eMI BYU 16

Solution: Acceleration of a Charged Dielectric Cylinder

As the cylinder rotates with angular acceleration \(\alpha\), its angular velocity is \(\omega = \alpha t\). The rotating surface charge density \(\sigma\) creates a surface current \(K = \sigma v = \sigma R \omega\). This surface current generates a magnetic field inside the cylinder (similar to a solenoid): $$ B = \mu_0 K = \mu_0 \sigma R \omega $$ The magnetic flux through the cross-section of the cylinder is: $$ \Phi = B (\pi R^2) = \mu_0 \sigma \pi R^3 \omega $$

The changing magnetic flux induces an electric field \(E\) along the tangential direction. Using Faraday’s Law \(\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi}{dt}\): $$ E (2\pi R) = \frac{d}{dt} (\mu_0 \sigma \pi R^3 \omega) = \mu_0 \sigma \pi R^3 \alpha $$ $$ E = \frac{\mu_0 \sigma R^2 \alpha}{2} $$ This electric field exerts a force on the surface charge. The retarding torque \(\tau_{mag}\) is: $$ \tau_{mag} = \text{Force} \times R = (E \cdot Q_{total}) R = E (\sigma \cdot 2\pi R l) R $$ $$ \tau_{mag} = \left( \frac{\mu_0 \sigma R^2 \alpha}{2} \right) (2\pi \sigma R l) R = \pi \mu_0 \sigma^2 R^4 l \alpha $$

For the block (mass \(m\)): \(mg – T = ma\)
For the cylinder (mass \(M\)): The torque from tension \(TR\) drives rotation, while \(\tau_{mag}\) opposes it. $$ TR – \tau_{mag} = I_{cyl} \alpha $$ Substituting \(I_{cyl} = MR^2\) (hollow cylinder) and the derived \(\tau_{mag}\): $$ T R – \pi \mu_0 \sigma^2 R^4 l \alpha = M R^2 \alpha $$ $$ T = (M + \pi \mu_0 \sigma^2 R^2 l) R \alpha $$

Using the constraint \(a = R\alpha\), we substitute \(T\) into the block’s equation: $$ mg – (M + \pi \mu_0 \sigma^2 R^2 l) a = ma $$ $$ mg = a (m + M + \pi \mu_0 \sigma^2 R^2 l) $$