Question

A circular coil of $25$ turns and radius $6.0\text{cm}$, carrying a current of $10\text{A}$ is suspended vertically in a uniform magnetic field of magnitude $1.2\text{T}$. The field lines run horizontally in the plane of the coil. What is the magnitude of the external balancing torque required to prevent turning of the coil?

• A. $3.59\text{Nm}$
• B. $3.39\text{Nm}$
• C. $3.79\text{Nm}$
• D. $3.19\text{Nm}$

Answer 1

The correct option is B $3.39\text{Nm}$

Using, $\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$,

$\tau =\mu B\sin \theta$

$=niAB\sin \theta$

Here, $\theta ={90}^{\circ }$ as magnetic moment vector is always perpendicular to the magnetic field.

So, $\tau =25\times 10\times (\pi \times {0.06}^2)\times 1.2\times \sin {90}^{\circ }$

$=3.39\text{Nm}$

Hence, the magnitude of the external balancing torque required to prevent turning of the coil is $3.39\text{Nm}$.