite wire Ideal solenoid Toroid

Figure

(1) Draw the Amperian loop

(2) Find the current enclosed by the Amperian loop

encI I= encI NI= encI NI=

(3) Calculate

along the loop

d⋅∫ B s (2 )d B rπ⋅ =∫ B s d Bl⋅ =∫ B s (2 )d B rπ⋅ =∫ B s

47

(4) Equate 0 encIµ with

to obtain d⋅∫ B s B0

2IBr

µπ

= 00

NIB nIl

µ µ= = 0

2NIBr

µπ

=

9.11 Solved Problems 9.11.1 Magnetic Field of a Straight Wire Consider a straight wire of length L carrying a current I along the +x-direction, as shown in Figure 9.11.1 (ignore the return path of the current or the source for the current.) What is the magnetic field at an arbitrary point P on the xy-plane?

Figure 9.11.1 A finite straight wire carrying a current I. Solution: The problem is very similar to Example 9.1. However, now the field point is an arbitrary point in the xy-plane. Once again we solve the problem using the methodology outlined in Section 9.10. (1) Source point From Figure 9.10.1, we see that the infinitesimal length dx′ described by the position vector ˆ' 'x=r i constitutes a current source ˆ( )I d Idx′=s i . (2) Field point As can be seen from Figure 9.10.1, the position vector for the field point P is ˆ ˆx y= +r i j . (3) Relative position vector The relative position vector from the source to P is ˆ' ( ')P

ˆx x y− = − +r = r r i j , with 2 2 1| | | ' | [( ) ]Pr x x′= = − = − +r r r 2y being the distance. The corresponding unit vector is

48

2 2 1

ˆ ˆ' ( )ˆ| ' | [( ) ]

P

P2

x x yr x x

′− − += = =

′− − +r rr irr r y

j

(4) Simplifying the cross product The cross product d can be simplified as ×s r ˆ ˆ ˆ ˆ( ' ) [( ') ] 'dx x x y y dx× − + =i i j k where we have used ˆ ˆ and ˆ ˆ× =i i 0 ˆ× =i j k . (5) Writing down dB Using the Biot-Savart law, the infinitesimal contribution due to Id s is

0 0 02 3 2 2

ˆ ˆ4 4 4 [( )

I I Id d y dxdr r x x y 3 2]

µ µ µπ π π

′× ×= = =

′− +s r s rB k (9.11.1)

Thus, we see that the direction of the magnetic field is in the ˆ+k direction. (6) Carrying out the integration to obtain B The total magnetic field at P can then be obtained by integrating over the entire length of the wire:

/ 2/ 2 0 0

2 2 3 2 2 2/ 2wire / 2

02 2 2 2

( )ˆ ˆ4 [( ) ] 4 ( )

( / 2) ( / 2) ˆ4 ( / 2) ( / 2)

LL

LL

Iy dx I x xdx x y y x x y

I x L x Ly x L y x L y

µ µπ π

µπ

−−

′ ′−= = = −

′− + ′− +

⎡ ⎤− += − −⎢ ⎥

⎢ − + + + ⎥⎣ ⎦

∫ ∫B B k

k

k

(9.11.2)

Let’s consider the following limits: (i) 0x = In this case, the field point P is at ( , ) (0, )x y y= on the y axis. The magnetic field becomes

49

0 02 2 2 2 2 2

/ 2 / 2 / 2ˆ ˆ cos4 2( / 2) ( / 2) ( / 2)

I IL L Ly yL y L y L y

µ µ 0 ˆ2

Iy

µ θπ π

⎡ ⎤− += − − = =⎢ ⎥

⎢ − + + + ⎥ +⎣ ⎦B k

πk k

(9.11.3) in agreement with Eq. (9.1.6). (ii) Infinite length limit Consider the limit where ,L x y . This gives back the expected infinite-length result:

0 / 2 / 2 ˆ4 / 2 / 2 2

0 ˆI IL Ly L L y

µπ π

− +⎡ ⎤= − − =⎢ ⎥⎣ ⎦B µk k (9.11.4)

If we use cylindrical coordinates with the wire pointing along the +z-axis then the magnetic field is given by the expression

0 ˆ2

Ir

µπ

=B φ (9.11.5)

where is the tangential unit vector and the field point P is a distance r away from the wire.

φ

9.11.2 Current-Carrying Arc Consider the current-carrying loop formed of radial lines and segments of circles whose centers are at point P as shown below. Find the magnetic field B at P.

Figure 9.11.2 Current-carrying arc Solution: According to the Biot-Savart law, the magnitude of the magnetic field due to a differential current-carrying element I d s is given by

50

0 0 02 2

ˆ ' '4 4 4

dI I r ddB dr r

µ µ µθ Ir

θπ π π

×= = =

s r (9.11.6)

For the outer arc, we have

0outer 0

'4 4

0I IB db b

θµ µ θθπ π

= =∫ (9.11.7)

The direction of is determined by the cross productouterB ˆd ×s r which points out of the page. Similarly, for the inner arc, we have

0inner 0

'4 4

0I IB da a

θµ µ θθπ π

= =∫ (9.11.8)

For , points into the page. Thus, the total magnitude of magnetic field is innerB ˆd ×s r

0inner outer

1 1 (into page)4

Ia b

µ θπ

⎛ ⎞= = −⎜ ⎟⎝ ⎠

B B + B (9.11.9)

9.11.3 Rectangular Current Loop

Determine the magnetic field (in terms of I, a and b) at the origin O due to the current loop shown in Figure 9.11.3

Figure 9.11.3 Rectangular current loop

51

For a finite wire carrying a current I, the contribution to the magnetic field at a point P is given by Eq. (9.1.5):

( )01 2cos cos

4IBr

µ θ θπ

= +

where 1 and 2θ θ are the angles which parameterize the length of the wire.

To obtain the magnetic field at O, we make use of the above formula. The contributions can be divided into three parts: (i) Consider the left segment of the wire which extends from ( , to

. The angles which parameterize this segment give co) ( , )x y a= − +∞

( , )a d− + 1s 1θ = ( 1 0θ = ) and 2

2cos /b b aθ = − + 2 . Therefore,

( )0 01 1 2 2 2

cos cos 14 4

I I bBa a b a

µ µθ θπ π

⎛ ⎞= + = −⎜

+⎝ ⎠⎟ (9.11.10)

The direction of is out of page, or1B ˆ+k . (ii) Next, we consider the segment which extends from ( , ) ( , )x y a b= − + to . Again, the (cosine of the) angles are given by

( , )a b+ +

1 2 2cos a

a bθ =

+ (9.11.11)

2 1 2 2cos cos a

a bθ θ= =

+ (9.11.12)

This leads to

02 2 2 2 2 2 24 2

I a aBb a b a b b a b

µ µπ π

⎛ ⎞= + =⎜ ⎟ 0Ia

+ + +⎝ ⎠ (9.11.13)

The direction of is into the page, or 2B ˆ−k . (iii) The third segment of the wire runs from ( , ) ( , )x y a b= + + to ( , )a+ +∞ . One may readily show that it gives the same contribution as the first one: 3 1B B= (9.11.14)

52

Solution:

The direction of is again out of page, or 3B ˆ+k . The magnetic field is

( )

0 01 2 3 1 2 2 2 2 2

2 2 2 202 2

ˆ ˆ2 12 2

ˆ2

I Iba a b b a b

I b a b b aab a b

µ µπ π

µ

π

⎛ ⎞= + + = + = − −⎜ ⎟

+⎝ ⎠

= + − −+

B B B B B B k k

k

a+ (9.11.15)

Note that in the limit , the horizontal segment is absent, and the two semi-infinite wires carrying currents in the opposite direction overlap each other and their contributions completely cancel. Thus, the magnetic field vanishes in this limit.

0a →

9.11.4 Hairpin-Shaped Current-Carrying Wire An infinitely long current-carrying wire is bent into a hairpin-like shape shown in Figure 9.11.4. Find the magnetic field at the point P which lies at the center of the half-circle.

Figure 9.11.4 Hairpin-shaped current-carrying wire Solution: Again we break the wire into three parts: two semi-infinite plus a semi-circular segments. (i) Let P be located at the origin in the xy plane. The first semi-infinite segment then extends from ( , ) ( , )x y r= −∞ − to (0, )r− . The two angles which parameterize this segment are characterized by 1cos 1θ = ( 1 0θ = ) and 2 2cos 0 ( / 2)θ θ π= = . Therefore, its contribution to the magnetic field at P is

( )0 01 1 2cos cos (1 0)

4 40

4I IBr r

Ir

µ µθ θ µπ π π

= + = + = (9.11.16)

The direction of is out of page, or 1B ˆ+k .

53

(ii) For the semi-circular arc of radius r, we make use of the Biot-Savart law:

02

ˆ4

I dr

µπ

×= ∫

s rB (9.11.17)

and obtain

02 204 4

0I IrdBr r

πµ µθπ

= =∫ (9.11.18)

The direction of is out of page, or2B ˆ+k . (iii) The third segment of the wire runs from ( , ) (0, )x y r= + to ( , r)−∞ + . One may readily show that it gives the same contribution as the first one:

03 1 4

IB Br

µπ

= = (9.11.19)

The direction of is again out of page, or 3B ˆ+k . The total magnitude of the magnetic field is

0 0 01 2 3 1 2

ˆ ˆ2 (2 4 4

I I Ir r r

ˆ2 )µ µ µ ππ π

= + + = + = + = +B B B B B B k k k (9.11.20)

Notice that the contribution from the two semi-infinite wires is equal to that due to an infinite wire:

01 3 1

ˆ22

Ir

µπ

+ = =B B B k (9.11.21)

9.11.5 Two Infinitely Long Wires Consider two infinitely long wires carrying currents are in the −x-direction.

54Figure 9.11.5 Two infinitely long wires

(a) Plot the magnetic field pattern in the yz-plane. (b) Find the distance d along the z-axis where the magnetic field is a maximum. Solutions: (a) The magnetic field lines are shown in Figure 9.11.6. Notice that the directions of both currents are into the page.

Figure 9.11.6 Magnetic field lines of two wires carrying current in the same direction. (b) The magnetic field at (0, 0, z) due to wire 1 on the left is, using Ampere’s law:

0 01 2 22 2

I IBr a z

µ µπ π

= =+

(9.11.22)

Since the current is flowing in the –x-direction, the magnetic field points in the direction of the cross product 1

ˆ ˆ ˆ ˆˆˆ( ) ( ) (cos sin ) sin cos ˆθ θ θ− × = − × + = −i r i j k j kθ (9.11.23) Thus, we have

( )01 2 2

ˆ ˆsin cos 2

Ia zµ θ θ

π= −

+B j k (9.11.24)

For wire 2 on the right, the magnetic field strength is the same as the left one: 1 2B B= . However, its direction is given by 2

ˆ ˆ ˆ ˆˆˆ( ) ( ) ( cos sin ) sin cos ˆθ θ θ− × = − × − + =i r i j k j+ kθ (9.11.25)

55

Adding up the contributions from both wires, the z-components cancel (as required by symmetry), and we arrive at

01 2 2 22 2

sin ˆ( )

Ia za z

0 ˆIzµ θ µππ

+ = =++

B = B B j j (9.11.26)

Figure 9.11.7 Superposition of magnetic fields due to two current sources To locate the maximum of B, we set /dB dz 0= and find

( )

2 20 0

22 2 2 2 2 2 2

1 2 0( )

I IdB z a zdz a z a z a z

µ µπ π

⎛ ⎞ −= − =⎜ ⎟+ +⎝ ⎠ +

2

= (9.11.27)

which gives z a= (9.11.28) Thus, at z=a, the magnetic field strength is a maximum, with a magnitude

0max 2

IBa

µπ

= (9.11.29)

9.11.6 Non-Uniform Current Density Consider an infinitely long, cylindrical conductor of radius R carrying a current I with a non-uniform current density J rα= (9.11.30) where α is a constant. Find the magnetic field everywhere.

