Assignment 13, solutions

Problem 1:

A hollow ball of radius R has a surface charge distribution which produces a potential on the surface of V(R,q) = k(cos2q + 1), where k is a constant and q is the usual polar angle relative to the z-axis.
(a)  Find the potential at all points in space.
(b)  Show that the charge distribution on the ball is s(R,q) = (ke0/R)(5cos2q - 1/3).

Solution:

Concepts, principles, relations that apply to the problem:
General solution to Laplace's equation in situations with azimuthal symmetry, boundary value problems
Why do they apply?
is the most general solution inside and outside of the sphere, since r = 0 inside and outside of the sphere.  To find the specific solution we apply boundary conditions.  The potential on the surface of the ball is given.
How do they apply?
(a)  Assume
f1(r,q) = A0 + A2r2P2(cosq) inside the sphere and
f2(r,q) = (B0/r) + (B2/r3)P2(cosq) outside the sphere.
The symmetry of the situation suggests this form of the solution.  If we can satisfy the boundary conditions, the uniqueness theorem guaranties that we have found the only solution.  The potential vanishes as r goes to infinity.
Boundary conditions:
f1(r,q)  is finite at r = 0 and f2(r,q)  goes to zero as r goes to infinity.
f is continuous across the boundary at r = R.
A0 = B0/R,  A2 = B2/R5. 
f(R,q) = A0 + A2R2P2(cosq) = A0 + A2R2(3cos2q -1)/2 = k(cos2q + 1).
3A2R2/2 = k, A0 - A2R2/2 = k,  A2 = 2k/(3R2),  A0 = 4k/3.
f1(r,q) = 4k/3 + (2k/(3R2))r2P2(cosq),  f2(r,q) = 4kR/(3r) + (2kR3/(3r3))P2(cosq) .
Details of the calculation:
(b)  , ¶f2/dr - ¶f1/dr |R = -s/e0.
¶f2/dr = -4kR/(3r2) - (2kR3/r4)P2(cosq),  ¶f1/dr = (4k/(3R2))rP2(cosq).
s(R,q)/e0 = 4k/(3R) + (2k/R)P2(cosq) + (4k/(3R))P2(cosq) 
= 4k/(3R) + 10k/(3R))P2(cosq).
s(R,q) = (ke0/R)(4/3 + (10/3)(3cos2q - 1)/2) = (ke0/R)(5cos2q - 1/3).

Problem 2:

An infinitely long solenoid is wound with one layer of very fine wire.  The solenoid has a radius r meters, has n turns per meter of length, and carries a current I.
 (a)  What is the magnetic field inside the solenoid?
(b)  What is the magnitude of force on a short length dl of wire in a turn, and what is its direction?
(c)  What is the tension in the wire?

Solution:

Concepts, principles, relations that apply to the problem:
Ampere's law, the Lorentz force
Why do they apply?
The problem has enough symmetry to let us calculate B from Ampere's law of B alone.
If the axis of the coil is the z-axis and the currents flow in the f-direction, then B = Bzk,  Bz = m0In inside the coil and B = 0 outside the coil.  The current in the wires may be treated as a surface current with surface current density k.  The field very close to a section of surface with current density k may be decomposed into the field B1 produced by the section itself and the field B2 produced by all other surrounding sections.  The force per unit length on each wire can be found by from F= IL´B2.
How do they apply?
(a)  The magnetic field inside the solenoid is  B = Bzk,  Bz = m0In.
(b)  Consider a closed loop of length L enclosing N wires of the coil as shown.  

The current flowing through the loop is NI, and very near the wires this current NI produces the same B1 field as a surface current  with density k = NI/L flowing in the same direction in a flat surface.  Ampere's law yields B1 = m0NI/2L = m0In/2 pointing into the z direction on the inside and into the -z direction on the outside of the coil.  So the field near the N wires that is produced by currents not flowing in the vicinity of the loop is B2 = m0In/2 k.  From F = IL´B2 we now find the force per unit length F/L = m0I2n/2 pointing in the positive r direction.

Alternate approach based on energy considerations:
The energy stored in the magnetic field in the coil in a section of unit length is
U = (1/(2m0))B2A = (1/(2m0))(m0In)2A = I2n2m0pR2/2.
If we change the radius of the coil while keeping the current constant then the stored energy increases at a rate
dU/dt = I2n2m0pRdR/dt.
The rate at which the battery does work to maintain constant current in the filementary circuit is
Ie = I d(In2m0pR2)/dt = 2I2n2m0pRdR/dt = 2dU/dt. (e is equal to the rate of change of the flux through the coil.)
The rate at which work is done by the circuit per unit length of coil therefore is dW/dt = I2n2m0pRdR/dt.
dW/dt = Ft×v = FtdR/dt.  Therefore Ft =  I2n2m0pR.  The force on a single wire per unit length is
F/L = Ft/(n2pR) = I2nm0/2.
Details of the calculation:
(c)  Consider a wire loop acted on by a radial force.  The magnitude of force per unit length is constant.  If the loop is cut into two equal pieces, then the total force pulling the pieces apart is .  

For the two pieces of the loop to not accelerate away from each other we need 2T = Fz, where T is the tension in the wire.
Therefore T = Fr/L.

