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C:\User\Teaching\505506\F10\HW11sol.doc
PHY 505
Classical Electrodynamics I
Fall 2010
Homework 11
(due Friday Dec. 3)
Problem 11.1
(to be graded of 15 points). A small, planar loop made of a thin wire, carrying current
I
, is
located far from a plane surface of a bulk superconductor. Within the idealdiamagnetic description of
superconductivity (
B
= 0), find:
(i) the energy of the loopsuperconductor interaction,
(ii) the force and torque acting on the loop,
(iii) the distribution of supercurrent on the superconductor’s surface.
Solution
: In the idealdiamagnetic approximation, all Cartesian components of the magnetic field
inside
a superconductor should be zero, including the component
B
n
normal to its surface. On the other hand,
according to Eq. (5.134) of the lecture notes, the normal component has to be continuous at any interface
between two media. Hence, so that as we approach the superconductor surface from
outside
, the field
has to satisfy the boundary condition
B
n
=
0
.
(
*
)
The field of a wire loop in free space, at distances much larger than the loop size, may be
described in the magnetic dipole approximation. An elementary generalization of Eq. (5.99) of the
lecture notes yields
5
2
0
3
4
)
(
'
'
'
'
'
r

r
r

r
m
m
r

r
r

r
r
B
,
where
r
’
is the position of the loop center, and
m
’
its magnetic moment (with
m’
=
IA
, where
A
is the
loop area). This expression shows that a simultaneous flipping of
signs of a certain Cartesian component of vectors
m
and (
r
–
r
’
)
reverses the sign of the same component of the field, but does not
affect its other components. From this symmetry, it is clear that
boundary condition (*) may be satisfied by presenting the total
magnetic field outside the superconductor as a sum of
B
’
and a
similar (freespace) field
B
” of a mirrorimage magnetic dipole
m
”
, which is located and oriented as shown in Figure on the right.
Selecting the coordinate axis as shown in the Figure (with axis
x
in
one vertical plane with vector
m
’
), we may write
,
,
0
,
0
d
z'
z''
y'
y''
x'
x''
cos
,
0
,
sin
m
m
m
m
m
m
m
m
'
z
''
z
'
y
''
y
'
x
''
x
.
Notice that the image dipole orientation is opposite to that