56

Figure 9.11.8 Non-uniform current density Solution: The problem can be solved by using the Ampere’s law: 0 encd Iµ⋅ =∫ B s (9.11.31) where the enclosed current Ienc is given by ( )( )enc ' 2 ' 'I d r r dα π= ⋅ = r∫ ∫J A (9.11.32) (a) For r , the enclosed current is R<

3

2enc 0

22 ' '3

r rI r dr παπα= =∫ (9.11.33)

Applying Ampere’s law, the magnetic field at P1 is given by

( )3

01

223

rB r µ παπ = (9.11.34)

or

201 3

B rαµ= (9.11.35)

The direction of the magnetic field 1B is tangential to the Amperian loop which encloses the current. (b) For , the enclosed current is r R>

3

2enc 0

22 ' '3

R RI r dr παπα= =∫ (9.11.36)

which yields

57

( )3

02

223

RB r µ παπ = (9.11.37)

Thus, the magnetic field at a point P2 outside the conductor is

3

02 3

RBr

αµ= (9.11.38)

A plot of B as a function of r is shown in Figure 9.11.9:

Figure 9.11.9 The magnetic field as a function of distance away from the conductor 9.11.7 Thin Strip of Metal Consider an infinitely long, thin strip of metal of width w lying in the xy plane. The strip carries a current I along the +x-direction, as shown in Figure 9.11.10. Find the magnetic field at a point P which is in the plane of the strip and at a distance s away from it.

Figure 9.11.10 Thin strip of metal

58

Consider a thin strip of width dr parallel to the direction of the current and at a distance r away from P, as shown in Figure 9.11.11. The amount of current carried by this differential element is

drdI Iw

⎛= ⎜⎝ ⎠

⎞⎟ (9.11.39)

Using Ampere’s law, we see that the strip’s contribution to the magnetic field at P is given by 0 enc 0(2 ) ( )dB r I dIπ µ µ= = (9.11.40) or

0 0

2 2dI I drdBr r w

µ µπ π

⎛= = ⎜⎝ ⎠

⎞⎟ (9.11.41)

Figure 9.11.11 A thin strip with thickness carrying a steady current dr I . Integrating this expression, we obtain

0 0 ln2 2

s w

s

I Idr s wBw r w s

µ µπ π

+ +⎛ ⎞ ⎛= =⎜ ⎟ ⎜⎝ ⎠ ⎝∫ ⎞

⎟⎠

(9.11.42)

Using the right-hand rule, the direction of the magnetic field can be shown to point in the +z-direction, or

0 ˆln 1 2

I ww s

µπ

⎛ ⎞= +⎜ ⎟⎝ ⎠

B k (9.11.43)

Notice that in the limit of vanishing width, , w s ln(1 / ) /w s w s+ ≈ , and the above expression becomes

0 ˆ2

Is

µπ

=B k (9.11.44)

which is the magnetic field due to an infinitely long thin straight wire.

59

Solution:

9.11.8 Two Semi-Infinite Wires A wire carrying current I runs down the y axis to the origin, thence out to infinity along the positive x axis. Show that the magnetic field in the quadrant with of the xy plane is given by

,x y > 0

02 2 2 2

1 14z

I x yBx y y x y x x y

µπ

⎛ ⎞⎜= + + +⎜ + +⎝ ⎠

⎟⎟

(9.11.45)

Solution: Let ( , )P x y be a point in the first quadrant at a distance from a point (01r , ')y on the y-axis and distance from on the x-axis. 2r ( ',0)x

Figure 9.11.12 Two semi-infinite wires Using the Biot-Savart law, the magnetic field at P is given by

0 0 01 1 2 22 2

1 2wire wire

ˆ ˆˆ4 4 4y x

I I Id dddr r 2r

µ µ µπ π π

× ××= = = +∫ ∫ ∫ ∫

s r s rs rB B (9.11.46)

Let’s analyze each segment separately. (i) Along the y axis, consider a differential element 1

ˆ'd dy= −s j which is located at a

distance 1ˆ ( ' ˆ)x y y= + −r i j

y

from P. This yields (9.11.47) 1 1

ˆ ˆ ˆ ˆ( ' ) [ ( ') ] 'd dy x y y x d× = − × + − =s r j i j k

60

(ii) Similarly, along the x-axis, we have 2

ˆ'd dx=s i and 2ˆ( ') ˆx x y= − +r i j which gives

2 2

ˆ'd y d× =s r kx (9.11.48) Thus, we see that the magnetic field at P points in the +z-direction. Using the above

results and 2 21 ( ')r x y y= + − and ( )2 2

2r x x′= − + y , we obtain

0 02 2 3/ 2 2 20 0

'4 [ ( ') ] 4 [ ( ') ]z

I Ix dy y dxBx y y y x x 3/ 2

'µ µπ π

∞ ∞= +

+ − + −∫ ∫ (9.11.49)

The integrals can be readily evaluated using

2 2 3/ 2 2 20

1[ ( ) ]

b ds ab a s b b a b

∞= +

+ − +∫ (9.11.50)

The final expression for the magnetic field is given by

02 2 2 2

1 1 ˆ4

I y xx yx x y y x y

µπ

⎡ ⎤= + + +⎢

⎢ + +⎣ ⎦B k⎥

⎥ (9.11.51)

We may show that the result is consistent with Eq. (9.1.5) 9.12 Conceptual Questions 1. Compare and contrast Biot-Savart law in magnetostatics with Coulomb’s law in electrostatics. 2. If a current is passed through a spring, does the spring stretch or compress? Explain. 3. How is the path of the integration of d⋅∫ B s chosen when applying Ampere’s law? 4. Two concentric, coplanar circular loops of different diameters carry steady currents in the same direction. Do the loops attract or repel each other? Explain. 5. Suppose three infinitely long parallel wires are arranged in such a way that when looking at the cross section, they are at the corners of an equilateral triangle. Can currents be arranged (combination of flowing in or out of the page) so that all three wires (a) attract, and (b) repel each other? Explain.

61

9.13 Additional Problems 9.13.1 Application of Ampere's Law The simplest possible application of Ampere's law allows us to calculate the magnetic field in the vicinity of a single infinitely long wire. Adding more wires with differing currents will check your understanding of Ampere's law. (a) Calculate with Ampere's law the magnetic field, | | ( )B r=B , as a function of distance r from the wire, in the vicinity of an infinitely long straight wire that carries current I. Show with a sketch the integration path you choose and state explicitly how you use symmetry. What is the field at a distance of 10 mm from the wire if the current is 10 A?

(b) Eight parallel wires cut the page perpendicularly at the points shown. A wire labeled with the integer k (k = 1, 2, ... , 8) bears the current 2k times 0I (i.e., 02kI k I= ). For those with k = 1 to 4, the current flows up out of the page; for the rest, the current flows down into the page. Evaluate d⋅∫ B s along the closed path (see figure) in the direction indicated by the arrowhead. (Watch your signs!)

Figure 9.13.1 Amperian loop

(c) Can you use a single application of Ampere's Law to find the field at a point in the vicinity of the 8 wires? Why? How would you proceed to find the field at an arbitrary point P? 9.13.2 Magnetic Field of a Current Distribution from Ampere's Law Consider the cylindrical conductor with a hollow center and copper walls of thickness

as shown in Figure 9.13.2. The radii of the inner and outer walls are a and b respectively, and the current I is uniformly spread over the cross section of the copper. b a−

62

(a) Calculate the magnitude of the magnetic field in the region outside the conductor, . (Hint: consider the entire conductor to be a single thin wire, construct an

Amperian loop, and apply Ampere's Law.) What is the direction ofr b>

B ?

Figure 9.13.2 Hollow cylinder carrying a steady current I.

(b) Calculate the magnetic field inside the inner radius, r < a. What is the direction of B ? (c) Calculate the magnetic field within the inner conductor, a < r < b. What is the direction of ? B (d) Plot the behavior of the magnitude of the magnetic field B(r) from r = 0 to . Is B(r) continuous at r = a and r = b? What about its slope?

4r = b

(e) Now suppose that a very thin wire running down the center of the conductor carries the same current I in the opposite direction. Can you plot, roughly, the variation of B(r) without another detailed calculation? (Hint: remember that the vectors d from different current elements can be added to obtain the total vector magnetic field.)

B

9.13.3 Cylinder with a Hole A long copper rod of radius a has an off-center cylindrical hole through its entire length, as shown in Figure 9.13.3. The conductor carries a current I which is directed out of the page and is uniformly distributed throughout the cross section. Find the magnitude and direction of the magnetic field at the point P.

63Figure 9.13.3 A cylindrical conductor with a hole.

9.13.4 The Magnetic Field Through a Solenoid A solenoid has 200 closely spaced turns so that, for most of its length, it may be considered to be an ideal solenoid. It has a length of 0.25 m, a diameter of 0.1 m, and carries a current of 0.30 A. (a) Sketch the solenoid, showing clearly the rotation direction of the windings, the current direction, and the magnetic field lines (inside and outside) with arrows to show their direction. What is the dominant direction of the magnetic field inside the solenoid? (b) Find the magnitude of the magnetic field inside the solenoid by constructing an Amperian loop and applying Ampere's law. (c) Does the magnetic field have a component in the direction of the wire in the loops making up the solenoid? If so, calculate its magnitude both inside and outside the solenoid, at radii 30 mm and 60 mm respectively, and show the directions on your sketch. 9.13.5 Rotating Disk A circular disk of radius R with uniform charge density σ rotates with an angular speed ω . Show that the magnetic field at the center of the disk is

012

B Rµ σω=

Hint: Consider a circular ring of radius r and thickness dr. Show that the current in this element is ( / 2 )dI dq r drω π ωσ= = . 9.13.6 Four Long Conducting Wires Four infinitely long parallel wires carrying equal current I are arranged in such a way that when looking at the cross section, they are at the corners of a square, as shown in Figure 9.13.5. Currents in A and D point out of the page, and into the page at B and C. What is the magnetic field at the center of the square?

64

Figure 9.13.5 Four parallel conducting wires 9.13.7 Magnetic Force on a Current Loop A rectangular loop of length l and width carries a steady current w 1I . The loop is then placed near an finitely long wire carrying a current 2I , as shown in Figure 9.13.6. What is the magnetic force experienced by the loop due to the magnetic field of the wire?

Figure 9.13.6 Magnetic force on a current loop. 9.13.8 Magnetic Moment of an Orbital Electron We want to estimate the magnetic dipole moment associated with the motion of an electron as it orbits a proton. We use a “semi-classical” model to do this. Assume that the electron has speed v and orbits a proton (assumed to be very massive) located at the origin. The electron is moving in a right-handed sense with respect to the z-axis in a circle of radius r = 0.53 Å, as shown in Figure 9.13.7. Note that 1 Å = . 1010 m−

Figure 9.13.7

65

(a) The inward force required to make the electron move in this circle is provided by the Coulomb attractive force between the electron and proton (me is the mass of the electron). Using this fact, and the value of r we give above, find the speed of the electron in our “semi-classical” model. [Ans: .]

2 /em v r

62.18 10 m/s× (b) Given this speed, what is the orbital period T of the electron? [Ans: .] 161.52 10 s−× (c) What current is associated with this motion? Think of the electron as stretched out uniformly around the circumference of the circle. In a time T, the total amount of charge q that passes an observer at a point on the circle is just e [Ans: 1.05 mA. Big!] (d) What is the magnetic dipole moment associated with this orbital motion? Give the magnitude and direction. The magnitude of this dipole moment is one Bohr magneton, Bµ . [Ans: along the −z axis.] 24 29.27 10 A m−× ⋅ (e) One of the reasons this model is “semi-classical” is because classically there is no reason for the radius of the orbit above to assume the specific value we have given. The value of r is determined from quantum mechanical considerations, to wit that the orbital angular momentum of the electron can only assume integral multiples of h/2π, where

is the Planck constant. What is the orbital angular momentum of the electron here, in units of

346.63 10 J/sh −= ×/ 2h π ?