Problem 3:

A permanent magnet has circular pole pieces of radius a separated by an arbitrary distance.  The magnetic field B is uniform between the pole pieces.  Calculate the force between the pole pieces in terms of B and a.

Solution:

Energy stored in the magnetic field
Why do they apply?
F = -dU/dx.
How do they apply?
The magnetic field is confined to the region between the pole pieces.  The energy stored in the field is U = (2m0)-1/2òall spaceB2dV in SI units.  Here U = (2m0)-1/2B2pa2x, where x denotes the distance between the pole pieces.  F = -dU/dx = -(2m0)-1/2B2pa2.  The minus sign indicates that the force is attractive.
Details of the calculation:
None

Problem 4:

Find the magnetic vector potential for the case of a long, straight wire carrying a steady current I.  Let R be the radius of the wire.

Solution:

Concepts, principles, relations that apply to the problem:
Ampere's law, B = Ñ´A
Why do they apply?
Ampere's law and symmetry are used to find the field B of a long straight wire.  We then can find a vector potential A so that B = Ñ´A. 
How do they apply?
Assume the current flows on the z-direction.  Then B is in the f-direction.
From Apere's law (SI units):
r > R:  B = m0I/(2pr),  r < R:  B = m0jpr2/(2pr) = m0Ipr2/(pR2 2pr) = m0Ir/(2pR2).
Bf = (Ñ´A)f = Ar/z - Az/¶r = - Az/¶r.  (We cannot have any z-dependence.)
r > R:  Az/¶r = -m0I/(2pr),  Az = -m0I/(2p)ln(r/r0)
r < R:  Az/¶r = -m0Ir/(2pR2),  Az = -m0I/(2pR2)(r2/2) + C.
Choose C = 0. Then at r = R we have
m0I/(4p) = m0I/(2p)ln(R/r0), ln(R/r0) = 1/2,  r0 = Re-1/2.
Details of the calculation:
None

Problem 5:

In Bohr's 1913 model of the hydrogen atom, the electron is in a circular orbit of radius 5.29´10-11 m and its speed is 2.19´106 m/s.

(a)  What is the magnitude of the magnetic moment due to the electron's motion?
(b)  If the electron orbits counterclockwise in a horizontal circle, what is the direction of this magnetic moment vector?

Solution:

Concepts, principles, relations that apply to the problem:
The magnetic moment m = IAn.
Why do they apply?
We are given the orbital radius and the speed of the electron.  The electron is a negatively charged particle and therefore its motion about the nucleus constitutes an average current in a current loop.
How do they apply?
(a)  m = IA
The current is defined as the charge that passes a stationary observer per unit time.
 I = q/T = qv/(2pR),  T = period.
m = qvR/2 = (1.6´1019)(2.19´106)(5.29´10-11) Am2 = 9.27´10-24 Am2.
We can also write m =  qL/(2m), where L is the angular momentum of the electron about the nucleus.
(b) Let the horizontal circle lie in the z = 0 plane and let the electron move in the f direction.  Then the current is flowing in the -f direction and m points in the -z direction.
m = -9.27´10-24 Am2 k.
Details of the calculation:
None

Problem 6:

A sphere of linear magnetic material is placed in an originally uniform magnetic field magnetic field B0.  Find the new field inside the sphere.

Solution:

Concepts, principles, relations that apply to the problem:
The uniqueness theorem, boundary conditions, B = mH.
Why do they apply?
The external magnetic field will magnetize the sphere.  The magnetic field of a uniformly magnetized sphere is uniform inside the sphere (B = 2Mm0/3) and a pure dipole field outside the sphere.  Assume that the total field inside the sphere is the superposition of the field of a uniform magnetized sphere and the constant external field B0.  The superposition of these two fields fulfills all the boundary conditions for B, [, ] on the surface of the sphere. 
We have a unique solution as long as inside B = mH = m0(1 + cm)H inside the sphere.
B = mH determines the magnetization surface current density.
How do they apply?
Bin = B0 + 2Mm0/3,  mHin = Bin,  M = cmHin = (cm/m)Bin.
(1 - 2m0cm/(3m))Bin = B0.  Bin = B0(3 + 3cm)/(3 + cm).
The magnetic moment of the sphere is m = (4pR3/3)M =(4pR3/3)(cm/m)Bin, where R is the radius of the sphere.  For r > R we have
Bout = B0 + (m0/4p)[3(m×r)r/r5 - m/r3].
Details of the calculation:
None

Problem 7:

An infinitely long cylinder of radius R carries a "frozen in" magnetization parallel to the axis, M = kr, where k is constant and r is the distance from the axis.  Find B and H inside and outside the cylinder.

Solution:

Concepts, principles, relations that apply to the problem:
Magnetization currents, Ampere's law for H
Why do they apply?
Let the symmetry axis of the cylinder be the z-axis.  Then the volume magnetization current density jm = Ñ´M has only a f component,
jmf = -M/r = -k.  The surface magnetization current density km = M´n  also has only a f component, kmf = kR.
The problem is equivalent to that of a set of infinitely-long, nested coils.  We therefore know that for r > R B = H = 0.  For r < R B and H can only have z-components, and we can solve the problem using Amperes law for H.
How do they apply?

The contour integral of the tangential component of H along the close loop G must be zero, H = 0 everywhere.
Inside the magnetized material  B = m0(H + M), B = m0M.
Details of the calculation:
None