A rectangular coil of length 0.12 m and width 0.1m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Weber/$m^2$.  The coil carries a current of 2  A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be

The required torque is 

$\tau = NIAB \sin \theta$ 
where N is the number of turns in the coil, I is the current through the coil, B is the uniform magnetic field, A is the area of the coil and $\theta$ is the angle between the direction of the magnetic field and normal to the plane of the coil. Here, N = 50, I = 2 A, A = 0.12 m ×  0.1 m= 0.012 $m^2$
B = 0.2 Wb /$m^2$ and  $\theta$ = 90° - 30° = 60° 
$\therefore \tau =\left(50\right)(2 \mathrm{A})\left(0.012{ \mathrm{m}}^2\right)\left(0.2 \mathrm{Wb}/{\mathrm{m}}^2\right)\sin 60°$

=0.20 Nm