Chapter 33. The Magnetic Field

1. Moving charges make a magnetic field

2. The magnetic field exerts a force on moving charges

Thus moving charges exert forces on each other (Biot Savart Law)

3. A collection of moving charges constitutes and electrical current

4. Currents exert forces and torques on each other

1. Moving charges (or equivalently electrical currents) create a magnetic field, rB(rr)=μ04πqrv×rr2,

Here rv i s the velocit y o f t hemoving char .ge

This is the analog to the express ionfor t he electric fiel d due t o a point charge. rE(rr)=q4πε0rr2Superpositi on applies if a number o f moving charges are present just as in t he ca se ofelectric fields. As in t he ca se of electric fields, most of you r grie f will come from tryingt oa pplythe principle of superposition.

*

* This isn’t really true but pretend for the moment that it is.

2. A charge moving through a magnetic field feels a force (called the Lorenz force)given by rF=qrv×rB.T hetota l electri c and magnetic force ona movi ng charged particl eis the su mof thecontributions due t oelectr icand magneti cfields, rF=q(rE+rv×rB).Just to beconfusi ngthis is sometimes called t heLorenz force t .oo

There are a host of consequenc 1. 2.es of and, which will be explored in thischapte .r An example is paralle l currents attract and antiparalle l currents repel.

A big complicati on when discussing magneti cfields is t heappearance of vectorcros sproducts. Don’t thi nktha t you can sli de by without learni ng how t oevaluat ethe .m

rB(

rr) =

μ0

4πqrv×rr2

rE(

rr) =

q4πε0

rr2

Electric Field Magnetic Field

q

r

rv

What are the magnitudes and directions of the electric and magnetic fields at this point? Assume q > 0

r

Comparisons: both go like r-2, are proportional to q, have 4π in the denominator, have funny Greek letters

Differences: E along r, B perpendicular to r and v

Electric Field points away from positive charge

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rE(

rr) =

q4πε0

rr2

rB(

rr)=

μ0

4πqrv×rr2

Magnetic field is perpendicular to

Greek Letters

rB(

rr) =

μ0

4πqrv×rr2

rE(

rr)=

q4πε0

rr2

μ0 = 4π ×10−7 Henries /meterε0 = 8.8542 ×10−12 Farads /meter

Huh?

These funny numbers are a consequence of us having decided to measure charge in Coulombs

Later we will find:

1 / ε0μ0 =c μ0 / ε 0 = 377 Ohms

Aside on cross products: rC=rA×rB

θ rA rB

rC

Let θ be the angle between rAand rB, always chose thi s to b e les s than 180°.

T hemagnitude of rCis gi ven by rC=rArBsinθT he directi on o f rC i sperpendicular t o bot h rAand rB, and determined by the “right handrul ”e

Right hand rule: put the fingers of your right hand in the direction of rA, then rotatethem through the angle θ (les st 180han °) to t he direction of rB. Yourthum b nowindicates the directi on of rC.T he order of rAand rB i s importan :t rC=rA×rB=-rB×rAIf rAand rB are parallel, then (θ=0) rC=0.

Let’s say we are given rAand rBin component form in Cartesian coordinates.

rA=Axi+Ayj+AzkrB=Bxi+Byj+Bzk.How do we fi ndthe components of rC? rC=Cxi+Cyj+Czk.Answer: Cx=AyBz−AzByCy=AzBx−AxBzCz=AxBy−AyBx(m y favorite)

i j k

Ax Ay AzBx By Bz

Same result as:

rB(

rr) =

μ0

4πqrv×rr2

The positive charge is moving straight out of the page. What is the direction of the magnetic field at the position of the dot?

A. LeftB. RightC. DownD. Up

rB(

rr) =

μ0

4πqrv×rr2

A. LeftB. RightC. DownD. Up

The positive charge is moving straight out of the page. What is the direction of the magnetic field at the position of the dot?

rB(

rr) =

μ0

4πqrv×rr2

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Field near a point charge

Field lines emerge from charges

Iron filings align parallel to magnetic fieldField lines do not end

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are needed to see this picture.

Electric field due to a dipole - two opposite charges separated by a distance

Magnetic field due to a single loop of current

Magnets have “poles” labeled north and south:

Like poles repel

NS N SUnlike poles attract

NS NS

Magnets are objects that have charges moving inside themElectrical currents

If you cut a magnet in half, both halves will have a north and south pole.

