Question

A uniform current carrying ring of mass m and radius R is connected to a massless string as shown. A uniform magnetic field $B_0$ exists in the region to keep the ring in equilibrium in horizontal position. The current in the ring is :
(Take acceleration due to gravity g)

• A. $\frac{mg}{\pi R{B}_0}$ clockwise as seen from above the ring
• B. $\frac{mg}{\pi R{B}_0}$ anticlockwise as seen from above the ring
• C. $\frac{mg}{2\pi R{B}_0}$ clockwise as seen from above the ring
• D. $\frac{mg}{2\pi R{B}_0}$ anticlociwise as seen from above the ring

Answer 1

Answer: (B)

$\frac{mg}{\pi R{B}_0}$ anticlockwise as seen from above the ring

The torque due to gravity balances the torque on coil due to uniform magnetic field.

Taking torque about point of suspension

$mgR=M\times B_{\circ }$

$mgR=i\pi R^2{B}_{\circ }$

$\Rightarrow i=\frac{mg}{\pi R{B}_{\circ }}$