Question

A uniform current carrying ring of mass $m$ and radius $R$ is connected by a massless string as shown in the figure. A uniform magnetic field $B_0$ exists in the region to keep the ring in horizontal position, then the current in the ring is ($l=$ length of string)

• A. $\frac{mg}{\pi R{B}_0}$
• B. $\frac{mg}{R{B}_0}$
• C. $\frac{mg}{\pi R{B}_0}$
• D. $\frac{mgl}{\pi R^2{B}_0}$

Answer 1

The correct option is C $\frac{mg}{\pi R{B}_0}$

Torque due to magnetic field ${\tau }_{mag}=M{B}_0=I\pi R^2{B}_0$
Torque due to weight about the point where string is connected
${\tau }_{weight}=mgR$
If ring remains horizontal , then ${\tau }_{mag}={\tau }_{weight}$
$I\pi R^2{B}_0=mgR$
$I=\frac{mg}{\pi R{B}_0}$