NS NS

NS

Cut

33.7 Force on a moving Charge

For a stationary charge we have rF=qrE(rr).For a moving charge there is an additional contribution, known as the Lorenz force, rF=q(rE+rv×rB).Here, rv i s the velocit y o f t hemoving char .ge

Some facts:

1. For a particle a t rest, rv=0the magneti cfiel dexerts no force on a cha rged particl .e2. T heLorenz for ceis proportional t othe charge, the magnetic field strengt h and theparticle’s velocit .y

3. The vector force is the result o f the vecto r cros s product of velocity andmagneticfiel .d

Motion of a charged particle in a magnetic field

Newton’s law: mra=rF=qrv×rB mra=mdrvdt=qrv×rBNote: Lorentz force is always perpendicular to velocit .y Ther efor ethe magnit ude ofvelocity will not cha . nge This implies the kinetic ener gy of the particl e isconstant, ddtKE=rv⋅rF=0, KE=m2rv2Note also, if velocity is parallel t omagneti cfiel dthe force is zero and vecto r velocit y isconstant.

Does the compass needle rotate clockwise (cw), counterclockwise (ccw) or not at all?

A. ClockwiseB. CounterclockwiseC. Not at all

A. ClockwiseB. CounterclockwiseC. Not at all

Does the compass needle rotate clockwise (cw), counterclockwise (ccw) or not at all?

Does the compass needle feel a force from the charged rod?

A. AttractedB. RepelledC. No forceD. Attracted but very weakly It’s complicated

Does the compass needle feel a force from the charged rod?

A. AttractedB. RepelledC. No forceD. Attracted but very weakly It’s complicated

B = 1T in the y direction

z

x

y

rv60° 106 m/s in x-y plane

rv =

What is the force on a moving electron?

rF =q(

rv×

rB)

rv ×

rB =

rv

rB sinθ What is θ in this problem?

A. 60°B. 30°C. 72 ° and sunny

rF =−1.6×10−19 (C)* 106 (m/ s)* 1(T )*

12

)k

rF =−0.8×10−13 )k Newtons

rF =q(

rv×

rB)

Cyclotron Motion

Newton’s Law mra =

rF

Both a and F directed toward center of circle

ra =

v2

rcyc

rF = qvB

rF =q(

rv×

rB)

rcyc

Equate

W=

v

rcyc=qB

m= 2p fcyc

rcyc =

v

W=

v

2p fcyc

rF =q(

rv×

rB)

rcyc

In this picture B is pointing into the page and q is positive. What changes if q is negative and B points out of the page

A. Rotation changes to counter-clockwise

B. There is no changeC. Why do things always have

to change?

rF =q(

rv×

rB)

rcyc

In this picture B is pointing into the page and q is positive. What changes if q is negative and B points out of the page

A. Rotation changes to counter-clockwise

B. There is no changeC. Why do things always have

to change?

Some Examples

Ionosphere at equator

fcyc =

qB

2pm rcyc =

v

2p fcyc

B ; 3.5¥ 10- 5T

Te ; 300oK

v ; 6¥ 104m / s

fcyc ; 106Hz

rcyc ; .01m

ITER fusion exp

B ; 6T

Te ; 108 oK

v ; .2c

fcyc ; 1.67 1011Hz

rcyc ; 5.7¥ 10- 5m

Motion of a charged particle in crossed electric and magnetic fields

. x

y

z. v

E

B

rv = vxi

rE = Ey j

rB = Bzk

Try these

md

rv

dt= q

rE +

rv¥

rB( ) = jq Ey - vxBz( )

If vx=Ey/Bz force is zero and no acceleration

Particle will move at a constant velocity in a direction perpendicular to both E and B

Where are you most likely to encounter E cross B motion?

A. Your doctor’s officeB. Your KitchenC. The bridge of the Starship Enterprise

Where are you most likely to encounter E cross B motion?

A. Your doctor’s officeB. Your KitchenC. The bridge of the Starship Enterprise

MagnetronElectrons spiral out from the cathode K toward anode A. Potential energy is converted into radiation, f=2.45 GHzRadiation then converted into popcorn

B-out of page

B, V0 and Slots in anode define radiation frequency

Current carrying wire

Inside the wire there are stationary charges and moving charges.

Metallic conductorMoving - free electronsStationary - positive ions and bound electrons

In any length of wire the number of positive and negative charges is essentially the same (unless the wire is charged).

The wire can carry a current because the free electrons move.

Current, number density and velocity

Consider a segment of length l, cross sectional area A

A

The segment has N free electrons

Number density, n, of free electrons is a property of the material.For copper n ; 1.1¥ 1029 m- 3

N = n lA( )volume

Suppose the free electrons have speed v. How long will it take all the electrons to leave the segment?

