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 $0.2Weber/m^2$. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of ${30}^o$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be

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

Answer 1

The correct option is C 0.20 Nm

The torque on a loop in uniform magnetic field is given by

$\tau =\overrightarrow{M}\times \overrightarrow{B}$

$\tau =MB\sin \theta$ [where $\theta$ is angle between normal and the plane of loop]

As the coil is inclined at an angle of ${30}^o$ with magnetic field so area vector will be inclined at an angle of ${60}^o$ with the field.

$M=NI\times \pi r^2=50\times 2\times .12\times .1=1.2$

$B=.2T$

$\tau =1.2\times .2\times \sin {60}^{\circ }$

$\tau =0.2Nm$