Question

A magnet of magnetic moment $\left(50\hat{i}\right){\text{A-m}}^2$ is placed in a uniform magnetic field given by, $\overrightarrow{B}=(0.5\hat{i}+0.3\hat{j})\text{T}$. The torque acting on the magnet is,

• A. $\left(25\hat{k}\right)\text{N - m}$
• B. $\left(15\hat{k}\right)\text{N - m}$
• C. $\left(15\hat{j}\right)\text{N - m}$
• D. $\left(25\hat{j}\right)\text{N - m}$

Answer 1

The correct option is B $\left(15\hat{k}\right)\text{N - m}$

The torque acting on the magnet is given by,

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

$\Rightarrow \overrightarrow{\tau }=\left(50i\right)\times (0.5\hat{i}+0.3\hat{j})$

$\because \hat{i}\times \hat{i}=0\text{and}\hat{i}\times \hat{j}=\hat{k}$

$\therefore \tau =\left(15\hat{k}\right)\text{N - m}$

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