The magnetic field generated by the wire will be azimuthal with its direction determined by the right hand rule. From the Lorentz force law,

$\begin{align*}\mathbf{F} = q \mathbf{v} \times \mathbf{B}\end{align*}$
it follows that there will be no magnetic force on the loop because $\mathbf{v}$ and $\mathbf{B}$ are parallel. Therefore, answer (E) is correct.

Solution to 2008 Problem 4