Question

A magnet of total magnetic moment ${10}^{-2}\hat{i}A-m^2$ is placed in a time varying magnetic field, $B\hat{i}\left(cos\omega t\right)$ where$B=1Tesla$ and $\omega =0.125rad/s$ . The work done for reversing the direction of the magnetic moment at $t=1s$, is

• A. $0.007J$
• B. $0.02J$
• C. $0.01J$
• D. $0.028J$

Answer 1

The correct option is B

$0.02J$

Step 1: Given data

Magnetic dipole moment, $\mu ={10}^{-2}\hat{i}A-m^2$

A time varying magnetic field, $B=B\hat{i}\left(cos\omega t\right)$,where$B=1Tesla$ and $\omega =0.125rad/s$

Step 2: Find work done for reversing the direction of the magnetic moment at $\mathit{t}\mathbf{=}\mathbf{1}\mathbf{}\mathit{s}$

Work done for reversing the direction of the magnetic moment, $W=\left(∆\mu \right).B$

$$\begin{gathered} =2\mu .B\left[As,\Rightarrow ∆\mu =\mu -(-\mu )=2\mu \right] \\ =2\times {10}^{-2}\hat{i}.\left[1\times \cos (0.125\times 1)\right]\hat{i} \\ =0.02J \end{gathered}$$

Hence, option (B) is correct.