9.13.9 Ferromagnetism and Permanent Magnets A disk of iron has a height and a radius 1.00 mmh = 1.00 cmr = . The magnetic dipole moment of an atom of iron is 23 21.8 10 A mµ −= × ⋅ . The molar mass of iron is 55.85 g, and its density is 7.9 g/cm3. Assume that all the iron atoms in the disk have their dipole moments aligned with the axis of the disk. (a) What is the number density of the iron atoms? How many atoms are in this disk? [Ans: ; .] 28 38.5 10 atoms/m× 222.7 10 atoms× (b) What is the magnetization in this disk? [Ans: , parallel to axis.] M 61.53 10 A/m× (c) What is the magnetic dipole moment of the disk? [Ans: 20.48 A m⋅ .]

(d) If we were to wrap one loop of wire around a circle of the same radius r, how much current would the wire have to carry to get the dipole moment in (c)? This is the “equivalent” surface current due to the atomic currents in the interior of the magnet. [Ans: 1525 A.]

66

9.13.10 Charge in a Magnetic Field A coil of radius R with its symmetric axis along the +x-direction carries a steady current I. A positive charge q moves with a velocity ˆv=v j when it crosses the axis at a distance x from the center of the coil, as shown in Figure 9.13.8.

Figure 9.13.8

Describe the subsequent motion of the charge. What is the instantaneous radius of curvature? 9.13.11 Permanent Magnets A magnet in the shape of a cylindrical rod has a length of 4.8 cm and a diameter of 1.1 cm. It has a uniform magnetization M of 5300 A/m, directed parallel to its axis. (a) Calculate the magnetic dipole moment of this magnet. (b) What is the axial field a distance of 1 meter from the center of this magnet, along its axis? [Ans: (a) , (b) , or .] 2 22.42 10 A m−× ⋅ 94.8 10 T−× 54.8 10 gauss−× 9.13.12 Magnetic Field of a Solenoid (a) A 3000-turn solenoid has a length of 60 cm and a diameter of 8 cm. If this solenoid carries a current of 5.0 A, find the magnitude of the magnetic field inside the solenoid by constructing an Amperian loop and applying Ampere's Law. How does this compare to the magnetic field of the earth (0.5 gauss). [Ans: 0.0314 T, or 314 gauss, or about 600 times the magnetic field of the earth]. We make a magnetic field in the following way: We have a long cylindrical shell of non-conducting material which carries a surface charge fixed in place (glued down) of

, as shown in Figure 9.13.9 The cylinder is suspended in a manner such that it is free to revolve about its axis, without friction. Initially it is at rest. We come along and spin it up until the speed of the surface of the cylinder is .

2 C/mσ

0v

67

Figure 9.13.9 (b) What is the surface current K on the walls of the cylinder, in A/m? [Ans: 0K vσ= .] (c) What is magnetic field inside the cylinder? [Ans. 0 0 0B K vµ µ σ= = , oriented along axis right-handed with respect to spin.] (d) What is the magnetic field outside of the cylinder? Assume that the cylinder is infinitely long. [Ans: 0]. 9.13.13 Effect of Paramagnetism A solenoid with 16 turns/cm carries a current of 1.3 A. (a) By how much does the magnetic field inside the solenoid increase when a close-fitting chromium rod is inserted? [Note: Chromium is a paramagnetic material with magnetic susceptibility .] 42.7 10χ −= × (b) Find the magnitude of the magnetization M of the rod. [Ans: (a) 0.86 µT; (b) 0.68 A/m.]

68

Chapter 11

Inductance and Magnetic Energy

11.1 Mutual Inductance ..................................................................................................2

Example 11.1 Mutual Inductance of Two Concentric Coplanar Loops .....................4

11.2 Self-Inductance .......................................................................................................4

Example 11.2 Self-Inductance of a Solenoid..............................................................5 Example 11.3 Self-Inductance of a Toroid .................................................................6 Example 11.4 Mutual Inductance of a Coil Wrapped Around a Solenoid .................7

11.3 Energy Stored in Magnetic Fields ..........................................................................9

Example 11.5 Energy Stored in a Solenoid ..............................................................10 Animation 11.1: Creating and Destroying Magnetic Energy..................................11 Animation 11.2: Magnets and Conducting Rings ...................................................12

11.4 RL Circuits ............................................................................................................14

11.4.1 Self-Inductance and the Modified Kirchhoff's Loop Rule.............................14 11.4.2 Rising Current................................................................................................17 11.4.3 Decaying Current ...........................................................................................19

11.5 LC Oscillations .....................................................................................................20

11.6 The RLC Series Circuit .........................................................................................25

11.7 Summary...............................................................................................................27

11.8 Appendix 1: General Solutions for the RLC Series Circuit ..................................29

11.8.1 Quality Factor ................................................................................................31

11.9 Appendix 2: Stresses Transmitted by Magnetic Fields ........................................32

Animation 11.3: A Charged Particle in a Time-Varying Magnetic Field ...............36

11.10 Problem-Solving Strategies ................................................................................37

11.10.1 Calculating Self-Inductance.........................................................................37 11.10.2 Circuits containing inductors .......................................................................38

11.11 Solved Problems .................................................................................................38

11.11.1 Energy stored in a toroid..............................................................................38 11.11.2 Magnetic Energy Density ............................................................................39 11.11.3 Mutual Inductance .......................................................................................40 11.11.4 RL Circuit.....................................................................................................41 11.11.5 RL Circuit.....................................................................................................43 11.11.6 LC Circuit.....................................................................................................44

11.12 Conceptual Questions .........................................................................................46

0

11.13 Additional Problems ...........................................................................................47

11.13.1 Solenoid .......................................................................................................47 11.13.2 Self-Inductance ............................................................................................47 11.13.3 Coupled Inductors........................................................................................47 11.13.4 RL Circuit.....................................................................................................48 11.13.5 RL Circuit.....................................................................................................49 11.13.6 Inductance of a Solenoid With and Without Iron Core ...............................49 11.13.7 RLC Circuit ..................................................................................................50 11.13.8 Spinning Cylinder ........................................................................................51 11.13.9 Spinning Loop..............................................................................................51

1

Inductance and Magnetic Energy 11.1 Mutual Inductance Suppose two coils are placed near each other, as shown in Figure 11.1.1

Figure 11.1.1 Changing current in coil 1 produces changing magnetic flux in coil 2. The first coil has N1 turns and carries a current I1 which gives rise to a magnetic field 1B . Since the two coils are close to each other, some of the magnetic field lines through coil 1 will also pass through coil 2. Let 21Φ denote the magnetic flux through one turn of coil 2 due to I1. Now, by varying I1 with time, there will be an induced emf associated with the changing magnetic flux in the second coil:

2121 2 1 2

coil 2

d dNdt dt

ε dΦ= − = − ⋅∫∫ B A (11.1.1)

The time rate of change of magnetic flux 21Φ in coil 2 is proportional to the time rate of change of the current in coil 1:

21 12 2

dN Mdt dt1

dIΦ= (11.1.2)

where the proportionality constant 21M is called the mutual inductance. It can also be written as

2 2121

1

NMIΦ

= (11.1.3)

The SI unit for inductance is the henry (H):

2

(11.1.4) 21 henry 1 H 1 T m /A= = ⋅ We shall see that the mutual inductance 21M depends only on the geometrical properties of the two coils such as the number of turns and the radii of the two coils. In a similar manner, suppose instead there is a current I2 in the second coil and it is varying with time (Figure 11.1.2). Then the induced emf in coil 1 becomes

1212 1 2 1

coil 1

d dNdt dt

ε dΦ= − = − ⋅∫∫ B A (11.1.5)

and a current is induced in coil 1.

Figure 11.1.2 Changing current in coil 2 produces changing magnetic flux in coil 1. This changing flux in coil 1 is proportional to the changing current in coil 2,

12 21 12

dN Mdt dt

dIΦ= (11.1.6)

where the proportionality constant 12M is another mutual inductance and can be written as

1 1212

2

NMIΦ

= (11.1.7)

However, using the reciprocity theorem which combines Ampere’s law and the Biot-Savart law, one may show that the constants are equal: 12 21M M M= ≡ (11.1.8)

3

Example 11.1 Mutual Inductance of Two Concentric Coplanar Loops Consider two single-turn co-planar, concentric coils of radii R1 and R2, with 1 2R R , as shown in Figure 11.1.3. What is the mutual inductance between the two loops?

Figure 11.1.3 Two concentric current loop Solution: The mutual inductance can be computed as follows. Using Eq. (9.1.15) of Chapter 9, we see that the magnetic field at the center of the ring due to I1 in the outer coil is given by

0 11

12IB

Rµ

= (11.1.9)

Since 1 2R R , we approximate the magnetic field through the entire inner coil by 1B . Hence, the flux through the second (inner) coil is

2

20 1 0 1 221 1 2 2

1 12 2I I RB A R

R Rµ µ ππ

⎛ ⎞Φ = = =⎜ ⎟

⎝ ⎠ (11.1.10)

Thus, the mutual inductance is given by

2

0 221

1 12RM

I Rµ πΦ

= = (11.1.11)

The result shows that M depends only on the geometrical factors, 1R and 2R , and is independent of the current 1I in the coil. 11.2 Self-Inductance Consider again a coil consisting of N turns and carrying current I in the counterclockwise direction, as shown in Figure 11.2.1. If the current is steady, then the magnetic flux through the loop will remain constant. However, suppose the current I changes with time,

4

then according to Faraday’s law, an induced emf will arise to oppose the change. The induced current will flow clockwise if , and counterclockwise if /dI dt > 0 / 0dI dt < . The property of the loop in which its own magnetic field opposes any change in current is called “self-inductance,” and the emf generated is called the self-induced emf or back emf, which we denote as Lε . All current-carrying loops exhibit this property. In particular, an inductor is a circuit element (symbol ) which has a large self-inductance.

Figure 11.2.1 Magnetic flux through the current loop Mathematically, the self-induced emf can be written as

BL

d dN Ndt dt

ε dΦ= − = − ⋅∫∫B A (11.2.1)

and is related to the self-inductance L by

LdILdt

ε = − (11.2.2)

The two expressions can be combined to yield

BNLIΦ

= (11.2.3)

Physically, the inductance L is a measure of an inductor’s “resistance” to the change of current; the larger the value of L, the lower the rate of change of current. Example 11.2 Self-Inductance of a Solenoid Compute the self-inductance of a solenoid with turns, length l , and radius N R with a current I flowing through each turn, as shown in Figure 11.2.2.

5

Figure 11.2.2 Solenoid Solution: Ignoring edge effects and applying Ampere’s law, the magnetic field inside a solenoid is given by Eq. (9.4.3):

00

ˆNI nIl

ˆµ µ= =B k k (11.2.4)

where is the number of turns per unit length. The magnetic flux through each turn is

/n N l=

2

0 0( )B2BA nI R nI Rµ π µ πΦ = = ⋅ = (11.2.5)

Thus, the self-inductance is

2 20

BNL n R lI

µ πΦ= = (11.2.6)

We see that L depends only on the geometrical factors (n, R and l) and is independent of the current I. Example 11.3 Self-Inductance of a Toroid Calculate the self-inductance of a toroid which consists of N turns and has a rectangular cross section, with inner radius a, outer radius b and height h, as shown in Figure 11.2.3(a).

