Topic 6 Magnetic Fields

8/13/2019 Topic 6 Magnetic Fields

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INTRODUCTION

In Topic 1, we learned that two electric charges at rest exert electrostatic forces oneach other according to Coulombs law. This Coulombic force depends on thevalue of the charges and also the distance between them. When the charges are inmotion, there are magnetic forces between the moving charges in addition to theelectric forces between them. The magnetic force depends on the velocity, v, ofthe charge, and the charge, Q.

MAGNETIC FIELD LINES

6.1.1 Bar Magnets

Surrounding any magnet is a magnetic field. This magnetic field represents theeffect of the magnet on its surroundings.

6.1

TTooppiicc66

MagneticFields

LEARNING OUTCOMESBy the end of this topic, you should be able to:

1. Describe magnetic fields around permanent magnets and between likeand unlike poles.

2.

Demonstrate the concept of magnetic fields by sprinkling iron filingsto observe the imaginary magnetic field lines around magnets;

3. Describe the magnetic field around a current-carrying conductor, loopof conductor and solenoid;

4. Calculate the magnitude and direction of the magnetic force on amoving charge in a magnetic field and on a current-carryingconductor

5. Define and apply Biot-Savarts Law and Amperes Law; and

6. Explain in simple terms the origin of magnetism in materials.


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TOPIC 6 MAGNETIC FIELDS 87

The presence of a magnetic field around a bar magnet can be shown by sprinklingiron filings around the magnet. First, put a bar magnet on a sheet of white paper.Then, sprinkle some iron filings around the magnet. You will observe that the iron

filings arrange themselves along lines that connect the two magnetic poles. Eachlong, thin iron filing becomes a small magnet by induction, and aligns itself alongwhat is called the magnetic field line. The magnetic field lines form closed loopsthat leave the N-pole of the bar magnet and enter the S-pole as shown inFigure. 6.1.

Figure 6.1:Magnetic field lines outside a bar magnets are closed loops leaving the N-pole of the

bar magnet and entering the S-pole.

A simulation of this can be found here:http://micro.magnet.fsu.edu/electromag/java/magneticlines/index.html

Therefore, a magnetic field can be visualised as magnetic lines of force, which areactually imaginary lines. These lines of force never intersect each other. The linesform continuous closed loops, originating from the N-pole and terminating at theS-pole. The tangent at any point on the on the loop is the direction of the magneticfield at that point. The lines are parallel, being from S-pole to N-pole inside thebar magnet. In addition, the field is stronger in regions where the field lines arecloser and weaker when the they are far further apart.

When we place two magnets near each other we find that: Unlike magnetic polesattract each other, and like magnetic poles repel each other.

How would the magnetic field lines look like when two poles are placed oppositeone another? A simulation of this situation can be found here:

http://micro.magnet.fsu.edu/electromag/java/magneticlines2/index.html


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TOPIC 6 MAGNETIC FIELDS88

6.1.2 Current-carrying wire

The magnetic field around a current-carrying wire can easily be shown by placinga wire conductor vertically through a horizontal piece of white cardboard, andthen sprinkling iron filings on the cardboard. When a current is applied to the

conductor, tap the cardboard. You will notice that the iron filings form a pattern ofconcentric circles around the conductor as shown in Fig. 6.2

Figure 6.2: The magnetic field lines around a long, straight current carrying conductor

But what produces this magnetic field in the first place? We must thank a Danishschool teacher named Hans Christian Oersted for providing us with the answer!Oersted discovered that it was the electric current that created the magnetic field.

We can determine the direction of the field by applying the right hand grip rule.

Grasp the wire with your right hand such that your thumb points in the directionof the current. The direction of the magnetic field generated by the current is thesame as the direction of the fingers.

Apart from magnitude, how is Coulombic force different from magneticforce?

SELF-CHECK 6.1


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Figure 6.3: The Right-Hand Grip Rule

6.1.3 Current-carrying loop of wire

Figure 6.4 (a) shows the magnetic field lines around a current-carrying loop ofwire.

Figure 6.4 (a):Magnetic field lines around a current-carrying loop of wire

6.1.4 Magnetic field lines in a plane through thecenter of a solenoid

A solenoid is made by winding a long wire into a tight coil with many circularcoils as shown in. If the radius of the loops is very small compared to the lengthLof the solenoid, a large magnetic field is produced parallel to the axis of thesolenoid. Outside the solenoid the field is almost zero. Inside the solenoid thefields from individual coils add together to form a very strong field along thecentre of the solenoid. See Figure 6.4(b).


