Question

Given a ring of mass $M$, radius $r$ resting on a rough horizontal plane having coefficient of friction $\mu$. A cylindrical vertical magnetic field passing through ring is switched on at $t=0$ which varies with time as $B=K{t}^2$. If the ring has charge $Q$ distributed uniformly over it. Find the time after which ring starts rotating.

• A. $\frac{\mu Mg}{QKr}$
• B. $\frac{2\mu Mg}{QKr}$
• C. $\frac{\mu Mg}{2QKr}$
• D. $\frac{\mu Mg}{4QKr}$

Answer 1

The correct option is A $\frac{\mu Mg}{QKr}$

Given $B=K{t}^2$
let $E$ be induced electric field due to change in magnetic flux
$\int \overrightarrow{E}.\overrightarrow{dl}=\frac{d\varphi }{dt}=\frac{AdB}{dt}$
$\Rightarrow E.2\pi r=\pi r^2.2Kt$
$\Rightarrow E=Krt$

As $E$ is tangential in nature
Electric force $F=QE=QKrt$(tangential direction)

Ring starts moving when torque due to electric field becomes greater than torque due to limiting friction
${\tau }_E={\tau }_f$
$QEr=\mu Mgr$
$\Rightarrow Q.Krt.r=\mu Mgr$
$\Rightarrow t=\frac{\mu Mg}{QKr}$