Question

A magnetic field B is confined to a region r $\leq$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 $\Delta$t. The angular velocity $\omega$ of the ring after the field vanishes, is

• A. $\frac{qBa^{2}}{2mb}$
• B. $\frac{qBa}{2mb}$
• C. $\frac{2mb^{2}}{qba^{2}}$
• D. $\frac{qBa^{2}}{2mb^{2}}$

Answer 1

Answer: (D)

$\frac{qBa^{2}}{2mb^{2}}$

Answer 2

Let E is the electric field generated around the charged ring of radius b, then

$\varepsilon\frac{d\varphi}{dt}$

$\oint_{}^{}{\overset{\rightarrow}{E}.d\overset{\rightarrow}{l}} = \frac{B\pi a^{2}}{\Delta t}$

Or $Eb = \frac{Ba^{2}}{2(\Delta t)}$ ……. (i)

Torque acting on the ring

$\tau = b \times force = bqE$

$= \frac{qBa^{2}}{2(\Delta t)}$ [Using (i)]

If $\Delta L$ is change in angular momentum of the charged ring then

$\tau = \frac{\Delta L}{\Delta t} = \frac{L_{2} - L_{1}}{\Delta t}$

$\therefore L_{2} - L_{1} = \tau(\Delta t)$

$= \frac{qBa^{2}\Delta t}{2\Delta t} = \frac{qBa^{2}}{2}$

As initial angular momentum $L_{1} = 0$

$\therefore L_{2} = \frac{qBa^{2}}{2} = I\omega = mb^{2}\omega\therefore\omega = \frac{qBa^{2}}{2mb^{2}}$