(a) (b)

Figure 11.2.3 A toroid with N turns

6

Solution: According to Ampere’s law discussed in Section 9.3, the magnetic field is given by (11.2.7) 0(2 )d Bds B ds B r Nπ µ⋅ = = = =∫ ∫ ∫B s I or

0

2NIBr

µπ

= (11.2.8)

The magnetic flux through one turn of the toroid may be obtained by integrating over the rectangular cross section, with dA h dr= as the differential area element (Figure 11.2.3b):

0 0 ln2 2

b

B a

NI NIh bd hdrr a

µ µπ π

⎛ ⎞ ⎛ ⎞Φ = ⋅ = = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∫∫ ∫B A (11.2.9)

The total flux is . Therefore, the self-inductance is BNΦ

2

0 ln2

B N hNL bI a

µπ

Φ ⎛ ⎞= = ⎜ ⎟⎝ ⎠

(11.2.10)

Again, the self-inductance L depends only on the geometrical factors. Let’s consider the situation where a b . In this limit, the logarithmic term in the equation above may be expanded as

a−

ln ln 1b b a ba a

aa

− −⎛ ⎞ ⎛ ⎞= + ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(11.2.11)

and the self-inductance becomes

2 2

0 0

2 2N h N A N Ab aL

a a lµ µ

π π−

≈ ⋅ = =2

0µ

)

(11.2.12)

where (A h b a= − is the cross-sectional area, and 2l aπ= . We see that the self-inductance of the toroid in this limit has the same form as that of a solenoid. Example 11.4 Mutual Inductance of a Coil Wrapped Around a Solenoid A long solenoid with length l and a cross-sectional area A consists of N1 turns of wire. An insulated coil of N2 turns is wrapped around it, as shown in Figure 11.2.4.

7

(a) Calculate the mutual inductance M , assuming that all the flux from the solenoid passes through the outer coil. (b) Relate the mutual inductance M to the self-inductances and of the solenoid and the coil.

1L 2L

Figure 11.2.4 A coil wrapped around a solenoid Solutions: (a) The magnetic flux through each turn of the outer coil due to the solenoid is

0 1 121

N IBAl

AµΦ = = (11.2.13)

where 0 1 1 /B N I lµ= is the uniform magnetic field inside the solenoid. Thus, the mutual inductance is

0 1 22 21

1

N N ANMI l

µΦ= = (11.2.14)

(b) From Example 11.2, we see that the self-inductance of the solenoid with N1 turns is given by

2

0 11 111

1

N ANLI l

µΦ= = (11.2.15)

where is the magnetic flux through one turn of the solenoid due to the magnetic field produced by

11Φ

1I . Similarly, we have 22 0 2 /L N A lµ= for the outer coil. In terms of and

, the mutual inductance can be written as 1L

2L 1 2M L L= (11.2.16) More generally the mutual inductance is given by 1 2 , 0 1M k L L k= ≤ ≤ (11.2.17)

8

where is the “coupling coefficient.” In our example, we have k 1k = which means that all of the magnetic flux produced by the solenoid passes through the outer coil, and vice versa, in this idealization. 11.3 Energy Stored in Magnetic Fields Since an inductor in a circuit serves to oppose any change in the current through it, work must be done by an external source such as a battery in order to establish a current in the inductor. From the work-energy theorem, we conclude that energy can be stored in an inductor. The role played by an inductor in the magnetic case is analogous to that of a capacitor in the electric case. The power, or rate at which an external emf extε works to overcome the self-induced emf

Lε and pass current I in the inductor is

extextL

dWPdt

Iε= = (11.3.1)

If only the external emf and the inductor are present, then ext Lε ε= − which implies

extL L

dW dIP Idt dt

ε= = − = +IL

0

0

(11.3.2)

If the current is increasing with , then which means that the external source is doing positive work to transfer energy to the inductor. Thus, the internal energy

of the inductor is increased. On the other hand, if the current is decreasing with , we then have . In this case, the external source takes energy away from

the inductor, causing its internal energy to go down. The total work done by the external source to increase the current form zero to

/dI dt > 0P >

BU/dI dt < 0P <

I is then

2ext ext 0

1' '2

IW dW LI dI L= = =∫ ∫ I (11.3.3)

This is equal to the magnetic energy stored in the inductor:

212BU L= I (11.3.4)

The above expression is analogous to the electric energy stored in a capacitor:

21

2EQUC

= (11.3.5)

9

We comment that from the energy perspective there is an important distinction between an inductor and a resistor. Whenever a current I goes through a resistor, energy flows into the resistor and dissipates in the form of heat regardless of whether I is steady or time-dependent (recall that power dissipated in a resistor is ). On the other hand, energy flows into an ideal inductor only when the current is varying with . The energy is not dissipated but stored there; it is released later when the current decreases with . If the current that passes through the inductor is steady, then there is no change in energy since .

2R RP IV I R= =

/dI dt > 0

0/dI dt <( / ) 0LP LI dI dt= =

Example 11.5 Energy Stored in a Solenoid A long solenoid with length l and a radius R consists of N turns of wire. A current I passes through the coil. Find the energy stored in the system. Solution: Using Eqs. (11.2.6) and (11.3.4), we readily obtain

2 2 20

1 12 2BU LI n I Rµ π= = 2l (11.3.6)

The result can be expressed in terms of the magnetic field strength 0B nIµ= :

2

2 2 20

0 0

1 ( ) ( ) (2 2B

BU nI R lµ π πµ µ

= = )R l (11.3.7)

Since 2R lπ is the volume within the solenoid, and the magnetic field inside is uniform, the term

2

02BBuµ

= (11.3.8)

may be identified as the magnetic energy density, or the energy per unit volume of the magnetic field. The above expression holds true even when the magnetic field is non-uniform. The result can be compared with the energy density associated with an electric field:

20

12Eu Eε= (11.3.9)

10

Animation 11.1: Creating and Destroying Magnetic Energy Let’s consider the process involved in creating magnetic energy. Figure 11.3.1 shows the process by which an external agent(s) creates magnetic energy. Suppose we have five rings that carry a number of free positive charges that are not moving. Since there is no current, there is no magnetic field. Now suppose a set of external agents come along (one for each charge) and simultaneously spin up the charges counterclockwise as seen from above, at the same time and at the same rate, in a manner that has been pre-arranged. Once the charges on the rings start to accelerate, there is a magnetic field in the space between the rings, mostly parallel to their common axis, which is stronger inside the rings than outside. This is the solenoid configuration.

Figure 11.3.1 Creating and destroying magnetic field energy. As the magnetic flux through the rings grows, Faraday’s law of induction tells us that there is an electric field induced by the time-changing magnetic field that is circulating clockwise as seen from above. The force on the charges due to this electric field is thus opposite the direction the external agents are trying to spin the rings up (counterclockwise), and thus the agents have to do additional work to spin up the charges because of their charge. This is the source of the energy that is appearing in the magnetic field between the rings — the work done by the agents against the “back emf.”

Over the course of the “create” animation associated with Figure 11.3.1, the agents moving the charges to a higher speed against the induced electric field are continually doing work. The electromagnetic energy that they are creating at the place where they are doing work (the path along which the charges move) flows both inward and outward. The direction of the flow of this energy is shown by the animated texture patterns in Figure 11.3.1. This is the electromagnetic energy flow that increases the strength of the magnetic field in the space between the rings as each positive charge is accelerated to a higher and higher speed. When the external agents have gotten up the charges to a pre-determined speed, they stop the acceleration. The charges then move at a constant speed, with a constant field inside the solenoid, and zero “induced” electric field, in accordance with Faraday’s law of induction. We also have an animation of the “destroy” process linked to Figure 11.3.1. This process proceeds as follows. Our set of external agents now simultaneously start to spin down the moving charges (which are still moving counterclockwise as seen from above), at the

11

same time and at the same rate, in a manner that has been pre-arranged. Once the charges on the rings start to decelerate, the magnetic field in the space between the rings starts to decrease in magnitude. As the magnetic flux through the rings decreases, Faraday’s law tells us that there is now an electric field induced by the time-changing magnetic field that is circulating counterclockwise as seen from above. The force on the charges due to this electric field is thus in the same direction as the motion of the charges. In this situation the agents have work done on them as they try to spin the charges down.

Over the course of the “destroy” animation associated with Figure 11.3.1, the strength of the magnetic field decreases, and this energy flows from the field back to the path along which the charges move, and is now being provided to the agents trying to spin down the moving charges. The energy provided to those agents as they destroy the magnetic field is exactly the amount of energy that they put into creating the magnetic field in the first place, neglecting radiative losses (such losses are small if we move the charges at speeds small compared to the speed of light). This is a totally reversible process if we neglect such losses. That is, the amount of energy the agents put into creating the magnetic field is exactly returned to the agents as the field is destroyed.

There is one final point to be made. Whenever electromagnetic energy is being created, an electric charge is moving (or being moved) against an electric field ( ). Whenever electromagnetic energy is being destroyed, an electric charge is moving (or being moved) along an electric field (

0q ⋅ <v E

0q ⋅ >v E ). This is the same rule we saw above when we were creating and destroying electric energy above. Animation 11.2: Magnets and Conducting Rings In the example of Faraday’s law that we gave above, the sense of the electric field associated with a time-changing magnetic field was always such as to try to resist change. We consider another example of Faraday’s law that illustrates this same tendency in a different way.

Figure 11.3.2 A perfectly conducting ring falls on the axis of a permanent magnet. The induced currents and the resulting magnetic field stresses are such as to slow the fall of the ring. If the ring is light enough (or the magnet strong enough), the ring will levitate above the magnet.

12

In Figure 11.3.2, we show a permanent magnet that is fixed at the origin with its dipole moment pointing upward. On the z-axis above the magnet, we have a co-axial, conducting, non-magnetic ring with radius a, inductance L, and resistance R. The center of the conducting ring is constrained to move along the vertical axis. The ring is released from rest and falls under gravity toward the stationary magnet. Eddy currents arise in the ring because of the changing magnetic flux and induced electric field as the ring falls toward the magnet, and the sense of these currents is to repel the ring when it is above the magnet. This physical situation can be formulated mathematically in terms of three coupled ordinary differential equations for the position of the ring, its velocity, and the current in the ring. We consider in Figure 11.3.2 the particular situation where the resistance of the ring (which in our model can have any value) is identically zero, and the mass of the ring is small enough (or the field of the magnet is large enough) so that the ring levitates above the magnet. We let the ring begin at rest a distance 2a above the magnet. The ring begins to fall under gravity. When the ring reaches a distance of about a above the ring, its acceleration slows because of the increasing current in the ring. As the current increases, energy is stored in the magnetic field, and when the ring comes to rest, all of the initial gravitational potential of the ring is stored in the magnetic field. That magnetic energy is then returned to the ring as it “bounces” and returns to its original position a distance 2a above the magnet. Because there is no dissipation in the system for our particular choice of R in this example, this motion repeats indefinitely.

What are the important points to be learned from this animation? Initially, all the free energy in this situation is stored in the gravitational potential energy of the ring. As the ring begins to fall, that gravitational energy begins to appear as kinetic energy in the ring. It also begins to appear as energy stored in the magnetic field. The compressed field below the ring enables the transmission of an upward force to the moving ring as well as a downward force to the magnet. But that compression also stores energy in the magnetic field. It is plausible to argue based on the animation that the kinetic energy of the downwardly moving ring is decreasing as more and more energy is stored in the magnetostatic field, and conversely when the ring is rising. Figure 11.3.3 shows a more realistic case in which the resistance of the ring is finite. Now energy is not conserved, and the ring eventually falls past the magnet. When it passes the magnet, the sense of the induced electric field and thus of the eddy currents reverses, and the ring is now attracted to the magnet above it, which again retards its fall. There are many other examples of the falling ring and stationary magnet, or falling magnet and stationary ring, given in the animations at this link. All of them show that the effect of the electric field associated with a time-changing magnetic field is to try to keep things the same. In the limiting case of zero resistance, it can in fact achieve this goal, e.g. in Figure 11.3.2 the magnetic flux through the ring never changes over the course of the motion.

13

Figure 11.3.3 A ring with finite resistance falls on the axis of a magnet. We show the ring after it has fallen past the magnet, when it is attracted to the magnet above it.

