Question

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

• A. $\frac{\text{mg}}{\pi {\text{RB}}_0}$
• B. $\frac{\text{mg}}{{\text{RB}}_0}$
• C. $\frac{\text{mg}}{3\pi {\text{RB}}_0}$
• D. $\frac{mg\ell }{\pi {\text{R}}^2{\text{B}}_0}$

Answer 1

Answer: (A)

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