Question

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

• A. $0.014\text{J}$
• B. $0.01\text{J}$
• C. None of the above
• D. $0.007\text{J}$
• E. $0.028\text{J}$

Answer 1

The correct option is C None of the above

Given:
Magnetic moment $\overrightarrow{M}={10}^{-2}\hat{i}{\text{A-m}}^2$
Magnetic field $\overrightarrow{B}=B\hat{i}\left(\cos \omega t\right)$
$B=1\text{Tesla}$
$\omega =0.125\text{rad/s}$
$t=1\text{second}$

Work done,
$W=-|\overrightarrow{M}||\overrightarrow{B}|(\cos {\theta }_2-\cos {\theta }_1)$

$W=-{10}^{-2}\times 1\times \cos (0.125\times 1)\times (\cos {180}^{\circ }-\cos 0^{\circ })$

$W=0.02\text{J}$