Magnetic Field LEcture notes

8/10/2019 Magnetic Field LEcture notes

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PHY2061 Enriched Physics 2 Lecture Notes Magnetic Fields

D. Acosta Page 1 10/17/2006

Magnetic Fields

Disclaimer: These lecture notes are not meant to replace the course textbook. Thecontent may be incomplete. Some topics may be unclear. These notes are only meant to

be a study aid and a supplement to your own notes. Please report any inaccuracies to theprofessor.

Magnetism

Is ubiquitous in every-day life!

Refrigerator magnets (who could live without them?) Coils that deflect the electron beam in a CRT television or monitor

Cassette tape storage (audio or digital) Computer disk drive storage Electromagnet for Magnetic Resonant Imaging (MRI)

Magnetic Field

Magnets contain two poles: north and south. The force between like-poles repels(north-north, south-south), while opposite poles attract (north-south). This is reminiscentof the electric force between two charged objects (which can have positive or negativecharge).

Recall that the electric field was invoked to explain the action at a distance effect of theelectric force, and was defined by:

elq=

FE

where qelis electric charge of a positive test charge and Fis the force acting on it.

We might be tempted to define the same for the magnetic field, and write:

magq=

FB

where qmagis the magnetic charge of a positive test charge and Fis the force acting onit.

However, such a single magnetic charge, a magnetic monopole, has never beenobserved experimentally! You cannot break a bar magnet in half to get just a north poleor a south pole. As far as we know, no such single magnetic charges exist in the universe,


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although we continue to look. Thus, we must look for other interactions with magneticforce to define the magnetic field.

It turns out that an electricallycharged object can also be accelerated by a magneticforce, and through that interaction we can define the magnetic field.

In fact, the electric and magnetic force share a much deeper relation. They are reallymanifestations of the same force, and can be shown to be related by transformations inEinsteins theory of Special Relativity. But here let us discuss the historical perspective.

Cathode Ray Tube

Consider a Crookes Tube, which is otherwise known as a Cathode Ray Tube (CRT) a primitive version of what is in a television. Such a CRT has an electron gun thataccelerates electrons between two electrodes with a large electric potential difference

between them (and a hole in the far plate). The resulting beam of electrons can berendered visible with a phosphorous screen, and then we can observe the effects on thedeflection of the beam in magnetic fields.

From such experiments we can determine several characteristics of electrically chargedparticles interacting with magnets:

The force depends on the direction of the magnetic field (i.e.whether it emanatesfrom a north pole or a south pole).

The force is perpendicular to both the velocity and magnetic field directions The force is zero if the particle velocity is zero (and depends on the sign of v) The force depends on the sign of the electric charge

Defini tion of the Magnetic Field

Thus, we will converge on the following relation for the magnitude of the magnetic forceon a charged object:

sinB q =F v B

or, turned around, allows us to define the magnitude of the magnetic field as:

sinB

q =

FB

v

e

B


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where is the angle between the velocity, v , and the magnetic field, B.

The units of the magnetic field areN N N

T (Tesla)m C A-mC ms s

= =

Another unit based on the cgs metric system is the Gauss, where 41 G = 10 T . TheEarths magnetic field has a magnitude of approximately 0.5 G.

Direction of the Magnetic Field

What direction do we assign to the magnetic field to insert into the force equation? Weimagine field lines the are directed outward from the north pole of a magnet, and inwardto the south pole.

Now in full vector form, we write the expression for the magnetic force acting on aelectrically charged particle as:

B q= F v B

We can see that it will satisfy all the empirical observations noted earlier.

