Question

A magnetic field $B$ is confined to a region $r \le\, a$ and points out of the paper (the $z$-axis), $r = 0$ being the centre of the circular region. A charged ring (charge = $q$) of radius $b(b \,> a)$ and mass $m$ lies in the $x-y$ plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time $\Delta$ The angular velocity $\omega\,t$ of the ring after the field vanishes, is

• A. $\frac{q\, Ba^{2}}{2\,mb}$
• B. $\frac{q\, Ba}{2\,mb^{2}}$
• C. $\frac{2\, mb^{2}}{qBa^{2}$
• D. $\frac{q\,Ba^{2}}{2\,mb^{2}$

Answer 1

Answer: (D)

Let $E$ is the electric field generated around the charged ring of radius $b$, then
$\varepsilon=\frac{d \phi}{dt}$
$\oint\vec{E}\cdot d \vec{l}=\frac{B\pi a^{2}}{\Delta t}$
or $ Eb =\frac{Ba^{2}}{-2\left(\Delta t\right)} \ldots\left(i\right)$
Torque acting on the ring
$\tau=b\times force = bqE$
$=\frac{qBa^{2}}{2\left(\Delta t\right)}$ [Using (i)]
If $\Delta\, L$ is change in angular momentum of the charged ring, then
$\tau=\frac{\Delta L}{\Delta t}=\frac{L_{2}-L_{1}}{\Delta t}$
$\therefore L_{2}-L_{1}=\tau\left(\Delta t\right)$
$=\frac{qBa^{2}\,\Delta t}{2\Delta t}=\frac{q\,Ba^{2}}{2}$
As initial angular momentum, $L_{1}=0$
$\therefore L_{2}=\frac{q\,Ba^{2}}{2}=I\omega=mb^{2}\omega$
$\therefore \omega=\frac{qBa^{2}}{2mb^{2}}$