Question

A current carrying uniform square frame is suspended from hinged supports as shown in the figure such that it can freely rotate about its upper side. The length and mass of each side of the frame is $2m$ and $4kg$ respectively. A uniform magnetic field $\overrightarrow{B}=(3\hat{i}+4\hat{j})$ is applied. When the wire frame is rotated to $45^{\circ }$ from vertical and released it remains in equilibrium. If the magnitude of current (in A) in the wire frame is I then find $\left(\frac{3}{5}\right)$ $I.$
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Answer 1

Answer: 6

$\overrightarrow{\mu }$ (Magnetic moment of loop) when it is
lifted by $45^{\circ }=i{\ell }^2\left(\frac{\hat{j}+\hat{k}}{\sqrt{2}}\right)$
$\therefore \overrightarrow{\tau }$ due to magnetic field
$=\overrightarrow{\mu }\times \overrightarrow{B}=\frac{i{\ell }^2}{\sqrt{2}}[(\hat{j}+\hat{k})\times (3\hat{i}+4\hat{j})]$
$\overrightarrow{\tau }$ due to $mg$ (about top edge)
$=4mg\frac{\ell }{2}cos45^{\circ }\hat{i}$
$\therefore$ For equilibrium net torque along X-axis $=0$
$\therefore \frac{4mg\ell }{2\sqrt{2}}=\frac{4i{\ell }^2}{\sqrt{2}}\Rightarrow i=\frac{mg}{2\ell }=10A$