Solution: Rotation of Charged Rod in Changing Magnetic Field
The two balls are located at the ends of the rod of length \(l\), so their distance from the center is \(r = l/2\). The magnetic field \(B\) is confined to the region between the cylindrical poles of diameter \(d\). From the diagram, the rod extends beyond the poles (\(l > d\)), meaning the charges are outside the magnetic field region.
To find the induced electric field at the position of the balls, we apply Faraday’s Law to a circular loop of radius \(r = l/2\) passing through the charges. However, the magnetic flux \(\Phi\) only exists within the pole area (radius \(R = d/2\)).
$$ \Phi = B \cdot (\text{Area of Field}) = B \cdot \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2 B}{4} $$Using Faraday’s Law \(\oint \vec{E} \cdot d\vec{l} = \left| \frac{d\Phi}{dt} \right|\):
$$ E(2\pi r) = \frac{d}{dt} \left( \frac{\pi d^2 B}{4} \right) $$Substituting \(r = l/2\):
$$ E \cdot 2\pi (l/2) = \frac{\pi d^2}{4} \left| \frac{dB}{dt} \right| $$ $$ E (\pi l) = \frac{\pi d^2}{4} \left| \frac{dB}{dt} \right| \implies E = \frac{d^2}{4l} \left| \frac{dB}{dt} \right| $$The force on each charge is \(F = qE\). The torque \(\tau\) about the center is:
$$ \tau_{total} = 2 \times (F \cdot r) = 2 \cdot \left( q \frac{d^2}{4l} \left| \frac{dB}{dt} \right| \right) \cdot \frac{l}{2} $$Simplifying the expression:
$$ \tau_{total} = \frac{q d^2}{4} \left| \frac{dB}{dt} \right| $$Using the angular impulse-momentum equation \(\int \tau dt = \Delta L\):
$$ \int \frac{q d^2}{4} \frac{dB}{dt} dt = I \omega $$ $$ \frac{q d^2}{4} \int_{B_0}^{0} dB = I \omega \implies \frac{q d^2 B_0}{4} = I \omega $$The moment of inertia \(I\) for two masses \(m\) at distance \(l/2\) is:
$$ I = 2 \times m \left(\frac{l}{2}\right)^2 = \frac{ml^2}{2} $$Substituting \(I\) back into the impulse equation:
$$ \frac{q d^2 B_0}{4} = \left( \frac{ml^2}{2} \right) \omega $$Solving for \(\omega\):