11.4 RL Circuits 11.4.1 Self-Inductance and the Modified Kirchhoff's Loop Rule The addition of time-changing magnetic fields to simple circuits means that the closed line integral of the electric field around a circuit is no longer zero. Instead, we have, for any open surface

dddt

d⋅ = − ⋅∫ ∫∫E s B A (11.4.1)

Any circuit where the current changes with time will have time-changing magnetic fields, and therefore induced electric fields. How do we solve simple circuits taking such effects into account? We discuss here a consistent way to understand the consequences of introducing time-changing magnetic fields into circuit theory -- that is, inductance. As soon as we introduce time-changing magnetic fields, the electric potential difference between two points in our circuit is no longer well-defined, because when the line integral of the electric field around a closed loop is nonzero, the potential difference between two points, say a and b, is no longer independent of the path taken to get from a to b. That is, the electric field is no longer a conservative field, and the electric potential is no longer an appropriate concept, since we can no longer write E as the negative gradient of a scalar potential. However, we can still write down in a straightforward fashion the equation that determines the behavior of a circuit.

Figure 11.4.1 One-loop inductor circuit

14

To show how to do this, consider the circuit shown in Figure 11.4.1. We have a battery, a resistor, a switch S that is closed at t = 0, and a “one-loop inductor.” It will become clear what the consequences of this “inductance” are as we proceed. For t > 0, current will flow in the direction shown (from the positive terminal of the battery to the negative, as usual). What is the equation that governs the behavior of our current ( )I t for t > 0? To investigate this, apply Faraday's law to the open surface bounded by our circuit, where we take d to be out of the page, and dA s right-handed with respect to that choice (counter-clockwise). First, what is the integral of the electric field around this circuit? There is an electric field in the battery, directed from the positive terminal to the negative terminal, and when we go through the battery in the direction of d s that we have chosen, we are moving against that electric field, so that 0d⋅ <E s . Thus the contribution of the battery to our integral is ε− . Then, there is an electric field in the resistor, in the direction of the current, so when we move through the resistor in that direction, is greater than zero, and that contribution to our integral is

d⋅E sIR+ . What about when we move

through our one-loop inductor? There is no electric field in this loop if the resistance of the wire making up the loop is zero. Thus, going around the closed loop clockwise against the current, we have d ε IR⋅ = − +∫ E s (11.4.2) Now, what is the magnetic flux BΦ through our open surface? First of all, we arrange the geometry so that the part of the circuit which includes the battery, the switch, and the resistor makes only a small contribution to BΦ as compared to the (much larger in area) part of the open surface which includes our one-loop inductor. Second, we know that

is positive in that part of the surface, because current flowing counterclockwise will produce a magnetic field B pointing out of the page, which is the same direction we have assumed for . Thus, the dot product

BΦ

dA 0d⋅ >B A . Note that B is the self magnetic field − that is, the magnetic field produced by the current flowing in the circuit, and not by any external currents. From Section 11.1, we also see that the magnetic flux BΦ is proportional to I, and may be written as , where L is the self-inductance which depends on the geometry of the circuit. The time rate of change of

B LIΦ =

BΦ is just ( / )L dI dt , so that we have from Faraday's law

Bd dId IR Ldt dt

ε Φ⋅ = − + = − = −∫ E s (11.4.3)

We can write the governing equation for ( )I t from above as

15

0dIV IR Ldt

ε∆ = − − = (11.4.4)

where the expression has been cast in a form that resembles Kirchhoff's loop rule, namely that the sum of the potential drops around a circuit is zero. To preserve the loop rule, we must specify the “potential drop” across an inductor.

(a) (b)

Figure 11.4.2 Modified Kirchhoff’s rule for inductors (a) with increasing current, and (b) with decreasing current. The modified rule for inductors may be obtained as follows: The polarity of the self-induced emf is such as to oppose the change in current, in accord with Lenz’s law. If the rate of change of current is positive, as shown in Figure 11.4.2(a), the self-induced emf

Lε sets up an induced current indI moving in the opposite direction of the current I to oppose such an increase. The inductor could be replaced by an emf | | | / | ( /L L dI dt L dI dt)ε = = + with the polarity shown in Figure 11.4.2(a). On the other hand, if , as shown in Figure 11.4.2(b), the induced current /dI dt < 0 indI set up by the self-induced emf Lε flows in the same direction as I to oppose such a decrease. We see that whether the rate of change of current in increasing ( ) or decreasing ( ), in both cases, the change in potential when moving from a to b along the direction of the current I is . Thus, we have

/dI dt > 00/dI dt <

( / )b aV V L dI d t− = −

Kirchhoff's Loop Rule Modified for Inductors: If an inductor is traversed in the direction of the current, the “potential change” is

( / )L dI dt− . On the other hand, if the inductor is traversed in the direction opposite of the current, the “potential change” is ( / )L dI dt+ . Use of this modified Kirchhoff’s rule will give the correct equations for circuit problems that contain inductors. However, keep in mind that it is misleading at best, and at some level wrong in terms of the physics. Again, we emphasize that Kirchhoff's loop rule was originally based on the fact that the line integral of E around a closed loop was zero.

16

With time-changing magnetic fields, this is no longer so, and thus the sum of the “potential drops” around the circuit, if we take that to mean the negative of the closed loop integral of , is no longer zero − in fact it is E ( / )L dI dt+ . 11.4.2 Rising Current Consider the RL circuit shown in Figure 11.4.3. At 0t = the switch is closed. We find that the current does not rise immediately to its maximum value / Rε . This is due to the presence of the self-induced emf in the inductor.

Figure 11.4.3 (a) RL Circuit with rising current. (b) Equivalent circuit using the modified Kirchhoff’s loop rule. Using the modified Kirchhoff’s rule for increasing current, , the RL circuit is described by the following differential equation:

/dI dt > 0

| | 0LdIIR IR Ldt

ε ε ε− − = − − = (11.4.5)

Note that there is an important distinction between an inductor and a resistor. The potential difference across a resistor depends on I, while the potential difference across an inductor depends on . The self-induced emf does not oppose the current itself, but the change of current .

/dI d t/dI d t

The above equation can be rewritten as

/ /

dI dtI R Lε

= −− R

(11.4.6)

Integrating over both sides and imposing the condition ( 0) 0I t = = , the solution to the differential equation is

( /( ) 1 tI t e )Rτε −= − (11.4.7)

where

17

LR

τ = (11.4.8)

is the time constant of the RL circuit. The qualitative behavior of the current as a function of time is depicted in Figure 11.4.4.

Figure 11.4.4 Current in the RL circuit as a function of time

Note that after a sufficiently long time, the current reaches its equilibrium value / Rε . The time constant τ is a measure of how fast the equilibrium state is attained; the larger the value of L, the longer it takes to build up the current. A comparison of the behavior of current in a circuit with or without an inductor is shown in Figure 11.4.5 below. Similarly, the magnitude of the self-induced emf can be obtained as

/| | tL

dIL edt

τε −= − = ε (11.4.9)

which is at a maximum when t = 0 and vanishes as t approaches infinity. This implies that a sufficiently long time after the switch is closed, self-induction disappears and the inductor simply acts as a conducting wire connecting two parts of the circuit.

Figure 11.4.5 Behavior of current in a circuit with or without an inductor

To see that energy is conserved in the circuit, we multiply Eq. (11.4.7) by I and obtain

2 dII I R LIdt

ε = + (11.4.10)

18

The left-hand side represents the rate at which the battery delivers energy to the circuit. On the other hand, the first term on the right-hand side is the power dissipated in the resistor in the form of heat, and the second term is the rate at which energy is stored in the inductor. While the energy dissipated through the resistor is irrecoverable, the magnetic energy stored in the inductor can be released later. 11.4.3 Decaying Current Next we consider the RL circuit shown in Figure 11.4.6. Suppose the switch S1 has been closed for a long time so that the current is at its equilibrium value / Rε . What happens to the current when at t = 0 switches S1 is opened and S2 closed? Applying modified Kirchhoff’s loop rule to the right loop for decreasing current,

, yields /dI dt < 0

| | 0LdIIR L IRdt

ε − = − − = (11.4.11)

which can be rewritten as

/

dI dtI L R

= − (11.4.12)

Figure 11.4.6 (a) RL circuit with decaying current, and (b) equivalent circuit. The solution to the above differential equation is

/( ) tI t eR

τε −= (11.4.13)

where /L Rτ = is the same time constant as in the case of rising current. A plot of the current as a function of time is shown in Figure 11.4.7.

19

Figure 11.4.7 Decaying current in an RL circuit 11.5 LC Oscillations Consider an LC circuit in which a capacitor is connected to an inductor, as shown in Figure 11.5.1.

Figure 11.5.1 LC Circuit Suppose the capacitor initially has charge . When the switch is closed, the capacitor begins to discharge and the electric energy is decreased. On the other hand, the current created from the discharging process generates magnetic energy which then gets stored in the inductor. In the absence of resistance, the total energy is transformed back and forth between the electric energy in the capacitor and the magnetic energy in the inductor. This phenomenon is called electromagnetic oscillation.

0Q

The total energy in the LC circuit at some instant after closing the switch is

2

21 12 2C L

QU U U LIC

= + = + (11.5.1)

The fact that U remains constant implies that

2

21 1 02 2

dU d Q Q dQ dILI LIdt dt C C dt dt

⎛ ⎞= + = +⎜ ⎟

⎝ ⎠= (11.5.2)

or

20

2

2 0Q d QLC dt

+ = (11.5.3)

where /I dQ dt= − (and ). Notice the sign convention we have adopted here. The negative sign implies that the current I is equal to the rate of decrease of charge in the capacitor plate immediately after the switch has been closed. The same equation can be obtained by applying the modified Kirchhoff’s loop rule clockwise:

2/dI dt d Q dt= − 2/

0Q dILC dt

− = (11.5.4)

followed by our definition of current. The general solution to Eq. (11.5.3) is 0 0( ) cos( )Q t Q tω φ= + (11.5.5) where Q0 is the amplitude of the charge and φ is the phase. The angular frequency 0ω is given by

01LC

ω = (11.5.6)

The corresponding current in the inductor is

0 0 0 0 0( ) sin( ) sin( )dQI t Q t I tdt

ω ω φ ω φ= − = + = + (11.5.7)

where 0 0 0I Qω= . From the initial conditions 0( 0)Q t Q= = and ( 0)I t 0= = , the phase φ can be determined to be 0φ = . Thus, the solutions for the charge and the current in our LC circuit are 0( ) cosQ t Q t0ω= (11.5.8) and 0( ) sin 0I t I tω= (11.5.9) The time dependence of and ( )Q t ( )I t are depicted in Figure 11.5.2.

21

Figure 11.5.2 Charge and current in the LC circuit as a function of time

Using Eqs. (11.5.8) and (11.5.9), we see that at any instant of time, the electric energy and the magnetic energies are given by

22

200

( ) cos2 2E

QQ tUC C

tω⎛ ⎞

= = ⎜ ⎟⎝ ⎠

(11.5.10)

and

2 2 2

2 2 20 0 0 00

( )1 ( ) sin sin sin2 2 2 2B

LI L Q QU LI t t tC

ω 20tω ω

⎛ ⎞−= = = = ⎜ ⎟

⎝ ⎠ω (11.5.11)

respectively. One can easily show that the total energy remains constant:

2 2

2 20 00 0cos sin

2 2E BQ QU U U t tC C

ω ω⎛ ⎞ ⎛ ⎞

= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

20

2QC

(11.5.12)

The electric and magnetic energy oscillation is illustrated in Figure 11.5.3.

Figure 11.5.3 Electric and magnetic energy oscillations The mechanical analog of the LC oscillations is the mass-spring system, shown in Figure 11.5.4.

Figure 11.5.4 Mass-spring oscillations

22

If the mass is moving with a speed v and the spring having a spring constant k is displaced from its equilibrium by x, then the total energy of this mechanical system is

2sp

1 12 2

U K U mv kx= + = + 2 (11.5.13)

where K and are the kinetic energy of the mass and the potential energy of the spring, respectively. In the absence of friction, U is conserved and we obtain

spU

2 21 1 02 2

dU d dv dxmv kx mv kxdt dt dt dt

⎛ ⎞= + = +⎜ ⎟⎝ ⎠

=

2

(11.5.14)

Using and , the above equation may be rewritten as /v dx dt= 2/ /dv dt d x dt=

2

2 0d xm kxdt

+ = (11.5.15)

The general solution for the displacement is 0 0( ) cos( )x t x tω φ= + (11.5.16) where

0km

ω = (11.5.17)

is the angular frequency and x0 is the amplitude of the oscillations. Thus, at any instant in time, the energy of the system may be written as

2 2 2 2 20 0 0 0 0

2 2 20 0 0

1 1sin ( ) cos ( )2 21 1sin ( ) cos ( )2 2

U mx t kx t

kx t t kx20

ω ω φ ω φ

ω φ ω φ

= + +

⎡ ⎤= + + +⎣ ⎦

+

= (11.5.18)

In Figure 11.5.5 we illustrate the energy oscillations in the LC Circuit and the mass-spring system (harmonic oscillator).

