Question

A current carrying loop of magnetic moment $16\text{J/T}$ is aligned at ${45}^{\circ }$ with a uniform external magnetic field of $4\text{T}$. What would be the change in potential energy in turning the loop to align its magnetic moment ${135}^{\circ }$ to the magnetic field?

• A. $32\sqrt{2}\text{J}$
• B. $-32\sqrt{2}\text{J}$
• C. $64\sqrt{2}\text{J}$
• D. $0$

Answer 1

The correct option is C $64\sqrt{2}\text{J}$

Given,
magnetic moment, $\mu =16\text{J/T}$,

magnetic field, $B=4\text{T}$,

Potential energy of a current carrying loop placed in a uniform magnetic field$$U=-\overrightarrow{\mu }.\overrightarrow{B}=-\mu B\cos \theta$$Thus, change in potential energy

$\Delta U=U_f-U_i=\mu B(\cos {\theta }_1-\cos {\theta }_2)$

As, ${\theta }_1={45}^{\circ }$ and ${\theta }_2={135}^{\circ }$

$\Rightarrow \Delta U=16\times 4(\cos {45}^{\circ }-\cos {135}^{\circ })$

$\therefore \Delta U=64\sqrt{2}\text{J}$

Hence, option (c) is the correct answer.