Dt = l / v

Current

During a time interval t = l/v, a net charge Q = -eN flows through any cross section of the wire.

The wire is thus carrying I

I =

Q

D t=eNv

lThe current is flowing in a direction opposite to that of v.

A

How big is v for typical currents in copper?

A. Close to the speed of lightB. SupersonicC. About as fast as a PriusD. Glacial

A

How big is v for typical currents in copper?

A. Close to the speed of lightB. SupersonicC. About as fast as a PriusD. Glacial

A

For a 1mm radius copper wire carrying 1 A of current

v =

I

Aen= 1.8¥ 10- 5m / s

What is the force on those N electrons?

Force on a single electron

rF = - e

rv¥

rB( )

Force on N electrons

rF = - eN

rv¥

rB( )

Now use eN

rv = I l

Lets make l a vector - eNrv = I

rl

points parallel to the wire in the direction of the current when I is positive

Force on wire segment:

rFwire = I

rl ¥

rB

rFwire = I

rl ¥

rB

Comments:Force is perpendicular to both B and lForce is proportional to I, B, and length of line segment

Superposition: To find the total force on a wire you must break it into segments and sum up the contributions from each segment

rFtotal = I

rli ¥

rB(

rri )

segments- iÂ

What is the force on a rectangular loop?

Net force is zero

Torque on a loop

Plane of loop is tilted by angle with respect to direction of B.The loop wants to move so that its axis and B are aligned.

Definition of torque

rt =

rr ¥

rF

sidesÂ

t y = 2

a

2

Ê

ËÁÁÁ

ˆ

¯˜˜ IbBsinq= i(ab)Bsinq

Nuclear magnetic resonance Basis of MRI magnetic resonance imaging

Magnetic Field due to a current

Magnetic Field due to a single charge

rB(

rr) =

μ0

4πqrv×rr2

If many charges use superposition

rB(

rr) =

μ0

4πqi

rvi ×ri

ri2

charges−i∑

#2 q2 , v2

#1 q1, v1

#3 q3 , v3

r3

r2

r1

Where I want to know what B is

r3

r2

r1

For moving charges in a wire, first sum over charges in each segment, then sum over segments

rB(

rr) =

μ0

4πqi

rvi ×ri

ri2

chargesineachsegment−i

∑⎛

⎝

⎜⎜

⎞

⎠

⎟⎟sgments−j

∑ =μ0

4πI

rlj ×rj

rj2

sgments−j∑

qirv i ×ri

ri2

chargesineachsegment−i

∑ =I

rlj ×rj

rj2

Summing over segments - integrating along curve

rB(

rr) ==

μ0

4πI

rlj ×rj

rj2

sgments−j∑ →

μ0

4πIdrl ×rr2—∫

drl

r

I r

Integral expression looks simple but…..you have to keep track of two position vectors

Biot Savart law

which is where you want to know B

which is the location of the line segment that is contributing to B. This is what you integrate over.

rr

r′r

rr

r′r

Magnetic field due to an infinitely long wire

x

z

I

drl = d ¢z k

r

Current I flows along z axis

I want to find B at the point

rB(

rr)==

μ0

4πIdrl ×rr2∫

rr = xi + 0 j + 0k

I will sum over segments at points r′r = 0i + 0 j + ′z k

r =

rr -

r¢r

rr -

r¢r

=xi - ¢z k

x2 + ¢z 2

r =

rr -

r¢r = x2 + ¢z 2

rB(

rr) ==

μ0

4πIdrl ×rr2∫

drl = d ¢z k

r =

rr -

r¢r

rr -

r¢r

=xi - ¢z k

x2 + ¢z 2

r =

rr -

r¢r = x2 + ¢z 2

rB(

rr) ==

μ0 I4π

d ′z xk×i - ′zk×k( )

x2 + ′z 2( )3/2

−∞

∞

∫

rB(

rr) ==

μ0 I4π

d ′z x j( )

x2 + ′z 2( )3/2

−∞

∞

∫ =μ0 I2πx

j 0

See lecture notes

rB =

μ0 I2πr

r

compare with E-field for a line charge

rE =

λ2πε0r

Forces on Parallel Wires

F2on1 = Irl1 ¥

rB2 (1)

Magnetic field due to #2

Evaluated at location of #1

F2on1 = Irl1 ¥

rB2 (1)

Magnetic field due to #2

Evaluated at location of #1

rB2 (1) =

m0I2

2pd

Direction out of page for positive

What is the current direction in the loop?

A. Out of the page at the top of the loop, into the page at the bottom.

B. Out of the page at the bottom of the loop, into the page at the top.

A. Out of the page at the top of the loop, into the page at the bottom.

B. Out of the page at the bottom of the loop, into the page at the top.

What is the current direction in the loop?

