Question

A circular wire loop of radius $R$, mass $m$, and current $I$ lies on a rough surface. There is a horizontal magnetic field $\overrightarrow{B}$. The minimum value of current $I$ for which one edge of the loop will lift off the surface ?

• A. $\frac{4mg}{\pi RB}$
• B. $\frac{3mg}{\pi RB}$
• C. $\frac{2mg}{\pi RB}$
• D. $\frac{mg}{\pi RB}$

Answer 1

The correct option is D $\frac{mg}{\pi RB}$

The loop will start to lift when the magnetic torque equals the gravitational torque.

$\therefore {\tau }_m={\tau }_g$

The magnetic torque acting on the loop is

${\overrightarrow{\tau }}_m=\overrightarrow{\mu }\times \overrightarrow{B}$

$\therefore {\tau }_m=\mu B\sin {90}^{\circ }=\left(I\pi R^2\right)B$

The gravitational torque exerted on the loop is given by

${\overrightarrow{\tau }}_g=\overrightarrow{R}\times m\overrightarrow{g}$

$\therefore {\tau }_g=mgR$

Torque of magnetic force acts in such a way that $\overrightarrow{\mu }$ vector tries to align with the magnetic field vector $\overrightarrow{B}$.


After equating the torques, we get

$I=\frac{mg}{\pi RB}$

Hence, option $\left(d\right)$ is the correct answer.