Question

Charge $Q$ is uniformly distributed on a thin insulating ring of mass $m$ which is initially at rest. To what angular velocity will the ring be accelerated when a magnetic field $B$, perpendicular to the plane of the ring, is switched on?

• A. $\frac{QB}{2m}$
• B. $\frac{3QB}{2m}$
• C. $\frac{QB}{m}$
• D. $\frac{QB}{4m}$
• E. answer required

Answer 1

The correct option is A $\frac{QB}{2m}$

Let r be the radius of ring.

${\int }^{}E.dl=E\times 2\times \pi r$

${\int }^{}\frac{B}{dt}\times dA=\left|\frac{B}{t}\right|\times \pi r^2$

Equating E and B:

$E=\frac{dB}{dt}\times r/2$

Now, $Q\times E=m\times r\times a$

where m is mass of ring and a is angular acceleration.

So, $a=(Q\times \frac{dB}{dt})/2m$

Let angular velocity be $w$, so

$w={\int }^{}a.dt$

=> $w=Q/2m\times {\int }^{}\left(dB/dt\right).dt$

=> $w=QB/2m$