Physics 304
Due: Monday, Dec. 5
Problem Set 11
(1) Consider a very long, thin wire which lies along the z axis and carries current I flowing in
the positive z direction. A small current loop which can be considered to be a point dipole is
located in the positive x direction at position xo. Find the force and torque on the loop in the
!
cases, (i) The moment m points along x̂ ; (ii) The moment points along ŷ ; (iii) The moment
points along ẑ .
(2) A toroid of linear magnetic material with permeability µ is configured as in fig. 5.38 of the
text, with a rectangular cross-section, inner radius a, outer radius b, and thickness d. The total
number of windings is N, very finely spaced so that we consider the cylindrical symmetry to be
preserved.
(a) Find H inside and outside the toroid, using the integral version of Ampere’s law for H.
(b) From your result for (a), determine B and M in the toroid. By what factor is B increased in
this case relative to the situation of example 5.10?
(c) Find the bound volume and surface current densities in the volume of the toroid and on all
the surfaces.
(3) Griffiths problem 6.10 (page 266).
(4) A thin disk magnet has thickness d and radius R, and has uniform magnetization M, pointing
along its axis (that is, perpendicular to the circular face of the disk).
(a) Find the distribution of bound currents in this magnet.
(b) Find the magnetic field a distance z along the axis, outside the magnet, assuming d is very
small compared to R.
(c) At z = 0, use the boundary conditions on B to determine the magnetic field in the magnet at
its center, and thereby determine H at this point. (Assume d is small enough that the fields do not
vary appreciably through the thickness of the magnet.)
(5) A very long wire running parallel to the surface of a linear magnetic material may be treated
by the method of images. Assume that the wire carries current I in the x direction, and is a
distance z = d above the surface of a semi-infinite magnetic material, whose surface coincides
with the x-y plane. In this case a parallel image wire below the surface will combine to give a
field at the surface that satisfies the boundary conditions. (Similar to the situation of the image
charge problems, in this case the actual currents are spread out on the surface of the material, but
can be represented by the image current for purposes of solving the problem.) Solve for the
required image current, and for the surface magnetic field vs. y, for the following situations:
(a) a material with µ = 0. (Superconductors have this property.)
(b) a material where µ approaches ! . (For example soft ferromagnetic iron such as used in
transformers can have very large permeability and behave in this way.) Note in case (b), for the
magnetization to remain finite, H must effectively be zero in this case just inside the surface.
(6) (a) A long cylinder of linear dielectric is placed in a uniform electric field Eo perpendicular to
its axis. The cylinder radius is R and permittivity is ε. Write down the general form for the
electric potential inside and out, which will be solutions of the Laplace equation, keeping only
the cos φ terms having the same symmetry as the “source” (applied field). From V, find E by
taking the gradient, and then from E obtain the corresponding form of D. You should have two
unknowns; solve by applying the appropriate boundary conditions. Finally write down the final
form of D inside and out, and make a sketch of the lines of D keeping in mind the boundary
conditions.
(b) Consider the corresponding solution for the magnetic field of a long cylinder of
paramagnetic material, linear permeability µ, placed in a uniform external magnetic field
! Bo!. As
indicated in class, D is the analog of B, and E is the analog of H. Also since we have B = µ H , µ
takes the role of ε in comparing these two cases. Since in part (a) the far-field D is ! o E , you can
replace ! o E " Bo , as well as ! o ! µo , and ! " µ , in your solutions for D. Do this, and show
that the resulting magnetic fields satisfy the perpendicular-component boundary condition at the
surface. This is the solution for the magnetic-field problem; you have a sketch of the field lines
in part (a).
! "
Note: the reason this works is!that" B is also a solution of a scalar potential B = !U as in
equation 5.65 in the text (or B = !" as I had in my slides). The traditional way would be to
solve the Laplace equation directly for U, and use this to find B by taking the gradient. The scalar
magnetic potential is often used in this way to solve practical problems in magnetism.