A thin uniform rod with negligible mass and length $l$ is attached to the floor by a frictionless hinge at point $P$ . A horizontal spring with force constant $k$ connects the other end to wall. The rod is in a uniform magnetic field $B$ directed into the plane of paper. What is extension in spring in equilibrium when a current $i$ is passed through the rod in direction shown. Assuming spring to be in natural length initially.
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Torque due to magnetic force
$|\mathrm{d} \vec{\tau}|=\int_{0}^{\mathrm{L}}(\mathrm{i} \mathrm{d} \mathrm{rB}) \mathrm{r}=\frac{\mathrm{i} L^{2} \mathrm{B}}{2}$
In equilibrium $\frac{\mathrm{i} \mathrm{L}^{2} \mathrm{B}}{2}=(\mathrm{kx}) \times \mathrm{L} \sin 30^{\circ}$
or $x=\frac{5 i L B}{8 k}$
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