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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a
Current flowing through the rod is 3Bωℓ24R
b
Magnetic force acting on the rod is 3B2ωℓ24R
c
Torque due to magnetic force acting on the rod is 3B2ωl48R
d
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.