23

LC Circuit

Mass-spring System

Energy

Figure 11.5.5 Energy oscillations in the LC Circuit and the mass-spring system

24

11.6 The RLC Series Circuit We now consider a series RLC circuit which contains a resistor, an inductor and a capacitor, as shown in Figure 11.6.1.

Figure 11.6.1 A series RLC circuit

The capacitor is initially charged to . After the switch is closed current will begin to flow. However, unlike the LC circuit energy will be dissipated through the resistor. The rate at which energy is dissipated is

0Q

2dU I Rdt

= − (11.6.1)

where the negative sign on the right-hand side implies that the total energy is decreasing. After substituting Eq. (11.5.2) for the left-hand side of the above equation, we obtain the following differential equation:

2Q dQ dILI IC dt dt

+ = − R (11.6.2)

Again, by our sign convention where current is equal to the rate of decrease of charge in the capacitor plates, /I dQ dt= − . Dividing both sides by I, the above equation can be rewritten as

2

2 0d Q dQ QL Rdt dt C

+ + = (11.6.3)

For small R (the underdamped case, see Appendix 1), one can readily verify that a solution to the above equation is 0( ) cos( ' )tQ t Q e tγ ω φ−= + (11.6.4) where

25

2RL

γ = (11.6.5)

is the damping factor and 2 2

0'ω ω γ= − (11.6.6) is the angular frequency of the damped oscillations. The constants and 0Q φ are real quantities to be determined from the initial conditions. In the limit where the resistance vanishes, , we recover the undamped, natural angular frequency0R = 0 1/ LCω = . There are three possible scenarios and the details are discussed in Appendix 1 (Section 11.8). The mechanical analog of the series RLC circuit is the damped harmonic oscillator system. The equation of motion for this system is given by

2

2 0d x dxm b kxdt dt

+ + = (11.6.7)

where the velocity-dependent term accounts for the non-conservative, dissipative force

dxF bdt

= − (11.6.8)

with b being the damping coefficient. The correspondence between the RLC circuit and the mechanical system is summarized in Table 11.6.1. (Note that the sign of the current I depends on the physical situation under consideration.)

RLC Circuit Damped Harmonic Oscillator

Variable s Q x

Variable ds/dt I± v

Coefficient of s 1/C k

Coefficient of ds/dt R b

Coefficient of d2s/dt2 L m

LI 2/2 mv2/2 Energy

Q2/2C kx2/2

Table 11.6.1 Correspondence between the RLC circuit and the mass-spring system

26

11.7 Summary

• Using Faraday’s law of induction, the mutual inductance of two coils is given by

12 12 1 2112 21

1 2

N NM M MI IΦ Φ

= = = =

• The induced emf in coil 2 due to the change in current in coil 1 is given by

12

dIMdt

ε = −

• The self-inductance of a coil with N turns is

BNLIΦ

=

where is the magnetic flux through one turn of the coil. BΦ

• The self-induced emf responding to a change in current inside a coil current is

LdILdt

ε = −

• The inductance of a solenoid with N turns, cross sectional area A and length l is

2

0N ALl

µ=

• If a battery supplying an emf ε is connected to an inductor and a resistor in series

at time t = 0, then the current in this RL circuit as a function of time is

( ) ( )/1 tI t eR

τε −= −

where /L Rτ = is the time constant of the circuit. If the battery is removed in the

RL circuit, the current will decay as

( ) /tI t eR

τε −⎛ ⎞= ⎜ ⎟⎝ ⎠

• The magnetic energy stored in an inductor with current I passing through is

27

212BU L= I

• The magnetic energy density at a point with magnetic field B is

2

02BBuµ

=

• The differential equation for an oscillating LC circuit is

2202 0d Q Q

dtω+ =

where 01LC

ω = is the angular frequency of oscillation. The charge on the

capacitor as a function of time is given by

( )0 0( ) cosQ t Q tω φ= +

and the current in the circuit is

( )0 0 0( ) sindQI t Q tdt

ω ω φ= − = + +

• The total energy in an LC circuit is, using 0 0 0I Qω= ,

2 22 20 0

0 0cos sin2 2E BQ LI QU U U t tC C

ω ω= + = + =20

2

• The differential equation for an RLC circuit is

2

202 2 0d Q dQ Q

dt d tγ ω+ + =

where 01LC

ω = and / 2R Lγ = . In the underdamped case, the charge on the

capacitor as a function of time is

( )0( ) cos 'tQ t Q e tγ ω φ−= + where 2 2

0'ω ω γ= − .

28

11.8 Appendix 1: General Solutions for the RLC Series Circuit In Section 11.6, we have shown that the LRC circuit is characterized by the following differential equation

2

2 0d Q dQ QL Rdt dt C

+ + = (11.8.1)

whose solutions is given by 0( ) cos( ' )tQ t Q e tγ ω φ−= + (11.8.2) where

2RL

γ = (11.8.3)

is the damping factor and 2 2

0'ω ω γ= − (11.8.4) is the angular frequency of the damped oscillations. There are three possible scenarios, depending on the relative values of γ and 0ω . Case 1: Underdamping When 0ω γ> , or equivalently, 'ω is real and positive, the system is said to be underdamped. This is the case when the resistance is small. Charge oscillates (the cosine function) with an exponentially decaying amplitude 0

tQ e γ− . However, the frequency of this damped oscillation is less than the undamped oscillation, 0'ω ω< . The qualitative behavior of the charge on the capacitor as a function of time is shown in Figure 11.8.1.

Figure 11.8.1 Underdamped oscillations

29

As an example, suppose the initial condition is 0( 0)Q t Q= = . The phase is then 0φ = , and 0( ) cos 'tQ t Q e tγ ω−= (11.8.5) The corresponding current is

[0( ) ' sin ' ( / ') cos 'tdQ ]I t Q e tdt

γ tω ω γ ω ω−= − = + (11.8.6)

For small R, the above expression may be approximated as

0( ) sin( ' )tQI t e tLC

γ ω δ−≈ + (11.8.7)

where

1tan'

γδω

− ⎛= ⎜⎝ ⎠

⎞⎟ (11.8.8)

The derivation is left to the readers as an exercise. Case 2: Overdamping In the overdamped case, 0ω γ< , implying that 'ω is imaginary. There is no oscillation in

this case. By writing ' iω β= , where 202β γ ω= − , one may show that the most general

solution can be written as ( ) ( )

1 2( ) tQ t Q e Q e tγ β γ− + − −= + β (11.8.9) where the constants and can be determined from the initial conditions. 1Q 2Q

Figure 11.8.2 Overdamping and critical damping

30

Case 3: Critical damping When the system is critically damped, 0ω γ= , ' 0ω = . Again there is no oscillation. The general solution is 1 2( ) ( ) tQ t Q Q t e γ−= + (11.8.10) where Q1 and Q2 are constants which can be determined from the initial conditions. In this case one may show that the energy of the system decays most rapidly with time. The qualitative behavior of in overdamping and critical damping is depicted in Figure 11.8.2.

( )Q t

11.8.1 Quality Factor When the resistance is small, the system is underdamped, and the charge oscillates with decaying amplitude 0

tQ e γ− . The “quality” of this underdamped oscillation is measured by the so-called “quality factor,” Q (not to be confused with charge.) The larger the value of Q, the less the damping and the higher the quality. Mathematically, Q is defined as

energy stored'average power dissipated | / |

UQdU dt

ω⎛ ⎞

= ⎜ ⎟⎝ ⎠

'ω= (11.8.11)

Using Eq. (11.8.2), the electric energy stored in the capacitor is

22

2 20( ) cos ( ' ) 2 2

tE

QQ tU e tC C

γ ω φ−= = + (11.8.12)

To obtain the magnetic energy, we approximate the current as

0

0

0

( ) ' sin( ' ) cos( ' )'

' sin( ' )

sin( ' )

t

t

t

dQI t Q e t tdt

Q e tQ e tLC

γ

γ

γ

γω ω φ ω φω

ω ω φ

ω φ

−

−

−

⎡ ⎤⎛ ⎞= − = + + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦≈ +

≈ +

(11.8.13)

assuming that 'ω γ and 2 2

0' 1/ LCω ω≈ = . Thus, the magnetic energy stored in the inductor is given by

2 2

2 2 2 2 2 20 01 ' sin ( ' ) sin ( ' ) 2 2 2

t tB

LQ QU LI e t e tC

γ γω ω φ ω φ− −= ≈ + ≈ + (11.8.14)

31

Adding up the two terms, the total energy of the system is

2 2 22 2 2 2 20 0 0cos ( ' ) sin ( ' )

2 2 2t t

E BQ Q QU U U e t e t e

C C Cγ γω φ ω φ− − ⎛ ⎞

= + ≈ + + + = ⎜ ⎟⎝ ⎠

tγ− (11.8.15)

Differentiating the expression with respect to t then yields the rate of change of energy:

2

202 2 2

tQdU edt C

γγ −⎛ ⎞= − = −⎜ ⎟

⎝ ⎠Uγ (11.8.16)

Thus, the quality factor becomes

' ''

| / | 2UQ

dU dt RLω ωω

γ= = = (11.8.17)

As expected, the smaller the value of R, the greater the value of Q, and therefore the higher the quality of oscillation. 11.9 Appendix 2: Stresses Transmitted by Magnetic Fields

“…It appears therefore that the stress in the axis of a line of magnetic force is a tension, like that of a rope…”

J. C. Maxwell [1861]. In Chapter 9, we showed that the magnetic field due to an infinite sheet in the xy-plane carrying a surface current ˆK=K i is given by

0

0

ˆ, 02

ˆ, 02

K z

K z

µ

µ

⎧− >⎪⎪= ⎨⎪ <⎪⎩

jB

j

(11.9.1)

Now consider two sheets separated by a distance d carrying surface currents in the opposite directions, as shown in Figure 11.9.1.

32

Figure 11.9.1 Magnetic field due to two sheets carrying surface current in the opposite directions Using the superposition principle, we may show that the magnetic field is non-vanishing only in the region between the two sheets, and is given by (11.9.2) 0

ˆ, /2 /2K d z dµ= − < <B j Using Eq. (11.3.8), the magnetic energy stored in this system is

22

20 0

0 0

( )( ) ( ) (2 2 2B

KBU Ad Ad Kµ µµ µ

= = = )Ad (11.9.3)

where A is the area of the plate. The corresponding magnetic energy density is

20

2B

BUu KAd

µ= = (11.9.4)

Now consider a small current-carrying element 1

ˆ( )Id K y x= ∆ ∆s i on the upper plate (Recall that K has dimensions of current/length). The force experienced by this element due to the magnetic field of the lower sheet is

20 021 1 2

ˆ ˆ ˆ( ) (2 2

d Id K y x K K x yµ µ⎛ ⎞= × = ∆ ∆ × = ∆ ∆⎜ ⎟⎝ ⎠

F s B i j )k (11.9.5)

The force points in the direction and therefore is repulsive. This is expected since the currents flow in opposite directions. Since

ˆ+k

21dF is proportional to the area of the current

element, we introduce force per unit area, 21f , and write

2021 1 2

ˆ2 B

ˆK uµ= × =f K B k = k (11.9.6)

33

using Eq. (11.9.4). The magnitude of the force per unit area, 21f , is exactly equal to the magnetic energy density . Physically, Bu 21f may be interpreted as the magnetic pressure

2

2102B

Bf P uµ

= = = (11.9.7)

The repulsive force experienced by the sheets is shown in Figure 11.9.2

Figure 11.9.2 Magnetic pressure exerted on (a) the upper plate, and (b) the lower plate Let’s now consider a more general case of stress (pressure or tension) transmitted by fields. In Figure 11.9.3, we show an imaginary closed surface (an imaginary box) placed in a magnetic field. If we look at the face on the left side of this imaginary box, the field on that face is perpendicular to the outward normal to that face. Using the result illustrated in Figure 11.9.2, the field on that face transmits a pressure perpendicular to itself. In this case, this is a push to the right. Similarly, if we look at the face on the right side of this imaginary box, the field on that face is perpendicular to the outward normal to that face, the field on that face transmits a pressure perpendicular to itself. In this case, this is a push to the left.

