A metallic rod of mass per unit length 0.5 kg m⁻¹ lies horizontally on a smooth inclined plane making an angle of 30° with the horizontal. A vertical magnetic field of induction 0.25 T acts on the rod. What current is required in the rod to keep it stationary?
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Question

A metallic rod of mass per unit length 0.5 kg m⁻¹ lies horizontally on a smooth inclined plane making an angle of 30° with the horizontal. A vertical magnetic field of induction 0.25 T acts on the rod. What current is required in the rod to keep it stationary?

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G1 Physics · Accepted Answer

Step 1: For a current-carrying rod on the smooth incline, balance the component of its weight down the slope with the magnetic-force component up the slope. Step 2: Use F(B) = ILB; with B vertical and slope angle θ, the force component along incline is ILB cosθ while gravity component is (λL)g sinθ. Step 3: Equilibrium gives ILBcosθ=λLg sinθ ⇒ I=λg tanθ/B. With λ=0.5kg/m, θ=30°, B=0.25T, I≈(0.5×9.8×0.577)/0.25=11.32A. Correct D. Final answer: (D) 11.32 A.