Irodov 3.323: A Superconductor Refuses New Flux – Amit & Deepali Sinha Foundation

Problem Irodov 3.323

A superconducting circular ring of radius $a$ and inductance $L$ lies in a uniform magnetic field $\vect B$ with its plane parallel to the field, and carries no current. The ring is then turned through $90^\circ$ , so that its plane becomes perpendicular to the field. Find:

the current induced in the ring after the turn
the work done in turning it

Before: the ring's plane parallel to B, its normal n square to the field, no current. — Before: the ring’s plane parallel to $\vect B$ , its normal $\vect n$ square to the field, no current.

After: the ring, of radius a, turned through 90°, its normal along B. — After: the ring, of radius $a$ , turned through $90^\circ$ , its normal along $\vect B$ .

Given

$a$
$L$
$B$ uniform
$I = 0$ at the start, plane parallel to $\vect B$
turned through $90^\circ$

Find

(a) the current

I

after the turn, (b) the work

A

done in the turn

Solution

Step 1A superconductor keeps its total flux

Let $\vect n$ be the ring’s normal and count the current $I$ positive when it circulates counterclockwise about $\vect n$ . The flux through the ring is its own, $LI$ , plus that of the external field, $\Phi_{\mathrm{e}} = \pi a^2 B\cos\theta$ with $\theta$ the angle between $\vect n$ and $\vect B$ . The ring has no resistance, so the induced emf must vanish at every moment, and with it the rate of change of the total flux:

\begin{align*} \mathcal{E}_{\mathrm{i}} = -\frac{d}{dt}\bigl(LI + \Phi_{\mathrm{e}}\bigr) &= RI = 0 \\ LI + \Phi_{\mathrm{e}} &= \text{const} = 0 \tag{1}\end{align*}

The constant is zero because at the start there is no current and the field lies in the ring’s plane, $\theta = 90^\circ$ .

Step 2(a) The current after the turn

After the turn $\theta = 0$ and the external flux is $\pi a^2 B$ , so (1) fixes the current:

LI = -\pi a^2 B

|I| = \frac{\pi a^2 B}{L}

(2)

The minus sign says the current circulates clockwise about $\vect n$ : its own field inside the ring points against $\vect B$ , and its flux cancels the external one exactly.

After the turn: the induced current I, whose own flux through the ring cancels that of B. — After the turn: the induced current $I$ , whose own flux through the ring cancels that of $\vect B$ .

Step 3(b) The work, from the force on the current

When the external flux through a circuit carrying the current $I$ changes by $d\Phi_{\mathrm{e}}$ , the Ampere forces of the external field do the work $I\,d\Phi_{\mathrm{e}}$ on it. The turn is slow, so the hand does the opposite work, and by (1) the current at each stage is $I = -\Phi_{\mathrm{e}}/L$ :

dA = -I\,d\Phi_{\mathrm{e}} = \frac{\Phi_{\mathrm{e}}\,d\Phi_{\mathrm{e}}}{L}

Integrating as $\Phi_{\mathrm{e}}$ grows from $0$ to $\pi a^2 B$ :

A = \frac{1}{L}\int_0^{\pi a^2 B}\Phi_{\mathrm{e}}\,d\Phi_{\mathrm{e}} = \frac{(\pi a^2 B)^2}{2L}

A = \frac{\pi^2 a^4 B^2}{2L}

(3)

This mechanical work is positive: the hand works against the field’s torque all the way.

Check

The work equals the energy $LI^2/2$ of the ring’s own current, which at constant flux is the ring’s potential energy in the field; with (2):

\frac{L}{2}\left(\frac{\pi a^2 B}{L}\right)^2 = \frac{\pi^2 a^4 B^2}{2L}

Units: $\pi a^2 B$ is a flux in Wb, and $\text{Wb}/\text{H} = \text{A}$ in (2), while $\text{Wb}^2/\text{H} = \text{J}$ in (3).

Remark

A superconducting ring cannot change the flux through it, so it behaves toward any change of field as a perfect diamagnet: turned into the field it carries just the current that keeps the field out. The work is stored, not lost: the position reached is an unstable balance, and nudged and let go, the ring turns back by itself, its current falls to zero, and the stored energy returns as the kinetic energy of the turning ring.