Question

A rectangular loop $\text{PQRS}$ made of a uniform wire ; has a length $a$ , width $b$ and mass $m$. It is free to rotate about arm $\text{PQ}$ which remains hinged along a horizontal line taken as $\text{Y-axis}$. Taking the $+ve$ $\text{Z-axis}$ as vertically upward direction. A uniform magnetic field $B=(3\hat{i}+4\hat{k})B_0$ exists in the region. The loop is held in $\text{XY-plane}$ and a current $I$ is passed through it. The loop is now released and found to stay in the horizontal position in equilibrium.
($B_0$ is a constant)

• A. $\frac{mg}{6b{B}_0}$
• B. $\frac{2mg}{3b{B}_0}$
• C. $\frac{4mg}{b{B}_0}$
• D. $\frac{mg}{3b{B}_0}$

Answer 1

The correct option is A $\frac{mg}{6b{B}_0}$

Let the direction of current in wire $\text{PQ}$ is from $\text{P}$ to $\text{Q}$ i.e in clockwise direction and magnitude of current be $I$.


The magnetic moment of the given loop is, $\overrightarrow{\mu }=ab\left(\widehat{-k}\right)$

$\because \overrightarrow{\mu }=I\overrightarrow{A}$ & $|\overrightarrow{A}|=ab$

Torque on the loop due to magnetic force is given by,

$\overrightarrow{{\tau }_m}=\overrightarrow{\mu }\times \overrightarrow{B}$

$\Rightarrow \overrightarrow{{\tau }_m}=(-Iab\hat{k})\times (3\hat{i}+4\hat{k})B_0$

$\Rightarrow \overrightarrow{{\tau }_m}=-3Iab{B}_0\hat{j}$

Torque of gravity on loop about axis $\text{PQ}$ is,

$\overrightarrow{{\tau }_{mg}}=\overrightarrow{r}\times \overrightarrow{F}$

$\Rightarrow \overrightarrow{{\tau }_{mg}}=\left(\frac{a}{2}\right)\hat{i}\times (-mg\hat{k})$

$\Rightarrow \overrightarrow{{\tau }_{mg}}=\frac{mga}{2}\hat{j}$

We can see that torque due to magnetic force and weight are in opposite direction.

Thus, for rotational equilibrium of loop,

${\overrightarrow{\tau }}_{mg}+{\overrightarrow{\tau }}_m=\overrightarrow{0}$

$\Rightarrow \frac{mga}{2}\hat{j}=3Iab{B}_0\hat{j}$

$\Rightarrow \frac{mga}{2}=3Iab{B}_0$

$\therefore I=\frac{mg}{6b{B}_0}$

Hence, option (a) is the correct answer.