Irodov 3.339: Spinning Charge Makes a Solenoid – Amit & Deepali Sinha Foundation

Problem Irodov 3.339

A long cylinder of radius $a$ , with a charge spread uniformly over its surface, rotates about its own axis at angular velocity $\omega$ . The charge per unit length of the cylinder is $\lambda$ , and the permeability everywhere is $\mu = 1$ . Find the energy of the magnetic field per unit length of the cylinder.

The long cylinder of radius a, side on, with its uniform surface charge (drawn positive), λ per unit length, rotating at ω about its axis (sense chosen). — The long cylinder of radius $a$ , side on, with its uniform surface charge (drawn positive), $\lambda$ per unit length, rotating at $\omega$ about its axis (sense chosen).

Given

$a$
$\omega$
$\lambda$
$\mu = 1$
uniform surface charge
long cylinder

Find

the magnetic energy per unit length

W_1

Solution

Step 1The rotating charge is a surface current

A strip of the cylinder’s surface of axial length $\ell$ carries the charge $\lambda\ell$ , and the rotation carries all of it past a fixed line along the surface $\omega/2\pi$ times per second. So the strip is a current $\lambda\ell\,\omega/2\pi$ circling the axis, and the current per unit length of the cylinder is

i = \frac{\lambda\omega}{2\pi} \tag{1}

Step 2The field of the current sheet

By symmetry the field of a long solenoid is axial, uniform inside and zero outside. Ampere’s circuital law on a rectangle of length $\ell$ with one long side inside and one outside, enclosing the current $i\ell$ , gives with (1)

\begin{aligned} B\ell &= \mu_0 i\ell\\ B &= \mu_0 i = \frac{\mu_0\lambda\omega}{2\pi} \end{aligned} \tag{2}

inside the cylinder, and no field outside it.

Step 3The energy per unit length

The field occupies only the cross-section $\pi a^2$ , where its energy density $B^2/2\mu_0$ is uniform. A unit length holds the volume $\pi a^2$ , so, with (2),

W_1 = \frac{B^2}{2\mu_0}\,\pi a^2 = \frac{1}{2\mu_0}\left(\frac{\mu_0\lambda\omega}{2\pi}\right)^2\pi a^2

Squaring and cancelling one $\mu_0$ and one $\pi$ :

W_1 = \frac{\mu_0\lambda^2\omega^2a^2}{8\pi}

(3)

The sign of $\lambda$ and the sense of rotation do not matter: reversing either reverses $\vect B$ and leaves $B^2$ unchanged.

Check

The same energy as the work of spinning the cylinder up. While $\omega$ grows, $B$ grows, and the eddy field on the surface, from Faraday’s law on the rim circle, $E\cdot 2\pi a = \pi a^2\,dB/dt$ , pushes back on the charge with a torque $\lambda E a$ per unit length. With (2), the work done against it from rest is

\int \lambda a\,\frac{a}{2}\,\frac{dB}{dt}\,\omega\,dt = \frac{\lambda a^2}{2}\,\frac{\mu_0\lambda}{2\pi}\int_0^{\omega}\omega'\,d\omega' = \frac{\mu_0\lambda^2\omega^2a^2}{8\pi},

the boxed $W_1$ . Units: $\mu_0\lambda^2\omega^2a^2$ is $(\text{H/m})(\text{C/m})^2(\text{s}^{-2})(\text{m}^2) = \text{H A}^2/\text{m} = \text{J/m}$ .

Remark

The electric field of the same cylinder stores $(\lambda^2/4\pi\varepsilon_0)\ln(R/a)$ per unit length out to a radius $R$ , so the ratio of the magnetic to the electric energy is $(\omega a/c)^2/2\ln(R/a)$ : the square of the surface’s speed over the speed of light. Magnetism is the small relativistic companion of the electric field of moving charge.