Question

A current carrying loop whose magnetic moment is at ${30}^{\circ }$ with a uniform external magnetic field of $0.16\text{T}$ experiences a torque of magnitude $0.032\text{Nm}$. If the coil is free to rotate, its potential energies when it is in stable and unstable equilibrium are respectively

• A. $-0.064\text{J},-0.064\text{J}$
• B. $+0.064\text{J},-0.064\text{J}$
• C. $+0.064\text{J},+0.064\text{J}$
• D. $-0.064\text{J},+0.064\text{J}$

Answer 1

The correct option is D $-0.064\text{J},+0.064\text{J}$

Torque on a current carrying loop in the presence of magnetic field is given by,

$\tau =\mu B\sin \theta$

Substituting the values, provided in the question

$0.032=\mu \times 0.16\times \sin {30}^{\circ }$

$\Rightarrow \mu =0.4{\text{Am}}^2$

The coil remains at the most stable position when its potential energy is minimum, i.e., when $\overrightarrow{\mu }$ is parallel to $\overrightarrow{B}$ or $\theta =0^{\circ }$.

So, potential energy at this position:

$U_{min}=-\mu B\cos 0^{\circ }=-0.4\times 0.16=-0.064\text{J}$

The coil remains at the most unstable position when its potential energy is maximum, i.e., when $\overrightarrow{\mu }$ is antiparallel to $\overrightarrow{B}$ or $\theta ={180}^{\circ }$.

So, potential energy at this position:

$U_{max}=-\mu B\cos {180}^{\circ }=-0.4\times 0.16\times (-1)=0.064\text{J}$

Hence, option (d) is the correct answer.