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A rod $O A$ of length $l$ is rotating (about end $O$ ) over a conducting ring in crossed magnetic field $B$ with constant angular velocity $ω$ as shown in figure.

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Current flowing through the rod is $\frac{3 B ω l^{2}}{4 R}$

Magnetic force acting on the rod is $\frac{3 B^{2} ω l^{3}}{4 R}$

Torque due to magnetic force acting on the rod is $\frac{3 B^{2} ω l^{4}}{8 R}$

Magnitude of external force that acts perpendicularly at the end of the rod to maintain the constant angular speed is $\frac{3 B^{2} ω l^{3}}{8 R}$

answer is A, B, C, D.

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Detailed Solution

$I = \frac{ε}{\frac{2 R}{3}} = \frac{3 ε}{2 R}$

$= \frac{3}{2 R} \times \frac{1}{2} B ω l^{2} = \frac{3 B ω l^{2}}{4 R}$

Magnetic force $F = \frac{3 B ω l^{2}}{4 R} \times l \times B = \frac{3 B^{2} ω l^{3}}{4 R}$

$τ = \frac{3 B^{2} ω l^{3}}{4 R} \times \frac{l}{2} = \frac{3 B^{2} ω l^{4}}{8 R}$
$∴$ Force to be applied at the end $= \frac{3 B^{2} ω l^{3}}{8 R}$ .

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