# HW11sol - PHY 505 Classical Electrodynamics I Fall 2010...

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1 C:\User\Teaching\505-506\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 ideal-diamagnetic description of superconductivity ( B = 0), find: (i) the energy of the loop-superconductor interaction, (ii) the force and torque acting on the loop, (iii) the distribution of supercurrent on the superconductor’s surface. Solution : In the ideal-diamagnetic 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 (free-space) field B ” of a mirror-image 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 in Homework Problem 6.1, due to the different boundary conditions we had to satisfy. Now, we may recycle the solution of that problem closely, just minding the signs and interaction constants, to get the following results:

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