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A thin uniform rod of negligible mass and length $l$ is attached to the floor by a hinge $P$ . The other end is connected to a spring of force constant $k$ . Rod is in a uniform magnetic field $B$ pointing inwards the plane of paper. A current $I$ is passed through the rod. Find the torque acting on the rod due to magnetic force when the rod makes an angle $53 °$ as shown in Fig.

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$I l^{2} B$

$\frac{I l^{2}}{2} B$

$\frac{3 I l^{2} B}{5}$

$\frac{4}{5} I l^{2} B$

answer is B.

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Detailed Solution

Magnetic force acting at the centre of the rod is F = IlB abd direction of this force is perpendicular to the rod.

Therefore magnetic moment of this force about the hinge $τ$ = F $\times \frac{l}{2} = \frac{1}{2} I B l^{2}$

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