Question

The figure represents four positions of a current carrying loop in a uniform magnetic field directed towards the right. $\hat{n}$ represents the direction of area vector of the loop. The correct order of potential energy is

• A. $I>II>IV>III$
• B. $I<III<II<IV$
• C. $III>II>IV>I$
• D. $I>III>II>IV$

Answer 1

The correct option is D $I>III>II>IV$

Potential energy of a current carrying loop placed in a uniform magnetic field is given by, $$U=-\overrightarrow{\mu }.\overrightarrow{B}=-\mu B\cos \theta$$
Where, $\theta$ is the angle between $\hat{n}$ and $\overrightarrow{B}$.

$\text{Case I}$ : $\theta ={180}^{\circ }$

$\therefore U_I=-\mu B\cos {180}^{\circ }=\mu B$

$\text{Case II}$ : $\theta ={90}^{\circ }$

$\therefore U_{II}=-\mu B\cos {90}^{\circ }=0$

$\text{Case III}$ : $\theta >{90}^{\circ }$

$\therefore U_{III}>0$ , But $<\mu B$

$\text{Case IV}$ : $\theta <{90}^{\circ }$

$\therefore U_{IV}<0$

So, $U_I>U_{III}>U_{II}>U_{IV}$

Hence, option (d) is the correct answer.