Question

A rectangular coil of length $0.12m$ and width $0.1m$ 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 $2A$. 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 $Nm$

$l=0.12m,b=0.1m,I=2A,N=50$
The torque on a loop in uniform magnetic field is given by

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

$|\overrightarrow{\tau }|=|\overrightarrow{M}||\overrightarrow{B}|\sin \theta$ [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.

$M=NI\times lb=50\times 2\times 0.12\times 0.1=1.2$

$B=0.2T$

$\tau =1.2\times 0.2\times \sin {60}^{\circ }$

$\tau =0.2Nm$