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Q.

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.
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