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A metallic rod of mass per unit length 0.5 kg m⁻¹ 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
- 7.14 A
- 5.98 A
- 11.32 A
- 14.76 A
Correct answer: 7.14 A
Solution
The force due to the magnetic field (F = BIL) must balance the component of gravitational force along the incline (mg sin θ). Using m/L = 0.5 kg/m, B = 0.25 T, and θ = 30°, we solve for I: I = (m/L)g sin θ / B = (0.5)(9.8)(0.5) / 0.25 = 7.14 A.
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