Energy to Spin Up a Charged Cylinder The angular velocity ω(t)zˆ of the cylindrical shell shown below increases from zero and smoothly approaches the steady value ω0. The shell has infinitesimal thickness and carries a uniform charge per unit length λ = 2πRσ, where σ is a uniform charge per unit area. Assume that the shell radius R L and that the spin-up is very slow so the displacement current may be neglected.
The Correct Answer and Explanation is:
To find the energy required to spin up a charged cylindrical shell from rest to an angular velocity $\omega_0$, we need to calculate the energy stored in the magnetic field generated during the process. Since the shell carries charge and rotates, it produces a current, which in turn creates a magnetic field. The energy used to spin it up is stored in this magnetic field.
Answer:
$$
\boxed{U = \frac{\mu_0 \lambda^2 \omega_0^2 R^2 L}{4\pi}}
$$
Where:
- $\mu_0$: permeability of free space
- $\lambda = 2\pi R \sigma$: charge per unit length
- $\omega_0$: final angular velocity
- $R$: radius of the shell
- $L$: length of the cylinder
We are dealing with a cylindrical shell of radius $R$, length $L$, and negligible thickness, carrying a uniform charge per unit area $\sigma$. As it spins up slowly (quasistatically) to a final angular velocity $\omega_0$, each bit of charge moves in a circle, thus constituting a circular current. The result is a magnetic field inside the shell.
Because the spin-up is slow, displacement currents are neglected, and the system can be analyzed using magnetostatics.
Step 1: Current due to rotation
The line charge density is:
$$
\lambda = \sigma \cdot 2\pi R
$$
Since this line charge is rotating with angular velocity $\omega$, the current per unit length (along the cylinder’s axis) is:
$$
I = \lambda \cdot v = \lambda \cdot R \omega = 2\pi R \sigma \cdot R \omega = 2\pi R^2 \sigma \omega
$$
Step 2: Magnetic field inside a spinning charged cylinder
Using Ampère’s law for a long cylinder, the magnetic field inside (along the axis) due to a surface current is:
$$
B = \frac{\mu_0 I}{2\pi R} = \frac{\mu_0 \lambda \omega R}{2\pi}
$$
Step 3: Magnetic energy stored
The energy density in a magnetic field is:
$$
u = \frac{B^2}{2\mu_0}
$$
Total magnetic energy stored:
$$
U = u \cdot \text{Volume} = \frac{B^2}{2\mu_0} \cdot \pi R^2 L
$$
Plugging in $B$:
$$
U = \frac{1}{2\mu_0} \left( \frac{\mu_0 \lambda \omega_0 R}{2\pi} \right)^2 \cdot \pi R^2 L = \frac{\mu_0 \lambda^2 \omega_0^2 R^2 L}{4\pi}
$$
This is the energy needed to spin up the cylinder.
