Question

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

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

Answer 1

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

${\tau }_{mg}$ about the left end (from where string is connected)= $|\overrightarrow{M}\times \overrightarrow{B}|=MBsin{90}^o$
or $\left(mgR\right)=\left(NiA\right)B_o=i\left(\pi R^2\right)B_o$
or $i=\frac{mg}{\pi R{B}_o}$