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 $30^{\circ }$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be

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

Answer 1

Answer: (D)

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 $\theta$ is the angle between the direction of the magnetic field and normal to the plane of the coil.
Here, $N=50,I=2A,A=0.12\times 0.1m=0.012m^2$
$B=0.2Wb/M^2$ and $\theta =90^{\circ }-30^{\circ }=60^{\circ }$
$\therefore t=(50)(2A)\left(0.012M^2\right)\left(0.2Wb/m^2\right)\sin 60^{\circ }$
$=0.20Nm$