Question

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

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

Answer 1

Answer: (A)

$\frac{mg}{\pi R{B}_0}$

Torque due to the magnetic field ${\overrightarrow{\tau }}_{mag}=\overrightarrow{\mu }\times \overrightarrow{B}$
Let the wire loop be in xy plane.
Let $\overrightarrow{\mu }=\mu \hat{k}$
$\overrightarrow{B}=B_0(-\hat{i})$
$\therefore \overrightarrow{\tau }=\mu \hat{k}\times B(-\hat{i})=\mu B(-\hat{j})=IA{B}_0(-\hat{j})$
Torque due to the weight mg is ${\tau }_{mg}=mgR$ where R is the radius of the circle.
If the loop does not tilt both the torques need to balance each other.
$\therefore IA{B}_0=mgR$

$\Rightarrow I=\frac{mgR}{A{B}_0}=\frac{mgR}{\pi R^2{B}_0}=\frac{mg}{\pi R{B}_0}$