As shown in the figure, a metallic rod of linear density 0.45 kg m⁻¹ is lying horizontally on a smooth inclined plane which makes an angle of 45° with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be :
Use g=10 m / s²

Solution

Physics sir · 4 steps

The magnetic push is horizontal, so match its component along the slope with gravity's component.

1. Direction of force. The rod is horizontal and carries current I along its length, and B⃗ is vertical, so F⃗ = Il⃗ × B⃗ is horizontal. It has magnitude F = IlB and points into the incline.
2. Balance along slope. The plane is smooth, so only the components along the incline must cancel. Gravity pulls down the slope with mg sin 45°. The horizontal magnetic force pushes up the slope with IlB cos 45°. The normal reaction takes care of the perpendicular direction.

   IlB cos 45° = mg sin 45° ⇒ I = ((m/l) g tan 45°)/B

3. Put in numbers. I = (0.45 × 10 × 1)/0.15 = 30 A. The length l cancels because only the mass per unit length is given.
4. So. The current needed is 30 A, which is option (a).

The idea to keep. On a smooth incline, resolve every force along the slope and ignore the normal direction. A horizontal force F holds a weight mg on an incline of angle θ when F = mg tan θ.

Answer from the NTA final answer key. Solution by PrepWiser.