Irodov 3.272: A Magnet Is a Ring Current – Amit & Deepali Sinha Foundation

Problem Irodov 3.272

A permanent magnet is a thin disc of radius $R = 1.0$ cm, magnetized along its axis. On that axis, at a distance $x = 10$ cm from the disc’s centre, the magnetic induction is $B = 30$ $\mu$ T. Estimate the molecular (magnetization) current $I'$ that flows round the rim of the disc.

The thin disc magnet edge-on (thickness exaggerated), radius R, magnetization J along its axis, and the point P on the axis at distance x. The distance is drawn shorter than its true x = 10R. — The thin disc magnet edge-on (thickness exaggerated), radius $R$ , magnetization $\vect{J}$ along its axis, and the point $P$ on the axis at distance $x$ . The distance is drawn shorter than its true $x = 10R$ .

Given

$R = 1.0$ cm
$x = 10$ cm
$B = 30$ $\mu$ T at $x$ on the axis
thin disc, magnetized along its axis
magnetization uniform (assumed)

Find

the molecular current

I'

along the rim

Solution

Step 1Where the molecular current flows

A magnetized body produces the same field as its molecular (magnetization) currents: a surface current of linear density $\vect{J}\times\vect{n}$ on its surface, with $\vect{n}$ the outward normal, and a volume current of density $\nabla\times\vect{J}$ inside. With the magnetization uniform the volume current vanishes. On the two flat faces $\vect{n}$ is parallel to $\vect{J}$ , so their surface current vanishes too. Only the rim, where $\vect{n}$ is perpendicular to $\vect{J}$ , carries current, and for a disc of thickness $h$ it totals

I' = Jh

For a thin disc this is a single ring current of radius $R$ , circulating round the axis by the right-hand rule about $\vect{J}$ .

The disc's magnetization is equivalent to the molecular current I' round its rim (out of the page at the top), and the rim current makes the field B at P along the axis. — The disc’s magnetization is equivalent to the molecular current $I'$ round its rim (out of the page at the top), and the rim current makes the field $\vect{B}$ at $P$ along the axis.

Step 2The ring’s field on the axis

By symmetry the transverse parts of the contributions of a ring current cancel on its axis and the axial parts add, giving the standard on-axis field

B = \frac{\mu_0 I' R^2}{2\,(R^2 + x^2)^{3/2}} \tag{1}

Step 3The number

With $R = 1.0\times10^{-2}$ m and $x = 0.10$ m, the geometric factor is

\begin{aligned} R^2 + x^2 &= 1.0\times10^{-4} + 1.0\times10^{-2} = 1.01\times10^{-2}\text{ m}^2\\ (R^2 + x^2)^{3/2} &= 1.015\times10^{-3}\text{ m}^3 \end{aligned}

Solving (1) for $I'$ and substituting the values,

I' = \frac{2\,(30\times10^{-6})(1.015\times10^{-3})}{(4\pi\times10^{-7})(1.0\times10^{-4})}\text{ A}

I' = \frac{2B\,(R^2 + x^2)^{3/2}}{\mu_0 R^2} \approx 0.48\text{ kA}

(2)

Because $x = 10R$ , the factor $(R^2 + x^2)^{3/2}$ is $x^3$ to within $1.5$ percent, and the result is $I' \approx 2Bx^3/(\mu_0R^2)$ , which gives the same $0.48$ kA to two figures.

Check

Read the other way, a rim current of $485$ A on a ring of radius $1.0$ cm gives at the centre $B = \mu_0I'/2R \approx 30$ mT, a thousand times the $30$ $\mu$ T at $x = 10R$ : the factor $(1 + x^2/R^2)^{3/2} = 101^{3/2} \approx 1015$ , as the on-axis formula requires. A field of tens of millitesla at the face is just what a small permanent magnet has, so the figure is reasonable.

Remark

The current is large, hundreds of amperes, yet nothing heats: no charge travels round the rim. The molecular current is the sum of the atoms’ circulating currents, which cancel between neighbours everywhere inside a uniformly magnetized body and fail to cancel only at its surface. Its size is $Jh$ , so for a disc of $1$ mm thickness this one corresponds to $J \approx 5\times10^5$ A/m, typical of a ferrite or alnico magnet.