Magnetic field due to a loop of current

Bz

Contribution from bottom

Contribution from top

rB(r = 0, z) =

m0I

2

R2

R2 + Z 2( )3/2

See derivation in book

Note: if z >> R

R

rB(r = 0, z) ;

m0I

2

R2

Z 3=m0

2p

IA( )

Z 3

Magnetic moment,Independent of loop shape

What is the current direction in this loop? And which side of the loop is the north pole?

A. To the right in front, north pole on bottomB. To the left in front; north pole on bottomC. To the right in front, north pole on topD. To the left in front; north pole on top

What is the current direction in this loop? And which side of the loop is the north pole?

A. To the right in front, north pole on bottomB. To the left in front; north pole on bottomC. To the right in front, north pole on topD. To the left in front; north pole on top

rE(

rr) =

q1

4πε0r2 r

Gauss’ Law:

rE⋅d

rA—∫ =

Qin

ε0 0 =

rE(

rr)⋅d

rs—∫

Coulombs Law implies Gauss’ Law and conservative E-field

But also, Gauss’ law and conservative E-field imply Coulombs law

rE(

rr) =

q1

4πε0r2 r

rB(

rr) =

μ0q1

4πr2

rv×r

Gauss’ Law:

rB⋅d

rA—∫ =0

rB(

rr)⋅d

rs=—∫ μ0 I through

Biot-Savart Law implies Gauss’ Law and Amperes Law

But also, Gauss’ law and Ampere’s Law imply the Biot -Savart law

Ampere’s Law

rB(

rr) =

μ0q1

4πr2

rv×r

Consequences of Coulomb’s Law

Gauss’ Law:

rE⋅d

rA—∫ =

Qin

ε0

Electric flux leaving a closed surface is equal to charge enclosed/εo

Always true for any closed surface! Surface does not need to correspond to any physical surface. It can exist only in your head.

Surface used to calculate electric field near a line charge

Electric field due to a single chargeMagnetic field due to a single loop of current

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Guassian surfaces

rB⋅d

rA—∫ =0

rE⋅d

rA—∫ =

Qin

ε0

Further consequences of CL

Existence ofa potential

V (rf )−V(ri ) =−

rE(

rr)⋅d

rs

ri

rf

∫Difference in potential is independent of path between ri and rf

V (ri ) ri

rE(

rr)

drs

V (rf ) rf

drs

rE(

rr)

0 =

rE(

rr)⋅d

rs—∫

Line integral around any closed loop is zero. Field is “conservative”

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are needed to see this picture.

drs

rE(

rr)

drs

drs

rB(

rr)

0 =

rE(

rr)⋅d

rs—∫

rB(

rr)⋅d

rs—∫ =μ0 I

Right hand rule: direction of ds determines sign of I

rB(

rr)⋅d

rs—∫ =???

For the path shown, what is

A: I1 + I2 + I3+I4

B: I2 - I3+I4

C: -I2 +I3-I4

D: something else

Does I1 contribute to B at points along the path?

A: YesB: No

rB(

rr)⋅d

rs—∫ =μ0 I

rB(

rr)⋅d

rs—∫ =2πd

rB(d)

Thus

rB(d) =

μoI2πd

Agrees with Biot-Savart

rB(

rr)⋅d

rs—∫ =μ0 I through

rB(

rr)⋅d

rs—∫ =2πr B(r)

Ithrough =I if r > R

Ithrough =r2

R2 I if r < R

rB(

rr)⋅d

rs=—∫ μ0 I through

rB(

rr)⋅d

rs=—∫ Bl

μ0Ithrough = μ 0NI

B=μ0NI

l=μ0 (N / l)I

# turns per unit length

rE(

rr) =

q1

4πε0r2 r

Gauss’ Law:

rE⋅d

rA—∫ =

Qin

ε0 0 =

rE(

rr)⋅d

rs—∫

Coulombs Law implies Gauss’ Law and conservative E-field

But also, Gauss’ law and conservative E-field imply Coulombs law

rE(

rr) =

q1

4πε0r2 r

rB(

rr) =

μ0q1

4πr2

rv×r

Gauss’ Law:

rB⋅d

rA—∫ =0

rB(

rr)⋅d

rs=—∫ μ0 I through

Biot-Savart Law implies Gauss’ Law and Amperes Law

But also, Gauss’ law and Ampere’s Law imply the Biot -Savart law

Ampere’s Law

rB(

rr) =

μ0q1

4πr2

rv×r