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Question:

A metallic rod of mass per unit length 0.5 kg/m is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is?

14.76A

11.32A

7.14A

5.98A

Solution:

B = 0.25 T
m = 0.5 kg/m
θ = 30°
F = Bil
Fcos30 balances mgsin30
∴ (Bil)cos30° = mgsin30
⇒ i = mgl/(Bsin30cos30) = (0.5 × 9.8)/(0.25 × sin30° × cos30°) = 11.32 A