Topic: Force on current-carrying conductor, force between parallel currents· Magnetic Effects of Current and Magnetism

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 induct…

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 $30^\circ$ 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

1. 7.14 A

2. 14.76 A

3. 5.98 A

4. 11.32 A

Doubtly Asked 30 Oct 2026 NEET UG 2018 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 4. 11.32 A

For the rod to stay at rest on a smooth incline, the horizontal magnetic force must balance the component of weight down the slope.

The field is vertical and the current is horizontal, so the force $BIl$ is horizontal. Along the incline: $BIl\cos\theta = mg\sin\theta$, so $BIl=mg\tan\theta$.

With $m/l=0.5$ kg/m: $$I=\frac{(m/l)\,g\tan30^\circ}{B}=\frac{0.5\times9.8\times0.577}{0.25}\approx11.3\ \mathrm{A}$$

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

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 induct…

1 answer

Your answer