Irodov 3.325: Flux Frozen into a Ring – Amit & Deepali Sinha Foundation

Problem Irodov 3.325

A ring of radius $a = 50$ mm, made of thin wire of radius $b = 1.0$ mm, lies in a uniform magnetic field of induction $B = 0.50$ mT with its plane perpendicular to $\vect B$ . The ring is cooled into the superconducting state, and the field is then switched off. Find the current in the ring afterwards. The inductance of a thin ring carrying a surface current is $L = \mu_0 a\left(\ln(8a/b) - 2\right)$ .

Given

$a = 50$ mm
$b = 1.0$ mm
$B = 0.50$ mT, square to the ring’s plane
$L = \mu_0 a\left(\ln(8a/b) - 2\right)$
no current before the cooling (assumed)

Find

the current

I

after the field is switched off

Solution

Step 1The superconductor keeps the flux it was cooled with

Before the cooling the ring is an ordinary conductor in a steady field, so it carries no current (assumed), and the flux through it is all external, $\pi a^2 B$ . Once superconducting it has no resistance, so the induced emf, and with it the rate of change of the total flux through the ring, the external flux plus its own $LI$ , is zero from then on:

\begin{align*} -\frac{d}{dt}\bigl(LI + \Phi_{\mathrm{e}}\bigr) &= RI = 0 \\ LI + \Phi_{\mathrm{e}} &= \pi a^2 B \tag{1}\end{align*}

Step 2The current after the field is gone

Switching the field off sets $\Phi_{\mathrm{e}} = 0$ , so the ring’s own flux $LI$ takes over the whole of $\pi a^2 B$ . With the given inductance:

I = \frac{\pi a^2 B}{L} = \frac{\pi a^2 B}{\mu_0 a\left(\ln(8a/b) - 2\right)}

Numerically, with $8a/b = 400$ :

I = \frac{\pi\,(0.050\text{ m})(0.50\times 10^{-3}\text{ T})}{\bigl(4\pi\times 10^{-7}\text{ H/m}\bigr)\left(\ln 400 - 2\right)}

I = \frac{\pi a B}{\mu_0\left(\ln(8a/b) - 2\right)} \approx 16\text{ A}

(2)

The current circulates in the sense that produced the original field through the ring: its own field inside the ring replaces the one that was switched off.

Check

The ring’s own field at its centre, from (2):

\frac{\mu_0 I}{2a} = \frac{\bigl(4\pi\times 10^{-7}\text{ H/m}\bigr)(16\text{ A})}{2(0.050\text{ m})} \approx 0.20\text{ mT}

This is smaller than the $0.50$ mT it replaces, as it must be: the field of a thin ring crowds near the wire, so a weaker field at the centre carries the same flux. Units: $aB/\mu_0$ is $\text{m}\cdot\text{T}/(\text{H/m}) = \text{T}\cdot\text{m}^2/\text{H} = \text{Wb}/\text{H} = \text{A}$ .

Remark

The current that was not there while the field was on appears when the field goes, because the superconductor holds the flux and not the current. Such a persistent current runs on without decay for years, which is how superconducting magnets in MRI machines are run, charged once and then disconnected from their supply. The logarithm makes the result depend only weakly on the thickness of the wire: halving $b$ changes $I$ by less than 20 percent.