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Figure 6.4 (b): Magnetic field lines inside a solenoid, which consists of current-carrying

loops of wire

MAGNETIC FIELDS

A distribution of electric charges at rest creates an electric field, E, in thesurrounding space, where the field exerts a force F= qEon any other charge qthat is present in the field. Like the electric field, the magnetic field is also avector quantity. A movingelectric charge or a current creates a magnetic fieldBin the surrounding space, in addition to its electric field. This field exerts a forceon any moving charge or current that is present in the field.

The unit of magnetic field is the Tesla T.

THE MAGNETIC FORCE

A stationary electric charge in a magnetic field will experience no magnetic force.But if the charge is moving in a magnetic field, then it will experience a magneticforce.

From experiment, it is found that the the magnitude of this forceFis given byF = qvBsin (6.1)

This implies that the force depends on the

(i) The magnitude, q,of the moving charge,

(ii) The speed, v,of the charge,

(iii) The magnitude,B, of the magnetic field,and

(iv) TThe angle between the magnetic field and the velocity v.

6.3

6.2


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The force on the charge is maximum when vis perpendicular toB.

Furthermore, the direction of the force is always perpendicular to the direction ofthe velocity and the magnetic field. This suggests that a cross product is involved.We can define the magnetic force magnetic, F, to be a vector quantity, such that:

= qF v B (6.2)

Figure 6.5(a): The vector cross product for the magnetic force

To find the direction of the force, we use the right-hand rule. Let the four fingersof your right hand curl from the first vector v to the secondvectorB.Your thumbwill then point in the direction of the magnetic force F.

Figure 6.5(b): Theright-hand rule

A charged particle moves through a region of space and does notexperience any magnetic force. Does this mean that the magnetic field iszero? Explain.

SELF-CHECK 6.2


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If q is a positive charge then this is the direction of F. If q is negative, your thumbpoints opposite to the direction of F.

Example 6.1

A beam of protons moves at the velocity of 5100.3 m/s through a uniformmagnetic field of 1.0 T. If the angle between the velocity and the magnetic field is30

o, find the magnitude of the force exerted on each particle.

Solution

Solving Equation 6.1 for the magnetic force, we have

-19 5 o -14(1.6010 C)(3.010 m / s)(1.0T)(sin30 ) = 2.410F = qvBsin= N.

Example 6.2An electron in a typical home television camera tube moves with speedv= 61.0 10 m/s in a magnetic field of magnitudeB= 80 mT.

(a) Determine the magnitude of the maximum magnetic force that can act on theelectron in the field

(b) At one point, the force on the electron is155 10

N. What is the anglebetween vand B at this point?

Solution

(a) The force on the electron is a maximum only when vis perpendicular toB,i.e., when 090= . The magnitude can then be found from Equation 6.1:

19 6 o

14

sin 1.60 10 1.0 10 1.0 sin90

1.28 10 N

= =

=

F qvB

(b)

( )

-15 6

-15

19 6

1 0

510 N, = 110 m / s 80mT 0 08T

510 0 391 6 10 110 0 08

0 39 23

F = v ,B .

Fsin .qvB . .

sin .

= =

= = =

= =


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MOTION OF A CHARGE IN MAGNETIC

FIELD

Lets now consider the case of charged particle q moving in a uniform magneticfieldBso that the direction of the particles velocity vis always perpendicular tothe field.

As a result, the charged particle moves in a circle of radius rand constant speed vin a plane perpendicular to the magnetic field. Thus the magnetic force F = qvB is in the radial direction and its magnitude is equal to the centripetal force

2F = mv /r.

From Newtons second law, we find that:2

mvF = qvB=r

(6.3)

where mis the mass of the particle.

Solving equation 6.3 for the radius rof the circular path, we find:

mvr =

qB (6.4)

The angular frequency of the rotating charged particle is:

v qB= =

r m (6.5)

6.4

EXERCISE 6.1

An ion (q = +3e) enters a magnetic field of 1 mT at a velocity o61.0 10 m/s perpendicular to the field. Determine the force acting on

the ion.


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Figure 6.6: Circular motion of a charge in a constant magnetic field. Notice that the magnetic field

is acting intothe plane of the page

Example 6.3

In an experiment, electrons are accelerated from rest through a potentialdifference 100 V in a magnetic field. The beam associated with the electrons ismeasured to have a radius of 5.0 cm. Calculate the magnitude of the magneticfield and the angular frequency of the electrons.

Solution

(a) We must calculate the speed of the electrons using their kinetic energy andpotential energy;

1

2

2mv = eV

As described above, a charged particle that is initially movingperpendicular to the magnetic field will travel in a circular path, with theplane of the circle being perpendicular to the direction of the field.

