(a) $B$ is along the z-axis and $A$ is normal to the $x-z$ plane.
$\vec{A}=-50 \times 10^{-4} \hat{j}$
$\vec{B}=0.3 \hat{k}$
$\vec{\tau}=12 \times\left(-50 \times 10^{-4}\right) \hat{j} \times 0.3 \hat{k}$
$=-1.8 \times 10^{-2} \hat{i} N m$
The net force is zero and the torque is $1.8 \times 10^{-2} \mathrm{~N} \mathrm{~m}$ along the negative $x$-direction.
(b) Torque, $\vec{\tau}=I \vec{A} \times \vec{B}$
It is clear from the diagram that is perpendicular to the coil. Because the coil forms a $-$ $30^{\circ}$ angle with the y axis, $A$ forms a $30^{\circ}$ angle with the positive x-axis in the negative y-direction, and $B$ is oriented along the z-axis. The angle formed by $A$ and B is $\theta=90^{\circ} .$. Therefore,
Magnitude of torque will be,
$\mathrm{T}=12 \times 50 \times 10^{-4} \times 0.3$
$=1.8 \times 10^{-2} \mathrm{~N} \mathrm{~m}$
The torque direction is $\left(90^{\circ}+30^{\circ}\right)$ from the negative x-axis or $360^{\circ}-120^{\circ}=240^{\circ}$ from the positive x-axis. The loop’s net force is zero.