NEET 2018 · PhysicsForce on wirePrevious Year Question
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:
Answer: (D) 11.32 A. Answer: (D) Solution: B is vertical and current is horizontal (along the rod), so the magnetic force F = BIl is horizontal.
- A.14.76 A
- B.5.98 A
- C.7.14 A
- D.11.32 A✓
Correct Answer
(D) 11.32 A
Solution & Explanation
Answer: (D) Solution: B is vertical and current is horizontal (along the rod), so the magnetic force F = BIl is horizontal. For equilibrium on the smooth incline, resolve along the slope: mg sin30° = (BIl) cos30°. ⟹ I = (m/l)·g·tan30°/B = (0.5 × 9.8 × tan30°)/0.25. I = (4.9 × 0.5774)/0.25 ≈ 11.32 A. → option D.
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