Question

A current carrying loop having magnetic moment $6\text{J/T}$ is placed in a uniform magnetic field of field strength $5\text{T}$ at its most unstable position. The work done in turning the loop to align at its most stable position is $(-x)\text{J}$. Find $x$. (Give integer value)

Answer 1

Given,

$\mu =6\text{J/T}$ and $B=5\text{T}$

The work done required will be equal to the change in potential energy,

$\therefore \Delta W=U_f-U_i$

Where,

The coil is at most unstable position when magnetic moment vector $\overrightarrow{\mu }$, is opposite to magnetic field vector $\overrightarrow{B}$.i.e., $\theta ={180}^{\circ }$

Thus, potential energy, $U_i$ at its most unstable position,

$U_i=-B\mu \cos {180}^o=\mu B$

The coil is at most stable position when magnetic moment vector $\overrightarrow{\mu }$, is parallel to magnetic field vector $\overrightarrow{B}$.i.e., $\theta =0^{\circ }$

Thus, potential energy, $U_f$ at its most stable position,

$U_f=-B\mu \cos 0^o=-\mu B$

$\Rightarrow W=-\mu B-\left(\mu B\right)$

$\Rightarrow W=-2\mu B$

$\therefore W=-2\times 6\times 5=-60J$

Negative work done means the magnetic moment is rotated towards the alignment with the field.

Accepted answer : $60$