## Solution to 1992 Problem 55

 The magnetic field a distance $r$ from an infinite current loop carrying current $I$ is\begin{align*}\frac{\mu_0 I}{2 \pi r}\end{align*}This follows immediately from Ampere's Law. Therefore, the magnetic field at the leftmost wire will be\begin{align*}\frac{\mu_0 I}{2 \pi r}\end{align*}The force on the leftmost wire will therefore be\begin{align*}\frac{\mu_0 I i b}{2 \pi r}\end{align*}As we found in Problem 54, this force is directed to the left. The magnetic field at the rightmost wire will be\begin{align*}\frac{\mu_0 I i b}{2 \pi (r + a)}\end{align*}As we found in Problem 54, this force is directed to the right. The forces on the bottom and the top of the wire will cancel, because the magnetic fields are the same, but the currents are in opposite directions. Therefore, the net force on the wire is\begin{align*}\frac{\mu_0 I i b}{2 \pi r} - \frac{\mu_0 I i b}{2 \pi (r + a)} = \boxed{\frac{\mu_0 I i a b }{r (r+a)}}\end{al...Therefore, answer (D) is correct.