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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Current flowing through the rod is 3Bωℓ24R

Magnetic force acting on the rod is 3B2ωℓ24R

Torque due to magnetic force acting on the rod is 3B2ωl48R

Magnitude of external force that acts perpendicularly at the end of the rod to maintain the constant angular speed is 3B2ωℓ48R

answer is A.

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Detailed Solution

Resistors R and 2R are in parallel.Current through rod , i=ε2R3=3ε2R=32R×12Bωℓ2=3Bωℓ24R Magnetic force F=iℓB=3Bωℓ24R×ℓ×B=3B2ωℓ34RTorque, τ=Fℓ2=3B2ωℓ34R×ℓ2=3B2ωℓ48R∴ Force to be applied at the end =τℓ=3B2ωℓ38R.

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