c
Magnitude of torque acting on the currentcarrying coil due to the magnetic field.
$\tau=n \operatorname{IAB} \sin \theta$
$=30 \times 6 \times \pi(0.08)^{2} \times 1 \times \sin 60^{\circ}$
$=30 \times 6 \times 3.14 \times 0.08 \times 0.08 \times \frac{\sqrt{3}}{2}=3.133\,\mathrm{Nm}$