The vector cross-product is defined by:

( ) ( ) ( )

det

sin

x y z y z y z x z x z x y x y

x y z

a a a a b b a a b b a a b b a

b b b

ab

= = = +

=

x y z

a b b a x y z

a b

From HRW 7/e


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Right-Hand Rule

If we consider the following simple example, that 0 v=v x and 0 B=B y , then we see that

the magnetic force direction is given by a right-hand rule: 0 0 F qv B= z

Magnetic Fields and Work

Lets calculate the work done by the magnetic force acts on a moving charged particlethat moves from point 1 to point 2:

2

1W d F s

We can re-writed dt=s v

to get:

( )2 2

1 1W dt q dt = F v v B v

Now with what we learned about the magnetic force, it is always perpendicular to the

velocity vector, so in fact the vector quantity ( )q v B is zero. So W=0 and no work is

done by the magnetic field! You can remember this simply as Magnetic fields do nowork. So apparently the magnetic force is on the physics welfare system!

This implies that there is no change in energy of a charged particle being accelerated by a

magnetic force, only a change in direction. We will come back to this when we discussmagnetic fields and circular motion.

Lorentz Force Equation

We can combine what we learned about the electric force to that we just learned about themagnetic force into one equation, the Lorentz force equation:

( )q= + F E v B

We can apply this to some historical work done by J.J. Thompson, who determined thatthe carrier of electricity was an electrically charged particle dubbed the electron.


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e/m Determination of the Electronthe Carrier ofElectricity

It was determined by J.J. Thompson in 1897 that cathode rays are chargedparticles emitted from a heated electrical cathode. It was known that such heatedcathodes lead to an electrical current, so Thompson determined that electricity wasquantized into individual charged particles (dubbed electrons). He deduced this byanalyzing the motion of cathode rays through perpendicular electric and magnetic fields,as shown below:

Here is a review of the procedure used to determined the charge-to-mass ratio forelectrons by application of the Lorentz Force equation, F E v B= + qb g . What J.J.Thompson assumed was that cathode rays were actually individually charged particles.

In the absence of a magnetic field, with an electric field aligned in the y-direction,we have:

F qE ma

aqE

m

y y y

y

y

= =

=

Let electric field occupies a region of length through which the cathode rays pass. Thecathode rays initially have a velocity v0 in the x-direction.

tv

v a tqE

mv

v

v

qE

mv

y y

y

y

x

y

= =

= =

0

0

02

Time to cross the electric field region

Velocity in y - direction upon exit

Tangent of exit angletan

The angle can be measured in the experiment, and the length of the electric plates isobviously known. The electric field can be determined from the electric potentialdifference Vapplied to the parallel plates divided by the separation:

V

d=E

+ + + +

E

x

ye

B

d


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However, we need a way to determine the initial velocity v0 . The trick is to turn on a

magnetic field such that the cathode beam is no longer deflected. Using the Lorentz Forceequation:

F E v B

E v B

E

= + =

= =

= =

q

v B v B

v E

Bv

x z x z

x

b g 0

0

Set to balance, = 0

and are perpendicular

| | | || |

Armed with this, we can determined the charge to mass ratio of the electron:

tan=qE

m

B

E

2

2

So the procedure is that we measure for a givenE, then we measureBfor =0. Thisyields a charge-to-mass ratio of:

What J.J. Thompson found in his laboratory was that this was a universal ratio. It didntmatter if the cathode was aluminum, steel, or nickel; and it didnt matter if the gas wasargon, nitrogen, or helium. The value always came out to be the same. Thus, there wasonly one type of charge carrier for electricitythe electron. It should be noted that thecharge of the electron is actually negative.

What else can we conclude? From the size of this ratio, either the charge of the electronis very large, or the mass of the electron is extremely small (or some combination ofboth).

q

m

E

B

e

m= =

tan " "

2

e

m= 176 1011. C / kg


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Review: Charge of the Electron

J.J. Thompson only determined the ratio of the electron charge to its mass, rather thaneach separately. It was Robert Millikan in 1911 who was able to measure the electron

charge directly. He did this by measuring the static electric charge on drops of oil, andfinding that it was always a multiple of a certain value. A schematic of the set up isshown below:

The oil drop is suspended when the acceleration a= 0

= =

=

F E gE q m

q mgd

V

Millikan measured the mass of the oil drops by turning off the electric field andmeasuring the terminal velocity. We will assume that the mass is known, as with thevoltage and the plate separation distance d. Through very precise measurements,Millikan found that the electric charge is a multiple of the following value:

Charge is quantized! From Thompsons e/mmeasurement, we can deduce:

The electron has a definite charge and mass which is the same for all electrons.

