HW 5 Solution

Note: Magnetic field by current loop

David Griffith, Introduction to Electrodynamics, example 5.6, p218

13-2

Find the polarization charge of a long solid dielectric cylinder, of radius a, with polarization

zzyyxxPrPP 22 00

aPrP

PP

ar 223

0

0

( is the volume charge density, is the surface charge density at r=a)

Since the cylinder with charge is rotating, current is formed. The volume current density J

at r is,

23 0rPrvJ

The surface current density K

at r=a is,

21 20aPavK

Magnetic field at center of current loop, radius r, with current I is,

zdIrzBd 20

For volume current, drrPdI 023

zaPzdrPzdIrzB a 43432 000 000

For surface current, 2021 aPdI

zaPzraPzB ar 44 00

2

00

Therefore, the total magnetic field is,

zaPzB 2 00

13-3

Using Amperes law, with cylindrical symmetry, one can easily verify 20 rIrB enc .

a)

For r

13-4

a)

Note from law of action and reaction, force on the loop has same magnitude as force on the wire.

Calculate the force on the loop.

Magnetic field by wire is,

2

0rIB

Therefore, the force on each of the segments of the loop, 1,2,3,4, is,

BIBvdqF l Where, the direction of l is given as the direction of current flow.

Note force on 1 & 3 cancel each other

The total force is,

waIaIF 222

02

0 ll

(Here, + sign is used for the attractive force)

1

2 3

4

b)

The problem is ill-posed.

For one to calculate the torque of the system, one has to specify the origin.

Note if we take the origin to be center of the loop, torque=0 for both loop and the wire, since

0424321 FFrFFFFrFr

14-2

Let current flows in x direction.

Let z direction to be the direction perpendicular to the plane.

(a)

From Biot-Savart law,

lxrxxIB

0 20

4 =0

Therefore, magnetic field arising from straight line segment is 0.

(b)

For semicircle,

20

0 20

4

4 rzIdr

zIB

(c)

Total magnetic field is,

204

rzIB

z

x

y

14-3

(a)

The magnitude of the magnetic field produced by a single loop on its axis is,

2/3222

02)( za

aIzB

where z is the distance from the center of the loop.

Therefore, for the given configuration,

2/32

2

2

2/322

20

222)(

xba

a

xba

aIxB

(The direction of the magnetic field is drawn at the figure.)

(b)

Keeping to order of 2x ,

4/815

4/2314

4

1442

22

22

22

22/322

2/3

22

22/322

2/32

22

2/322

baxb

baxxba

babxxbaxbxbaxba

B

Using this approximation, magnetic field can be approximated to order of 2x as,

2

22

222/3220

4

23

142)( xba

abbaIxB

(c)

For a=b, 2x term vanish.

Therefore, we have almost uniform magnetic field at the center.

(d)

For a=b, from d),

20

2/3

2/320

58

45

2)( caaIxB

14-6

Consider the conducting sphere of radius a and potential V.

(1) Find surface charge density and volume charge density of the sphere.

To maintain the potential, assume sphere is charged by charge Q.

aQiV

04

aVQ 04

Since charges will be distributed only at the surface of the conductor,

aV0 , 0

(2) For rotating conducting sphere with potential V, find the volume current density J

and surface current density K

.

0 vJ

sinsin 00 VaaVvK

(3) Find magnetic field at the center of the sphere.

Since there are only surface currents, we can decompose sphere into collection of current loops.

The magnitude of the magnetic field produced by current loop with radius r and current dI is,

2/3222

02 zr

rdIdB

For our problem, cosaz , sinar dI Kad

Therefore, the magnetic field at the center is,

2

000

300

2/32222

3200

2/322

20

32

32sin2

cossinsin

22

cV

VdV

daa

aaVzrrdIdB

(4) Find magnetic field for distance much longer than a

Since there are only surface currents, we can decompose sphere into collection of current loops.

Each loop can be treated as ideal magnetic dipoles.

Each loop has dipole moment

zKdazdIrm sin 232 Therefore, the total sphere can be treated as magnetic dipole.

zVazdVazKdam 34sinsin 030

30

30

23

ar

z

(5) Exact solution for the magnetic field inside the sphere can be solved by calculating the magnetic field of rotating sphere with surface charge .

David Griffith, Introduction to Electrodynamics, example 5.11, p236-p.237

(6) Exact solution for the magnetic field outside the sphere can be solved by calculating the magnetic field of sphere with uniform magnetization M,

Note for sphere with magnetic dipole moment zaV 34 3

0

,

zVM 0

sin0VrMK a

0 MJ Therefore solving the magnetic field of sphere with uniform magnetization zVM 0

will give

same magnetic field to our solution.

For a sphere with uniform magnetization, taking the curl of eqn 5.67 at (5) gives the field equivalent to the magnetic field of for sphere with magnetic dipole moment

zaV 34 3

0

15-1

B=E=0 inside the superconducting metal.

(a)

From 0 B , calculating volume integral for 0 B over cylinder with infinitesimal height,

0dABadB above Where aboveB denotes the magnetic field perpendicular to the interface, above superconductor.

Therefore, 0B at the surface.

(b)

Define image magnetic dipole inside superconductor as

Note due to symmetry, this image magnetic dipole will give 0B at the surface.

dA B0

B=0

d

d Image magnetic dipole

Therefore, solving the equivalent problem with image magnetic dipole will give same magnetic field.

(c)

Torque on the dipole

=Torque on the dipole 1 by magnetic field produced by image magnetic dipole 2 .

3

20

221210

2220

1

21

162sin

4

23

423

4

d

dzz

dzz

BN

Equilibrium angles is when torque is 0, i.e., 23,,2,0 .

d

d

Image magnetic dipole

z

d

d

Image magnetic dipole

z

x y

To find stability of equilibrium angles, calculate energy of the configuration,

0 32

0 2cos324 dNdW

For stable equilibrium, 022

dWd , for unstable equilibrium, 02

2d

Wd

Therefore,

23,2 : stable equilibrium

,0 : unstable equilibrium

(d)

Force on the dipole

=Force on the dipole 1 by magnetic field produced by image magnetic dipole 2 .

3

220

221210

2220

1

21

81cos

4

23

423

4

d

dzz

dzz

B

Calculating )( 21 BF gives the force on the magnetic dipole.

d

d

Image magnetic dipole

z

x y