Irodov 3.311: Spun Up by Switching On – Amit & Deepali Sinha Foundation

Problem Irodov 3.311

A thin non-conducting ring of mass $m$ carries a charge $q$ and is free to rotate about its own axis. Initially the ring is at rest and there is no magnetic field. A practically uniform magnetic field is then switched on, perpendicular to the plane of the ring, and grows with time according to some law $\vect{B}(t)$ . Find the ring’s angular velocity $\vect{\omega}$ as a function of $\vect{B}(t)$ .

Given

ring mass $m$ , charge $q$
free to turn about its axis
at rest at $B = 0$
$\vect{B}(t)$ uniform, normal to the plane
charge spread evenly round the ring (assumed)

Find

\vect{\omega}

as a function of

\vect{B}(t)

Solution

Step 1The torque of the eddy field

Call the ring’s radius $r$ and take the axis $z$ along the ring’s axis, so $\vect B = B\,\vect k$ . The changing field induces an eddy electric field, and its circulation round the ring is fixed by Faraday’s law whatever the field’s detailed shape:

\oint E_{\varphi}\,dl = -\frac{d\Phi}{dt} = -\pi r^2\,\frac{dB}{dt} \tag{1}

Here $E_{\varphi}$ is the component along the ring, positive counter-clockwise seen from the tip of $\vect k$ . Each element $dq = q\,dl/(2\pi r)$ of the evenly spread charge (assumed) feels the tangential force $E_{\varphi}\,dq$ at the lever arm $r$ , so the torque about the axis, using (1), is

N_z = \oint r E_{\varphi}\,\frac{q\,dl}{2\pi r} = \frac{q}{2\pi}\oint E_{\varphi}\,dl = -\frac{q}{2\pi}\,\pi r^2\,\frac{dB}{dt}

N_z = -\frac{q r^2}{2}\,\frac{dB}{dt}

(2)

Drawn for q > 0 and B out of the page and growing: the eddy field E along the ring of radius r runs clockwise and drives the charged ring round that way. — Drawn for $q > 0$ and $\vect B$ out of the page and growing: the eddy field $\vect E$ along the ring of radius $r$ runs clockwise and drives the charged ring round that way.

Step 2The angular momentum the torque delivers

All the ring’s mass is at radius $r$ , so its moment of inertia about the axis is $mr^2$ . Equating the rate of change of angular momentum to (2):

\frac{d}{dt}\bigl(mr^2\omega_z\bigr) = -\frac{q r^2}{2}\,\frac{dB}{dt}

Both sides are exact time derivatives. Integrating from the start, where $\omega_z = 0$ and $B = 0$ :

mr^2\omega_z = -\frac{qr^2}{2}\,B \quad\Longrightarrow\quad \omega_z = -\frac{q}{2m}\,B

\vect{\omega} = -\frac{q}{2m}\,\vect{B}(t)

(3)

For $q > 0$ the ring turns opposite to the sense that would carry its charge along a current making $\vect B$ by the right-hand rule: its rotation’s own field opposes the growth of $\vect B$ (Lenz), clockwise when $\vect B$ points at the viewer. For $q < 0$ it turns the other way.

Check

Units: $qB/m$ is C $\,\cdot\,$ T/kg, and since $qvB$ is a force, T = kg/(C s), so $qB/m$ is s $^{-1}$ , an angular velocity. Neither the radius nor the history $B(t)$ appears: the torque (2) is a time derivative of $B$ , so only the final $B$ matters, as it must for an angular momentum built up from rest. Switching the field back off returns $\omega$ to zero.

Remark

The result $\omega = qB/(2m)$ is the Larmor frequency, half the cyclotron frequency $qB/m$ : switching on a field spins up a charged body until its charges circulate at exactly this rate, which is the classical picture of diamagnetism in atoms. Because only the final $B$ enters, the spin cannot be avoided by switching on slowly: a gentle switch-on gives the same $\omega$ as a sudden one, the weaker eddy field simply acting for longer.