Question

A coil of $50$ turns is placed in a magnetic field of magnitude $B=0.25\text{Weber}$ as shown in figure. A current of $2\text{A}$ is flowing in the coil. The torque acting on the coil will be:

• A. $0.15\text{N}$
• B. $0.30\text{N}$
• C. $0.45\text{N}$
• D. $0.60\text{N}$

Answer 1

The correct option is B $0.30\text{N}$

Using Right-hand thumb rule the direction of area vector $\overrightarrow{A}$ is perpendicular to the magnetic field.

The magnetic moment $\overrightarrow{M}$ is perpendicular to $\overrightarrow{B}$

$\Rightarrow \theta ={90}^o$

Torque on the coil will be,

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

or, $\tau =MB\sin \theta$

$\tau =\left(niA\right)B\sin {90}^o$

$\tau =[50\times 2\times (12\times 10\times {10}^{-2}\left)\right]\times 0.25\times 1$

$\therefore \tau =0.3\text{N}$

Hence, option $\left(B\right)$ is the correct answer.