Question

A rectangular coil of length 0.12 m and width 0.1 m 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$^{\circ}$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be

• A. 0.24 Nm
• B. 0.12Nm
• C. 0.15Nm
• D. 0.20 Nm

Answer 1

Answer: (D)

0.20 Nm

Answer 2

The required torque is $t = NIAB \sin \theta;$
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 $?$ is the angle between the direction of the magnetic field and normal to the plane of the coil.
Here, $N = 50, I =2A , A = 0.12 \times 0.1 m = 0.012 m^2$
$B = 0.2 Wb / M^2 and \, ? = 90^\circ - 30^\circ = 60^\circ$
$\therefore t = (50) (2A) (0.012 M^2) (0.2 W b / m^2) sin60^\circ$
= 0.20 Nm