Solution: Rotation of Ring inside a Charging Cylinder
When the long hollow cylinder of radius \(R\) rotates with angular velocity \(\omega_0\), it behaves like a solenoid. The surface charge density is \(\sigma\).
The surface current per unit length (linear current density) \(K\) is:
$$ K = \sigma v = \sigma (R \omega_0) $$The magnetic field \(B\) inside a long solenoid is given by \(\mu_0 n I\), where \(nI\) corresponds to \(K\). Thus:
$$ B = \mu_0 \sigma R \omega_0 $$As the cylinder spins up from rest to \(\omega_0\), the magnetic field changes from \(0\) to \(B\). This changing flux induces an electric field inside. Consider the ring of radius \(r\) placed coaxially. Using Faraday’s Law:
$$ \oint \vec{E} \cdot d\vec{l} = \left| \frac{d\Phi}{dt} \right| $$ $$ E(2\pi r) = \pi r^2 \left| \frac{dB}{dt} \right| $$ $$ E = \frac{r}{2} \left| \frac{dB}{dt} \right| $$The electric field exerts a tangential force \(F = qE\) on the ring. The resulting torque \(\tau\) is:
$$ \tau = F \cdot r = \left( q \frac{r}{2} \frac{dB}{dt} \right) \cdot r = \frac{q r^2}{2} \frac{dB}{dt} $$Using the angular impulse-momentum equation \(\int \tau dt = \Delta L\):
$$ \int \frac{q r^2}{2} \frac{dB}{dt} dt = I_{ring} \omega_{ring} $$ $$ \frac{q r^2}{2} \int_{0}^{B} dB = (m r^2) \omega_{ring} $$Note that the moment of inertia of the ring is \(I_{ring} = m r^2\).
Substituting the total change in magnetic field \(\Delta B = \mu_0 \sigma R \omega_0\):
$$ \frac{q r^2}{2} (\mu_0 \sigma R \omega_0) = m r^2 \omega_{ring} $$Canceling \(r^2\) from both sides:
Direction: According to Lenz’s Law, the induced torque acts to oppose the change in magnetic flux. Therefore, the ring acquires an angular velocity opposite to the angular velocity of the cylinder.