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. 0.12 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, $N=50, I=2 A, A=0.12 \times 0.1 m=0.012 m^{2}$
$B=0.2 W b / M^{2}$ and $\theta=90^{\circ}-30^{\circ}=60^{\circ}$
$\therefore t=(50)(2 A)\left(0.012 M^{2}\right)\left(0.2 W b / m^{2}\right) \sin 60^{\circ}$
$=0.20\, Nm$