Answer
(a) clockwise
(b) $\frac{2I}{3\pi}$
Work Step by Step
(a) We know that according to the right hand rule, the current in the loop should flow clockwise in the loop to oppose the filed due to the straight wire.
(b) We know that
$B_{straight\space wire}=\frac{\mu_{\circ}I}{3\pi R}$
and $B_{loop}=\frac{\mu_{\circ}I_{loop}}{2R}$
Now $B_{straight\space wire}=B_{loop}$
$\frac{\mu_{\circ}I}{3\pi R}=\frac{\mu_{\circ}I_{loop}}{2R}$
This simplifies to:
$I_{loop}=\frac{\mu_{\circ}I_{loop}}{3\pi R}\times \frac{2R}{\mu_{\circ}}$
$I_{loop}=\frac{2I}{3\pi}$
