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 $B=1$ Tesla and $\omega =0.125rad/s$. The work done for reversing the direction of the magnetic moment at $t=1$ second, is__?

• A. $0.007J$
• B. $0.014J$
• C. $0.0198J$
• D. $0.028J$

Answer 1

The correct option is C $0.0198J$

Given: $\overrightarrow{M}={10}^{-2}$ $\hat{i}$ $A{m}^2$

$\overrightarrow{B}=B\hat{i}cos(\omega \ast t)$

$B=1T$

$\omega =0.125rad/s$

To find: $work$ $done$ $for$ $reversing$ $di{r}^n$

$of$ $magnetic$ $moment$ at $t=1second$ $=?$

Solution: Let required work done be W.

as we know that,

$W=-||\overrightarrow{M}||\ast ||\overrightarrow{B}||(cos{\theta }_2-cos{\theta }_1)$ $......\left(1\right)$

now ,${\theta }_2={180}^{\circ }$

${\theta }_1=0^{\circ }$

$||\overrightarrow{M}||={10}^{-2}$

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

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

$W=-{10}^{-2}\ast cos\left(0.125\right)\ast (-1-1)$

$W=2\ast {10}^{-2}\ast 0.99$

$W=0.0198$ $J$

hence,

The correct option is not present.