E = V/d

+

FE

mg

e q= = | | .1 6022 10 19 C

me= 911 10 31. kg


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Magnetic Force and Circular Motion

Recall that the magnetic force is perpendicular to the direction of motion. This impliesthat the trajectory in a uniform magnetic field is a circle in the plane transverse to the

direction of the field (a helix in 3 dimensions, generally).

Consider a positively charged particle with velocityv (taken to be toward the right)entering a region of uniform magnetic field pointing into the plane of the page. Theparticle will be deflected by a force that is perpendicular to the field and the initialvelocity. In the example below, this will be in the upward direction. But whenrecalculating the force at another point along the trajectory, we will find that the particleis continually deflected with an equal magnitude force, and the net effect is a circularorbit. Since the magnetic field does no work, the radius of this circle, r, will be aconstant.

We can solve for the radius of this orbit:

q q

m

= ==

F v B v B r

F a

where the magnitude of the centripetal acceleration is2v

ar

=

2mvq

r

mvr

qB

=

=

v B

In other words, since momentum is defined asp mv= , we have

pr

qB=

This gives the radius of the orbit in terms of the particle momentum, charge, and themagnetic field magnitude. This latter form of the equation is even correct with therelativistic definition of momentum.

B

F

v

rF

F

F


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To see this, recall that the relativistic definition of momentum is:

2 2

1 where and is the particle velocity

1 /p mu u

u c = =

The magnitude of the velocity in circular motion is constant, so the force can be written:

( ) since constantd d d

m mdt dt dt

m

= = = =

=

p uF u

F a

Setting this force equal to the magnetic force, and plugging in for centripetal acceleration:

2vm qvB

r

mv p qBr

pr

qB

=

= =

=

Orbital Frequency

The frequency of orbit in uniform circular motion is given by

2

vf

r=

The relativistic momentum is p mu= , so

2

pf

mr=

Now plugging in that p qBr= , we get

2

21

2

qB uf

m c=

So we see that the frequency is constant provided u c , but when we approach speedsnear that of light, the frequency slows down.

Thus, while the early particle accelerator, cyclotrons, used a constant frequency tomaintain circular motion, higher energy machines must synchronize the orbital frequencyaccording to Special Relativity.


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D. Acosta Page 10 10/17/2006

Cyclotrons and Synchrotrons

The previously derived equation is extremely useful for experimental particlephysicists! It can be used in two main ways. First, it can be used to bend charged particlesinto a circular orbit. This is how physicists maintain circular beams of electrons and

protons in a fixed tunnel. A series of magnets in the shape of a ring, all with the samefield, bend the particles so that they can be brought into collision over and over again.The largest accelerator, the Large Hadron Collider, is 4.3 km in radius and is illustratedbelow!

The other way to use the equation is to measure the radius of curvature of a chargedparticle in a known magnetic field in order to determine its momentum. In other words,

once the beam particles have collided, we must measure the momentum or energy of thecollision products in order to determine what reaction took place. For example, thepicture below shows the reconstructed trajectories of charged particles emanating from aproton collision. The curvature is inversely proportional the momentum.


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D. Acosta Page 11 10/17/2006

Hall Effect

Another example of the Lorentz Force is the magnetic effect on a current of electrons in aconductor. In the example below, a uniform magnetic field is directed into the plane ofthe paper and the current moves from top to bottom.

The electrons drifting upward in the shown conductor (electric current going down) willinitially feel a force due to the v B magnetic interaction (left picture). This will causethem to drift toward one side of the conductor and build up charge. Very soon, this willset up an electric field that balances the magnetic force (right picture). At equilibrium, theforces balance.

