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
$\tau_{\text {mag }}= MB _{0}= I \pi R ^{2} B _{0} \ldots($ i)
Torque due to weight about the point where string is connected
$\tau_{\text {weight }}= mgR$...(ii)
If ring remains horizontal, then $\tau_{\text {mag }}=\tau_{\text {weight }}$
$I \pi R ^{2} B _{0}= mgR$
$\Rightarrow I =\frac{ mg }{\pi RB _{0}}$