Solution
Physics Principle:
As the charged ring enters the magnetic field, the changing magnetic flux through the ring induces an electric field along the ring. This electric field exerts a torque on the charged ring, setting it into rotation. Since the table is frictionless, translational and rotational motions are linked via conservation laws involving the impulse provided by the field.
1. Angular Impulse and Momentum:
The torque is $\tau = \frac{q}{2\pi} (-\frac{d\Phi}{dt})$. The total angular impulse is:
$$ \Delta L = \int \tau dt = \frac{q}{2\pi} \Delta \Phi $$Total flux change $\Delta \Phi = B \pi r^2$.
$$ \Delta L = \frac{q}{2\pi} (\pi r^2 B) = \frac{1}{2} q B r^2 $$This equals the final angular momentum $I\omega$. For a ring, $I = mr^2$.
$$ \frac{1}{2} q B r^2 = m r^2 \omega \quad \text{…(1)} $$2. Energy/Velocity Relation:
The problem states the ring starts rolling ($v = \omega r$). The work done by the induced field converts translational kinetic energy into rotational kinetic energy. Using energy conservation for the transition:
$$ \frac{1}{2} m v_0^2 = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 $$Substituting $v = \omega r$ and $I = mr^2$:
$$ \frac{1}{2} m v_0^2 = \frac{1}{2} m v^2 + \frac{1}{2} (mr^2) \left(\frac{v}{r}\right)^2 = m v^2 $$ $$ v = \frac{v_0}{\sqrt{2}} $$So the final angular velocity is $\omega = \frac{v}{r} = \frac{v_0}{r\sqrt{2}}$.
3. Solving for Charge q:
Substitute $\omega$ back into equation (1):
$$ \frac{1}{2} q B r^2 = m r^2 \left( \frac{v_0}{r\sqrt{2}} \right) $$ $$ \frac{1}{2} q B = \frac{m v_0}{r\sqrt{2}} $$ $$ q = \frac{2 m v_0}{\sqrt{2} B r} $$