Question

A non-conducting insulating ring of mass $m$ and radius $R$, having charge $Q$ uniformly distributed over its circumference is hanging by a insulated thread with the help of a small smooth ring (not rigidly fixed with bigger ring). A time varying magnetic field $B=B_0\sin \omega t$ is switched on at t $=0$ and the ring is released at the same time.The average magnetic moment of Ring in time interval $0$ to $\frac{\pi }{\omega }$ is

• A. $\frac{B_0{q}^2{R}^2}{2\pi m}$
• B. $\frac{B_0{q}^2{R}^2}{4\pi m}$
• C. $\frac{B_0{q}^2{R}^2}{\pi m}$
• D. zero

Answer 1

The correct option is C $\frac{B_0{q}^2{R}^2}{\pi m}$

C. $\frac{B_0{q}^2{R}^2}{\pi m}$

The varying magnetic field,

$B=B_{0}sinwt$. . . . . . . . . . . . . .(1)

The radius of the ring,

$R=\frac{mv}{qB}$. . . . . . . . . . . . . . . . . . . (2)

velocity, $v=\frac{qBR}{m}$. . . . . . . . . . . . (3)

Angular momentum of the circular ring,

$L=mvR$. . . . . . . ... . . . . . . . . . . . (4)

The magnetic moment per unit angular momentum is,

$\frac{\mu }{L}=\frac{q}{2m}$

$\mu =\frac{qL}{2m}$

$\mu =\frac{qmvR}{2m}$. . . . . . . . . . . . . . . . . . .(5)

Put equation (1) , (2) and (3) in equation (5), we get,

$\mu =\frac{q^2{B}_0{R}^2}{2m}sinwt$

The average magnetic moment of ring in time interval $0$ and $\frac{\pi }{w}$,

$<\mu >=\frac{q^2{B}_0{R}^2}{2m}<sinwt>$

$<\mu >=\frac{q^2{B}_0{R}^2}{2m}\frac{1}{T}{\int }_0^{\pi /w}sinwtdt$

$<\mu >=\frac{q^2{B}_0{R}^2}{\pi m}$

Thus, the correct option is C.