What about a particle that is initially moving at some angle between

parallel and perpendicular to the field? Describe the path that this particle

will take.

SELF-CHECK 6.3


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-19

6

n31

2(1.60210 C)(100 V)= 5.9310 m/s

9.1110 kg

2eVv = =

m

Solving equation (6.4) for the magnetic field, we have

-31 6

-4

-19

(9.1110 kg)(5.9310 m/s)= = 4.2210

(1.60210 C)(0.08m)

mvB =

erT

(b) The angular frequency

6

75.9310 m/s

7.45100.08m

v= = =

rrad/s.

THE BIOT-SAVART LAW

Figure 6.7:Magnetic field at point P produced by a current-carrying wire

Figure 6.7 shows a wire of arbitrary shape carrying current I. The current elementIdsshown in this figure produces a magnetic field dB atP. Let rbe the distancefrom the current element toP.

6.5

EXERCISE 6.2

1. Electrons are moving with a speed of7

5 10 m/s at right angles to amagnetic field of magnitude 0.5T.

(a) What is the magnetic force on the electrons?

(b) What is the radius of the circle in which the electrons move?

2. Electrons are accelerated from rest through a p.d of 800V. They thenmove perpendicularly to magnetic field of 3mT. Find the radius othe orbit.


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The magnitude of the field dBproduced at a point P a distance rfrom the current-length elementIdsis given by

0

2

Idssin

dB = 4 r (6.6)

where is the angle between the directions of ds and r. 0 is called thepermeability of free space or the permeability constant. The value of 0 is

74 10 H/m. This equation is called the Biot-Savart law.

The magnetic field is found to be perpendicular to the plane containing r and ds sothat we may write Biot-Savarts law as

0

24

=

I dd

r

s rB (6.7)

where ris a unit vector that points toward P. The direction of dB,found using theright hand rule, points into the plane of the page .

To find the total magnetic field field at P due to the whole wire, we must integrateover the entire wire. However, this can be quite lengthy. As a result, we will quotesome of the results here without deriving them:

(a) The magnetic field a distance rfrom a long straight wire carrying a currentI:

2

0 I

B =r

(6.8)

(b) The magnetic field qt the center of a plane circular coil, radius r ofNturnsand carrying currentI

0

2

NIB

r= (6.9)

(c) The magnetic field at the centre of a long solenoid of N turns and length l

carrying a currentI

0NIBl

= (6.10)

The expressionN

nl

= is known as the number of turns per meter. We may

rewrite Equation (6.10) as 0B nI=


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Example 6.4

The figure below shows two long wires carrying currents of magnitude 10Ain the direction indicated by the arrows. What is the magnitude and direction

of the magnetic field at P?

Figure 6.8

Solution:

Consider wire 1. The magnitude of the magnetic field at P due to this wire is7

50

1

4 10 104 10 40

2 2 0 05

IB T

r .

= = = =

. From the right hand rule, the

direction of the magnetic field is into the paper,

Consider wire 2 now. The magnitude of the magnetic field at P due to this

wire is7

501

4 10 104 10 40

2 2 0 05

IB T

r .

= = = =

. The direction is also

Therefore, the magnitude of the resultant is 1 2 80B B B T .= + = Thedirection of the resultant magnetic field at P is also .

Example 6.5

A solenoid is 1 m long and consists of 100 turns of wire. At its center, thesolenoid produces a magnetic field of strength 4 mT. Find the current in thecoil.

3

7

0

Given that 1 100 4

4 10 1From Equation (7 10 : 31 8

4 10 100

= = =

= = =

l m,N ,B mT

Bl. ) i . A

N


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AMPERES LAW

Amperes law states that the line integral B.dsaround any closed path of a wirecarrying currentIis equal to I0

I == 0cos. dsBdsB (6.11)

where is the angle between the magnetic fieldBformed and the path element dsaround the wire. Figure 6.9 shows the magnetic field formed around a straightcurrent-carrying wire.

Figure 6.9:A concentric circle magnetic field (line) produced by a straight current-carrying

current wire.