Figure 11.9.3 An imaginary box in a magnetic field (blue vectors). The short vectors indicate the directions of stresses transmitted by the field, either pressures (on the left or right faces of the box) or tensions (on the top and bottom faces of the box). If we want to know the total electromagnetic force transmitted to the interior of this imaginary box in the left-right direction, we add these two transmitted stresses. If the electric or magnetic field is homogeneous, this total electromagnetic force transmitted to

34

the interior of the box in the left-right direction is a push to the left and an equal but opposite push to the right, and the transmitted force adds up to zero. In contrast, if the right side of this imaginary box is sitting inside a long vertical solenoid, for which the magnetic field is vertical and constant, and the left side is sitting outside of that solenoid, where the magnetic field is zero, then there is a net push to the left, and we say that the magnetic field exerts a outward pressure on the walls of the solenoid. We can deduce this by simply looking at the magnetic field topology. At sufficiently high magnetic field, such forces will cause the walls of a solenoid to explode outward. Similarly, if we look at the top face of the imaginary box in Figure 11.9.3, the field on that face is parallel to the outward normal to that face, and one may show that the field on that face transmits a tension along itself across that face. In this case, this is an upward pull, just as if we had attached a string under tension to that face, pulling upward. (The actual determination of the direction of the force requires an advance treatment using the Maxwell stress tensor.) On the other hand, if we look at the bottom face of this imaginary box, the field on that face is anti-parallel to the outward normal to that face, and Faraday would again have said that the field on that face transmits a tension along itself. In this case, this is a downward pull, just as if we had attached a string to that face, pulling downward. Note that this is a pull parallel to the outward surface normal, whether the field is into the surface or out of the surface, since the pressures or tensions are proportional to the squares of the field magnitudes.

If we want to know the total electromagnetic force transmitted to the interior of this imaginary box in the up-down direction, we add these two transmitted stresses. If the magnetic field is homogeneous, this total electromagnetic force transmitted to the interior of the box in the up-down direction is a pull upward plus an equal and opposite pull downward, and adds to zero. The magnitude of these pressures and tensions on the various faces of the imaginary surface in Figure 11.9.3 is given by 2

0/ 2B µ , as shown in Eq. (11.9.7). Our discussion may be summarized as follows:

Pressures and Tensions Transmitted by Magnetic Fields Electromagnetic fields are mediators of the interactions between material objects. The fields transmit stresses through space. A magnetic field transmits a tension along itself and a pressure perpendicular to itself. The magnitude of the tension or pressure transmitted by a magnetic field is given by

212B

o

P u Bµ

= =

35

Animation 11.3: A Charged Particle in a Time-Varying Magnetic Field As an example of the stresses transmitted by magnetic fields, consider a moving positive point charge at the origin in a rapidly changing time-dependent external field. This external field is uniform in space but varies in time according to the equation

40

2 ˆsin tBTπ⎛ ⎞= − ⎜ ⎟

⎝ ⎠B k (11.9.8)

We assume that the variation of this field is so rapid that the charge moves only a negligible distance in one period T. Figure 11.9.4 shows two frames of an animation of the total magnetic field configuration for this situation. Figure 11.9.4(a) is at t = 0, when the vertical magnetic field is zero, and we see only the magnetic field of the moving charge (the charge is moving out of the page, so the field circulates clockwise). Frame 11.9.4(b) is at a quarter period later, when the vertically downward magnetic field is at a maximum. To the left of the charge, where the field of the charge is in the same direction as the external magnetic field (downward), the magnetic field is enhanced. To the right of the charge, where the field of the charge is opposite that of the external magnetic field, the magnetic field is reduced (and is zero at one point to the right of the charge).

(a) (b) Figure 11.9.4 Two frames of an animation of the magnetic field around a positive charge moving out of the page in a time-changing magnetic field that points downward. The blue vector is the magnetic field and the white vector is the force on the point charge. We interpret the field configuration in Figure 11.9.4(b) as indicating a net force to the right on the moving charge. This occurs because the pressure of the magnetic field is much higher on the left as compared to the right. Note that if the charge had been moving into the page instead of out of the page, the force would have been to the left, because the magnetic pressure would have been higher on the right. The animation of Figure 11.9.4 shows dramatically the inflow of energy into the neighborhood of the charge as the external magnetic field grows, with a resulting build-up of stress that transmits a sideways force to the moving positive charge.

36

We can estimate the magnitude of the force on the moving charge in Figure 11.9.4(b) as follows. At the time shown in Figure 11.9.4(b), the distance to the right of the charge at which the magnetic field of the charge is equal and opposite to the constant magnetic field is determined by

0r

00 2

04qvBr

µπ

= (11.9.9)

The surface area of a sphere of this radius is 2

0 04 0/A r q vπ µ= = B . Now according to Eq. (11.9.7) the pressure (force per unit area) and/or tension transmitted across the surface of this sphere surrounding the charge is of the order of 2

0/ 2P B µ= . Since the magnetic field on the surface of the sphere is of the order 0B , the total force transmitted by the field is of order

2 2

20 0 00

0 0 0

(4 )2 2B B qvF PA r qvB

Bµπ

µ µ= = = ⋅ ≈ 0 (11.9.10)

Of course this net force is a combination of a pressure pushing to the right on the left side of the sphere and a tension pulling to the right on the right side of the sphere. The exact expression for the force on a charge moving in a magnetic field is B q= ×F v B (11.9.11) The rough estimate that we have just made demonstrates that the pressures and tensions transmitted across the surface of this sphere surrounding the moving charge are plausibly of the order 2

0/ 2B µ . In addition, this argument gives us some insight in to why the magnetic force on a moving charge is transverse to the velocity of the charge and to the direction of the background field. This is because of the side of the charge on which the total magnetic pressure is the highest. It is this pressure that causes the deflection of the charge. 11.10 Problem-Solving Strategies 11.10.1 Calculating Self-Inductance The self-inductance L of an inductor can be calculated using the following steps: 1. Assume a steady current I for the inductor, which may be a conducting loop, a solenoid, a toroid, or coaxial cables. 2. Choose an appropriate cross section S and compute the magnetic flux through S using

37

B

S

dΦ = ⋅∫∫B A

If the surface is bounded by N turns of wires, then the total magnetic flux through the surface would be . BNΦ 3. The inductance may be obtained as

BNLIΦ

=

11.10.2 Circuits containing inductors Three types of single-loop circuits were examined in this chapter: RL, LC and RLC. To set up the differential equation for a circuit, we apply the Kirchhoff’s loop and junction rules, as we did in Chpater 7 for the RC circuits. For circuits that contain inductors, the corresponding modified Kirchhoff’s rule is schematically shown below.

Note that the “potential difference” across the inductor is proportional to , the rate of change of current. The situation simplifies if we are only interested in the long-term behavior of the circuit where the currents have reached their steady state and

/dI d t

/ 0dI dt = . In this limit, the inductor acts as a short circuit and can simply be replaced by an ideal wire. 11.11 Solved Problems 11.11.1 Energy stored in a toroid A toroid consists of N turns and has a rectangular cross section, with inner radius a, outer radius b and height h (see Figure 11.2.3). Find the total magnetic energy stored in the toroid.

38

Solution: In Example 11.3 we showed that the self-inductance of a toroid is

2

0 ln2

B N hN bLI a

µπ

Φ ⎛ ⎞= = ⎜ ⎟⎝ ⎠

Thus, the magnetic energy stored in the toroid is simply

2 2

2 01 ln2 4B

N I h bU LIa

µπ

⎛ ⎞= = ⎜ ⎟⎝ ⎠

(11.11.1)

Alternatively, the energy may be interpreted as being stored in the magnetic field. For a toroid, the magnetic field is (see Chapter 9)

0

2NIBr

µπ

=

and the corresponding magnetic energy density is

2 22

02 2

0

12 8B

N IBur

µµ π

= = (11.11.2)

The total energy stored in the magnetic field can be found by integrating over the volume. We choose the differential volume element to be a cylinder with radius r, width dr and height h, so that 2dV rh drπ= . This leads to

2 2 2 2

0 02 2 2

8 4b

B B a

N I N I h bU u dV rh drr a

µ µππ π

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

∫ ∫ ln (11.11.3)

Thus, both methods yield the same result. 11.11.2 Magnetic Energy Density A wire of nonmagnetic material with radius R and length l carries a current I which is uniformly distributed over its cross-section. What is the magnetic energy inside the wire? Solution: Applying Ampere’s law, the magnetic field at distance r R≤ can be obtained as:

39

( ) 20 0 22 ( ) (I 2 )B r J r r

Rπ µ π µ π

π⎛ ⎞= = ⎜ ⎟⎝ ⎠

(11.11.4)

or

022

IrBR

µπ

= (11.11.5)

Since the magnetic energy density (energy per unit volume) is given by

2

02BBuµ

= (11.11.6)

the total magnetic energy stored in the system becomes

( )2 22 4

30 04 40 0

0

22 4 4 4

R R

B

20

16I l I lB RU rl dr r drR R

µ µπ I lµµ π π

⎛ ⎞= = = ⎜ ⎟

⎝ ⎠∫ ∫ π

= (11.11.7)

11.11.3 Mutual Inductance An infinite straight wire carrying current I is placed to the left of a rectangular loop of wire with width and length l, as shown in the Figure 11.11.3. Determine the mutual inductance of the system.

w

Figure 11.11.3 Rectangular loop placed near long straight current-carrying wire Solution: To calculate the mutual inductance M, we first need to know the magnetic flux through the rectangular loop. The magnetic field at a distance r away from the straight wire is

0 / 2B I rµ π= , using Ampere’s law. The total magnetic flux BΦ through the loop can be obtained by summing over contributions from all differential area elements dA =l dr:

40

0 0 ln2 2

s w

B B s

IL Ildr s wd dr s

µ µπ π

+ +⎛Φ = Φ = ⋅ = = ⎜⎝ ⎠∫ ∫ ∫B A ⎞

⎟ (11.11.8)

Thus, the mutual inductance is

0 ln2

B l s wMI s

µπ

Φ +⎛= = ⎜⎝ ⎠

⎞⎟ (11.11.9)

11.11.4 RL Circuit Consider the circuit shown in Figure 11.11.4 below.

Figure 11.11.4 RL circuit Determine the current through each resistor (a) immediately after the switch is closed. (b) a long time after the switch is closed. Suppose the switch is reopened a long time after it’s been closed. What is each current (c) immediately after it is opened? (d) after a long time? Solution: (a) Immediately after the switch is closed, the current through the inductor is zero because the self-induced emf prevents the current from rising abruptly. Therefore, 03 =I . Since 1 2 3I I I= + , we have 1 2I I=

41

Figure 11.11.5 Applying Kirchhoff’s rules to the first loop shown in Figure 11.11.5 yields

1 21 2

I IR R

ε= =

+ (11.11.10)

(b) After the switch has been closed for a long time, there is no induced emf in the inductor and the currents will be constant. Kirchhoff’s loop rule gives 1 1 2 2 0I R I Rε − − = (11.11.11) for the first loop, and 2 2 3 3 0I R I R− = (11.11.12) for the second. Combining the two equations with the junction rule 1 2 3I I I= + , we obtain

( )2 31

1 2 1 3 2 3

32

1 2 1 3 2 3

23

1 2 1 3 2 3

R RI

R R R R R R

RIR R R R R R

RIR R R R R R

ε

ε

ε

+=

+ +

=+ +

=+ +

(11.11.13)

(c) Immediately after the switch is opened, the current through R1 is zero, i.e., 01 =I . This implies that . On the other hand, loop 2 now forms a decaying RL circuit and I

2 3 0I I+ =

3 starts to decrease. Thus,

23 2

1 2 1 3 2 3

RI IR R R R R R

ε= − =

+ + (11.11.14)

(d) A long time after the switch has been closed, all currents will be zero. That is,

. 1 2 3 0I I I= = =

42

11.11.5 RL Circuit In the circuit shown in Figure 11.11.6, suppose the circuit is initially open. At time 0t = it is thrown closed. What is the current in the inductor at a later time t?

