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 B₀ 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 $τ_{\text{mag}} = \text{MB}_{0} = \text{I} \pi \text{R}^{2} \text{B}_{0}$ ...(i)
Torque due to weight about the point where string is connected $τ_{\text{weight}} = \text{mgR}$ ...(ii)
If ring remains horizontal, then $τ_{\text{mag}} = τ_{\text{weight}}$
$\text{I} \pi \text{R}^{2} \text{B}_{0} = \text{mgR} ⇒ \text{I} = \frac{\text{mg}}{\pi \text{RB}_{0}}$