Question

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

• 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

The correct option is A $\frac{mg}{\pi R{B}_0}$

Let ${}^{\prime }{O}^{\prime }$ be the point of contact of string and ring and $i$ be the current in the ring in anti clockwise direction.

It is given that, in the presence of magnetic field ring will be in rotational equilibrium in horizontal position.

Therefore, Torque due to weight of the ring about ${}^{\prime }{O}^{\prime }$ is equal to torque acting on current carrying ring due to magnetic field.

$\Rightarrow mgR=MB\sin {90}^{\circ }$

$\Rightarrow mgR=NiA{B}_0$

Thus, $mgR=i\left(\pi R^2\right)B_0$

$\Rightarrow i=\frac{mg}{\pi R{B}_0}$

Hence, option (a) is the correct answer.