Figure 11.11.6 RL circuit Solution: Let the currents through 1R , 2R and L be 1I , 2I and I , respectively, as shown in Figure 11.11.7. From Kirchhoff’s junction rule, we have 1 2I I I= + . Similarly, applying Kirchhoff’s loop rule to the left loop yields ( )2 1 2 2 0I I R I Rε − + − = (11.11.15)

Figure 11.11.7 Similarly, for the outer loop, the modified Kirchhoff’s loop rule gives

( )2 1dII I R Ldt

ε − + = (11.11.16)

The two equations can be combined to yield

2 2 22

dI L dII R L Idt R dt

= ⇒ = (11.11.17)

Substituting into Eq. (11.11.15) the expression obtained above for 2I , we have

43

1 21 1

2 2

0R RL dI dI dII R L IR LR dt dt R dt

ε ε⎛ ⎞ ⎛ ⎞+

− + − = − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(11.11.18)

Dividing the equation by 1 2( ) / 2R R R+ leads to

' ' dIIR Ldt

ε 0− − = (11.11.19)

where

1 2 2

1 2 1 2

' , 'R R RRR R R R

εε= =+ +

(11.11.20)

The differential equation can be solved and the solution is given by

( ) ( ' /' 1'

R t LI t e )Rε −= − (11.11.21)

Since

2 1 2

1 2 1 2 1

/( )'' /( )

R R RR R R R R R

εε ε+=

+= (11.11.22)

the current through the inductor may be rewritten as

( ) ( ) (' / /

1 1

1 1R t L tI t e eR R )τε ε−= − = − −

'

(11.11.23)

where /L Rτ = is the time constant. 11.11.6 LC Circuit Consider the circuit shown in Figure 11.11.8. Suppose the switch which has been connected to point a for a long time is suddenly thrown to b at t = 0.

Figure 11.11.8 LC circuit

44

Find the following quantities: (a) the frequency of oscillation of the LC circuit. (b) the maximum charge that appears on the capacitor. (c) the maximum current in the inductor. (d) the total energy the circuit possesses at any time t. Solution: (a) The (angular) frequency of oscillation of the LC circuit is given by

2 1/f Lω π= = C . Therefore, the frequency is

12

fLCπ

= (11.11.24)

(b) The maximum charge stored in the capacitor before the switch is thrown to b is Q Cε= (11.11.25) (c) The energy stored in the capacitor before the switch is thrown is

212EU Cε= (11.11.26)

On the other hand, the magnetic energy stored in the inductor is

212BU L= I (11.11.27)

Thus, when the current is at its maximum, all the energy originally stored in the capacitor is now in the inductor:

20

1 12 2

C Lε = 2I (11.11.28)

This implies a maximum current

0CIL

ε= (11.11.29)

45

(d) At any time, the total energy in the circuit would be equal to the initial energy that the capacitance stored, that is

212E BU U U Cε= + = (11.11.30)

11.12 Conceptual Questions 1. How would you shape a wire of fixed length to obtain the greatest and the smallest inductance? 2. If the wire of a tightly wound solenoid is unwound and made into another tightly wound solenoid with a diameter 3 times that of the original one, by what factor does the inductance change? 3. What analogies can you draw between an ideal solenoid and a parallel-plate capacitor? 4. In the RL circuit show in Figure 11.12.1, can the self-induced emf ever be greater than the emf supplied by the battery?

Figure 11.12.1 5. The magnetic energy density 2

0/ 2Bu B µ= may also be interpreted as the magnetic pressure. Using the magnetic pressure concept, explain the attractive (repulsive) force between two coils carrying currents in the same (opposite) direction. 6. Explain why the LC oscillation continues even after the capacitor has been completely discharged. 7. Explain physically why the time constant /L Rτ = in an RL circuit is proportional to L and inversely proportional to R.

46

11.13 Additional Problems 11.13.1 Solenoid A solenoid with a length of 30 cm, a radius of 1.0 cm and 500 turns carries a steady current . 2.0 AI = (a) What is the magnetic field at the center of the solenoid along the axis of symmetry? (b) Find the magnetic flux through the solenoid, assuming the magnetic field to be uniform. (c) What is the self-inductance of the solenoid? (d) What is the induced emf in the solenoid if the rate of change of current is

? / 100 A/dI dt = s 11.13.2 Self-Inductance Suppose you try to wind a wire of length d and radius a into an inductor which has the shape of a cylinder with a circular cross section of radius r. The windings are tight without wires overlapping. Show that the self-inductance of this inductor is

0 4rdL

aµ=

11.13.3 Coupled Inductors (a) If two inductors with inductances and are connected in series, show that the equivalent inductance is

1L 2L

eq 1 2 2L L L M= + ± where M is their mutual inductance. How is the sign chosen for M ? Under what condition can M be ignored? (b) If the inductors are instead connected in parallel, show that, if their mutual inductance can be ignored, the equivalent inductance is given by

eq 1 2

1 1 1L L L

= +

47

How would you take the effect of M into consideration? 11.13.4 RL Circuit The LR circuit shown in Figure 11.13.1 contains a resistor 1R and an inductance L in series with a battery of emf 0ε . The switch S is initially closed. At t = 0, the switch S is opened, so that an additional very large resistance 2R (with 2 1R R ) is now in series with the other elements.

Figure 11.13.1 RL circuit (a) If the switch has been closed for a long time before t = 0, what is the steady current

0I in the circuit? (b) While this current 0I is flowing, at time t = 0, the switch S is opened. Write the differential equation for ( )I t that describes the behavior of the circuit at times t ≥ 0. Solve this equation (by integration) for ( )I t under the approximation that 0 0ε = . (Assume that the battery emf is negligible compared to the total emf around the circuit for times just after the switch is opened.) Express your answer in terms of the initial current 0I , and 1R , 2R , and L. (c) Using your results from (b), find the value of the total emf around the circuit (which from Faraday's law is ) just after the switch is opened. Is your assumption in (b) that

/LdI dt−0ε could be ignored for times just after the switch is opened OK?

(d) What is the magnitude of the potential drop across the resistor 2R at times t > 0, just after the switch is opened? Express your answers in terms of 0ε , 1R , and 2R . How does the potential drop across 2R just after t = 0 compare to the battery emf 0ε , if

2 1100R R= ?

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11.13.5 RL Circuit In the circuit shown in Figure 11.13.2, 100 Vε = , 1 10 R = Ω , 2 20 R = Ω , , and the inductance L in the right loop of the circuit is 2.0 H. The inductance in the left loop of the circuit is zero.

3 30 R = Ω

Figure 11.13.2 RL circuit (a) Find 1I and 2I immediately after switch S is closed. (b) Find 1I and 2I a long time later. What is the energy stored in the inductor a long time later? (c) A long, long time later, switch S is opened again. Find 1I and 2I immediately after switch S is opened again. (d) Find 1I and 2I a long time after switch S is opened. How much energy is dissipated in resistors 2R and 3R between the time immediately after switch S is opened again, and a long time after that? (e) Give a crude estimate of what “a long time” is in this problem. 11.13.6 Inductance of a Solenoid With and Without Iron Core (a) A long solenoid consists of N turns of wire, has length l, and cross-sectional area A. Show that the self-inductance can be written as 2

0 /L N A lµ= . Note that L increases as N2, and has dimensions of 0µ times a length (as must always be true). (b) A solenoid has a length of 126 cm and a diameter of 5.45 cm, with 1870 windings. What is its inductance if its interior is vacuum? (c) If we now fill the interior with iron with an effective permeability constant κm = 968, what is its inductance?

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(d) Suppose we connect this iron core inductor up in series with a battery and resistor, and that the total resistance in the circuit, including that of the battery and inductor, is

. How long does it take after the circuit is established for the current to reach 50% of its final value? [Ans: (b) 8.1 mH; (c) 7.88 H; (d) 0.55 s]. 10 Ω

11.13.7 RLC Circuit An RLC circuit with battery is set up as shown in Figure 11.13.3. There is no current flowing in the circuit until time 0t = , when the switch is closed. 1S

Figure 11.13.3 (a) What is the current I in the circuit at a time after the switch is closed? 0t > 1S (b) What is the current I in the circuit a very long time (t >> L/R) after the switch is closed?

1S

(c) How much energy is stored in the magnetic field of the solenoid a very long time (t L R>> ) after the switch is closed? For the next two questions, assume that a very long time ( )t L R>> after the switch was closed, the voltage source is disconnected from the circuit by opening the switch and that the solenoid is simultaneously connected to a capacitor by closing the switch . Assume there is negligible resistance in this new circuit.

1S

1S

2S

Figure 11.13.4

(d) What is the maximum amount of charge that will appear on the capacitor, in terms of the quantities given?

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(e) How long will it take for the capacitor to first reach a maximal charge after the switch

has been closed? 2S 11.13.8 Spinning Cylinder

Two concentric, conducting cylindrical shells are charged up by moving from the outer to the inner conductor, so that the inner conductor has a charge of spread uniformly over its area, and the outer conductor is left with

Q+Q+

Q− uniformly distributed. The radius of the inner conductor is a; the radius of the outer conductor is b; the length of both is l; and you may assume that l >> a, b.

(a) What is the electric field for r < a, a < r < b, and r > b? Give both magnitude and direction. (b) What is the total amount of energy in the electric field? (Hint: you may use a variety of ways to calculate this, such as using the energy density, or the capacitance, or the potential as a function of Q . It never hurts to check by doing it two different ways.) (c) If the cylinders are now both spun counterclockwise (looking down the z axis) at the same angular velocity ω (so that the period of revolution is 2 /T π ω= ), what is the total current (magnitude and sign) carried by each of the cylinders? Give your answer in terms of ω and the quantities from the first paragraph, and consider a current to be positive if it is in the same direction as ω . (d) What is the magnetic field created when the cylinders are spinning at angular velocity ω ? You should give magnitude and direction of B in each of the three regions: r < a,

, r > b. (Hint: it’s easiest to do this by calculating Ba r b< < from each cylinder independently and then getting the net magnetic field as the vector sum.) (e) What is the total energy in the magnetic field when the cylinders are spinning at ω ? 11.13.9 Spinning Loop A circular, conducting loop of radius a has resistance R and is spun about its diameter which lies along the y-axis, perpendicular to an external, uniform magnetic field ˆB=B k . The angle between the normal to the loop and the magnetic field is θ , where tθ ω= . You may ignore the self-inductance of the loop. (a) What is the magnetic flux through the loop as a function of time?

(b) What is the emf induced around the loop as a function of time?

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(c) What is the current flowing in the loop as a function of time? (d) At an instant that the normal to the loop aligns with the x-axis, the top of the loop lies on the +z axis. At this moment is the current in this piece of loop in the ˆ+ j or ˆ−j direction? (e) What is the magnitude of the new magnetic field Bind (as a function of time) created at the center of the loop by the induced current? (f) Estimate the self-inductance L of the loop, using approximation that the magnetic field Bind is uniform over the area of the loop and has the value calculated in part (e). (g) At what angular speed ω will the maximum induced magnetic field Bind equal the external field B (therefore thoroughly contradicting the assumption of negligible self-inductance that went into the original calculation of Bind)? Express your answer in terms of R and L.

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