0E B

de e v+ =

=F F

We can define a Hall Voltage as the electric potential difference between the left andright sides of the shown conductor:

HV d= E

When combined with the force balance equation we get:

Hd

V v Bd

=

Now from Ohms Law, the current density is related to the electron drift velocity:

dnev=j

i

d

FB

Vd

d

FB

VdFE

++++++

++++

i

B


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D. Acosta Page 12 10/17/2006

So

where = current and = cross-sectional aread

H

j iv i A

ne neA

idV B

neA

= =

=

Or in other words, we can determine the magnetic field from the measured Hall voltageand current passing through the conductor:

H

neAB V

id=

Such a device used to measure magnetic fields is called a Hall Probe.


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D. Acosta Page 13 10/17/2006

Magnetic Force on a Current-Carrying Wire

Consider a length of conducting wire carrying a current, i, in a uniform magnetic field B.The drift-velocity of the electrons is constant and is denoted by dv . The time it takes for

an electron to drift across a length of wire,L, is given by:

d

Ltv

=

The amount of charge passing through the end of wire in that time is given by:

d

Lq it i

v= =

This must equal the amount of free charge contained in a length of wireLat any instant.

The force on that amount of charge is:

dF q= v B

and this must be the force acting on a length of wire L carrying a current. We can write interms of the current:

sin sin sind d

d

Lqv i v B iLB

v = = =F B

Or alternatively, if we define a length vector Las pointing in the direction of the currentand having a magnitude equal to the length, we can write for the force on a current-carrying wire in a magnetic field:

i= F L B

The force depends on the directions and magnitudes of the current and the magnetic field.If we haveNloops of wire carrying the same current, we multiply the above equation bythe number of wiresN.

For an infinitesimal length of wire, ds, we have: d id= F s B

ivd e

L B


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D. Acosta Page 14 10/17/2006

Torque on a Current Loop

Consider a rectangular current loop inside a uniform magnetic field (i.e.the basis of anelectric motor). It has a spindle about which it is free to rotate.

There are 4 straight segments of current-carrying wire, each of which will feel a force dueto the magnetic field.

Wires 2 and 4 are not always perpendicular to the magnetic field, so

( )2 42

4

sin 90 cos

cos

cos

ibB ibB

ibB

ibB

= = =

=

= +

F F

F y

F y

The forces balance, and there is no torque because the loop is not allowed to rotate about

the horizontal axis.

Wires 1 and 3 are always perpendicular to the magnetic field, so

1 3

1

3

iaB

iaB

iaB

= =

=

=

F F

F z

F z

B = B x

i

y

x

y

x

z

i

a

b

13

2

4


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D. Acosta Page 15 10/17/2006

The forces thus balance, although the torque about the vertical axis will not.

In total, all forces balance, but the two vertical sides 1 and 3 will create a torque. Thelever arm length is / 2r b= . Viewed from the top, the situation looks like:

The torque is:

( )

( )

1

3

1 3

sin2

sin2

sin

biaB

biaB

iabB

=

=

=

= + =

r F

They both add in the same direction. The product ab A= , the area of the loop.

siniAB =

We can define an area vector as having its magnitude equal to the loop area, and adirection perpendicular to the loop (and in a sense given by the right-hand rule and thecurrent direction). Thus, a vector form of the torque is:

i= A B

If there areNwindings of the wire around the loop, the formula becomes Ni= A B

If we further define the magnetic dipole moment of the loop to be i= A , then

= B

This formula holds for any current loop shape.

x

zF1

F3 B


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D. Acosta Page 16 10/17/2006

To determine the direction of use the right-hand rule (fingers in direction of current,then thumb points in direction of ) :

The effect of the torque on the current loop is to try to line up the magnetic dipolemoment vector with the magnetic field:

Once the two vectors are aligned, there will be no torque. However, if we reverse thefield direction, the will want to flip the orientation of the loop. If we keep doing this, weform the basis of an electric motor.

i

B

i

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Magnetic Field LEcture notes