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

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

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

Magnitude of external force that acts perpendicularly at the end of the rod to maintain the constant angular speed is $\frac{3 B^{2} ω ℓ^{4}}{8 R}$

answer is A, B, C, D.

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Detailed Solution

Resistors R and 2R are in parallel.

$\begin{array}{l} C u r r e n t t h r o u g h r o d, i = \frac{ε}{\frac{2 R}{3}} = \frac{3 ε}{2 R} \\ = \frac{3}{2 R} \times \frac{1}{2} B ω ℓ^{2} = \frac{3 B ω ℓ^{2}}{4 R} \\ Magnetic force F = iℓB = \frac{3 B ω ℓ^{2}}{4 R} \times ℓ \times B \\ = \frac{3 B^{2} ω ℓ^{3}}{4 R} \\ T o r q u e, τ = F \frac{ℓ}{2} = \frac{3 B^{2} ω ℓ^{3}}{4 R} \times \frac{ℓ}{2} = \frac{3 B^{2} ω ℓ^{4}}{8 R} \\ ∴ Force to be applied at the end = \frac{τ}{ℓ} = \frac{3 B^{2} ω ℓ^{3}}{8 R} . \end{array}$