Phys 742 - EM I In-class test 2 - 4 April 1996

1.

Solve the boundary-value problem for a spontaneously magnetized sphere of radius a. Assuming that M is constant in the sphere, find the H and B fields everywhere.

The magnetization is equivalent to a surface current K. Choose polar coordinates in such a way that K has only the component K_{ϕ} and give the explicit expression for K_{ϕ}.

Find a vector potential A that describes B everywhere. Give explicitly the components of A in the coordinates of your choice.

2.

A dielectric slab of thickness a and width b is partially inserted between the plates of a capacitor (see sketch on blackboard). Is the slab attracted or ejected? Assuming a linear dielectric constant ϵ, find the force on the slab when the potential difference between the plates is V. What is a typical value of ϵ and what is the corresponding numerical value of the force when a = 1 cm, b = 10 cm and V = 1000 Volts?

3.

(a) A solenoid consists of N turns of area A and carries a current I. Find the field B at large distances, far outside the solenoid.

(b) Find the torque on the solenoid when its center is at a distance d from the surface of a medium of linear permeability μ and its axis makes an angle θ with the normal to the surface. Assume that d is much larger than the dimensions of the solenoid.

Y_{00} = \frac{1}{\sqrt{4π}}  Y_{10} = \sqrt{\frac{3}{4π}} cos θ  Y_{11} = −\sqrt{\frac{3}{8π}} sin θ e^{iϕ}

Answer 1

The dipole moment is m = (4π/3)Ma^{3}. The field for r > a is

B^{out}= − m∇(z/r^{3})
B_{r}^{out} = 2m cos θ /r^{3} B_{θ}^{out} = m sin θ/r^{3}

From the continuity conditions, the fields for r < a are B^{in} = (8π/3)M and H^{in} = −(4π/3)M, in agreement with B^{in} = H^{in} + 4πM.

From K = cn × M get K_{ϕ} = cM sin θ. The vector potential in the Coulomb gauge is

A(r) = \frac{1}{c} ∫ \frac{J(r′)}{|r − r′|} d^{3}r′ = M ∫ \frac{e′_{ϕ} sin θ′}{|r − r′|} dΩ′

where e′_{ϕ} has cartesian components (− sin ϕ′, cos ϕ′, 0). As usual, it is convenient to compute A_{y} for x = 0, which is the same as A_{ϕ}. We obtain

A_{ϕ}(r) = M ∫ \frac{cos ϕ′ cos θ′}{|r − r′|} dΩ′

Answer 2

As the slab moves in an amount δx, the enegy increases by \frac{1}{2}(ϵ − 1)(bδx/a)V^{2}. The force is in the direction of motion (rmember that at constant V one has F_{x} = +∂U/∂x) and has magnitude

F = (ϵ − 1)\frac{bV^{2}}{2a} ≃ 50(ϵ − 1) dynes