Q.A magnetic field $\mathbf{B}$ 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 $\Delta t$. Find the angular velocity $\omega$ of the ring after the field vanishes.

The collapsing flux (confined to $r\le a$) induces an azimuthal electric field at the ring ($r=b$). From Faraday's law round the ring,

This field exerts a tangential force $QE$ on the ring's charge, giving a torque

$\tau = (QE)\,b = \frac{Q a^2 B_0}{2\,\Delta t}.$ …

Induced electric field at the ring

The field $B$ lives only in $r\le a$, but a changing flux induces an electric field everywhere, including at the ring's radius $r=b>a$. By Faraday's law round a circle of radius $b$,

Taking magnitudes as the field falls from $B_0$ to $0$ in $\Delta t$,

Torque on the charged ring

Each charge element feels a tangential force $dF = E\,dq$; its lever arm about the centre is $b$, so

(The $b$ from the lever arm cancels the $b$ in $E$, so $\tau$ does not depend on $b$.)

Angular impulse $\to$ final angular velocity …

Method: Induced Electric Field from a Confined Changing Flux, then Torque via Impulse

Use this when a changing magnetic field is confined to some region, and you need the effect (force, torque) on a charge that sits outside that region — there is no physical wire loop carrying a current here, only an induced electric field acting directly on free charge.

Steps

Step 1: Identify the enclosed flux through an imaginary circular path

Draw an imaginary circle at the radius where the charge sits. The flux through that circle is only whatever passes through the actual field region inside it (not the full circle's own area) — here, $\Phi=B\cdot(\pi a^2)$ since the field only exists for $r\le a$.

Step 2: Apply Faraday's law in integral form to get E

By the problem's circular symmetry, $\vec E$ has the same magnitude everywhere on the circle and points tangentially (azimuthally), so the loop integral collapses to $E\,(2\pi r)$. Solve for $E$.

Step 3: Find the force/torque on the charge

Each bit of charge $dq$ on the ring feels a tangential force $dF=E\,dq$. Since $E$ is the same at every point of the ring (same radius), the total tangential force is $F=EQ$, and the torque about the centre is $\tau=(EQ)\times(\text{radius})$.

$\text{Formula}:\quad \tau = Q E\, b \quad (b=\text{radius of the charged ring})$ …