Using Amperes law we can calculate the magnetic field around a long (infinite)straight wire carrying a current Iand at a perpendicular distance rfrom the wire.We have already seen from Figure 6.2 that the magnetic field forms a pattern ofconcentric circles around a wire. If we choose a circular loop of radius r with wireat the center, then B and dsare in the same direction. Also the magnitude ofB isalso constant around the loop. So

6.6

EXERCISE 6.3

1. Calculate the magnetic field:

(i) 2 cm from a long straight wire carrying a current of 1A.(ii) at the center of a short coil of 100 turns and 10 cm in diameter

carrying a current of 2A

2. Two very long and parallel wires, A and B, are 10cm apart and carrycurrents of 40 and 20A respectively in opposite directions. Calculate themagnitude magnetic field at a point P 8cm from A and 18cm from B.


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I == 0)2( rBdsB

rB

=

2

0I

MAGNETISM

You may wonder how it is possible for a bar magnet to create a magnetic fieldwhen it carries no current. What then is the source of the magnetism produced bythe bar magnet? Magnetism originates in the atom itself. In fact, the magneticproperties of substances like the bar of magnet is due to the motion of electronsin atoms. The circular motion of the electrons around the nucleus of the atom

produces a magnetic moment at the center of the atom, om . In addition to orbitingaround the nucleus each electron also spins on its own axis while it moves alongits orbit. This movement creates another dipole moment, called the spin moment,ms. See Figure 6.10.

Figure 6.10: Dipole moments created by an electron in motion

In most atoms, the magnetic moment is zero. This is because the electrons usuallypair up with their spins in opposite directions, so that the magnetic momentsarising from the spin and orbital motion cancel out each other.

However, in nature there are a small number of elements whose magneticmoments dont cancel out one another. We call such elements ferromagneticelements. Examples of ferromagnetic materials include iron, nickel and cobalt.

In ferromagnetic materials, a strong interaction exists between neighbouring spinsand atoms. As a result of this interaction, a large cluster of atoms called magneticdomains are created.

In an unmagnetised ferromagnetic material, the domains are aligned randomlywith respect to one another as shown in Figure 6.11 (a).

6.7


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However, when we apply an external magnetic field to this material, the domainsstart to align with one another. See Figure 6.11(b). Notice that within eachdomain, nearly all the atomic magnetic moments are parallel. In the presence of

the external magnetic field, the domains that are aligned in the direction of themagnetic tend to grow while those that are aligned in other directions began toshrink. As a result, the material becomes magnetised.

Figure 6.11: (a) Unmagnetised and (b) magnetised domains in a ferromagnetic material

In hard magnetic materials, the domains maintain their alignment even after weremove the external magnetic field. This results in a permanent magnet. Alloys of

iron, cobalt and nickel can become permanent magnets such that they retain their

magnetism for a long time. However, they lose their magnetism when they are

heated above a certain temperature, called the Curie temperature. The Curie

temperature for iron is about 770C.

On the other hand, in soft magnetic materials like iron, once the external field is

removed, the material quickly returns to its unmagnetised state.

When we compare electricity and magnetism, we can easily findsimilarities between them. For example, with electricity, there arepositive and negative charges. With magnetism, there are north and southpoles. Like charges repel, while unlike charges attract. Like poles repel,while unlike poles attract.

However, you can separate positive and negative charges, but you cannotisolate magnetic north and south poles. Why is this so?

SELF-CHECK 6.4


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Magnetic fields can be represented by magnetic field lines of force, which areimaginary lines in space. These lines never intersect each other and arecontinuous closed loops, being from the N-pole to the S-pole of the barmagnet.

A magnetic field exerts a force on a moving charge. The direction of the forceis perpendicular to both the magnetic field and the velocity of the chargedparticle. The magnitude of the force is given by F = qvBsin.

The motion of charged particles in magnetic fields is in circular path and canbe analysed using the magnetic force expression with Newtons second law ofmotion.

The magnetic properties of substances is due to the motion of electrons inatoms.

The Biot-Savart law can be used to find the magnetic field created a conductorcarrying a current. Amperes law provides an alternative formulation to findmagnetic field created around a current-carrying conductor.

Amperes lawBiot-savart law

Magnetic field

Magnetic forceMagnetism

1. Negative ions of charge 2e, accelerated through a pd of 100V, are injected

into a uniform magnetic field perpendicular to its direction. They then travel

in a circular orbit of diameter 8cm.(a) What is the magnitude of the magnetic field?

(b) What is the angular frequency of their motion?


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2. The particle shown below is a proton moving in the direction indicated.

Predict the direction of the force on it due to the magnetic field B.

1. The figure below shows two long wires carrying currents of magnitude 10A inthe direction indicated by the arrows. What is the magnitude and direction of

the magnetic field at Q?

2. In the equation = qF v B ,

(i) Which pairs of vectors are always perpendicular to each other?

(ii) Which pair may have an angle between them?

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Topic 6 Magnetic Fields