Irodov 3.280: A Winding That Silences a Magnet – Amit & Deepali Sinha Foundation

Problem Irodov 3.280

A permanent magnet in the shape of a cylinder of length $l = 15$ cm carries a uniform winding of $N = 300$ turns of thin wire. When a current $I = 3.0$ A is passed through the winding, the magnetic field outside the magnet vanishes. Find the coercive force $H_c$ of the magnet’s material.

The cylindrical permanent magnet of length l in longitudinal section, with its uniform winding of N turns carrying the current I. — The cylindrical permanent magnet of length $l$ in longitudinal section, with its uniform winding of $N$ turns carrying the current $I$ .

Given

$N = 300$
$l = 15$ cm
$I = 3.0$ A
no field outside
magnetization uniform (assumed)

Find

the coercive force

H_c

Solution

Step 1No field outside means no induction inside

The lines of $\vect B$ never end: a line running along the magnet would have to leave through an end face and close outside, where there is no field. The normal component of $\vect B$ is continuous across each end face, and it is zero just outside, so no line crosses. With the magnetization uniform (assumed), the induction inside is therefore zero everywhere:

\vect B = \mu_0(\vect H + \vect J) = 0 \tag{1}

The field strength in the magnet is then $\vect H = -\vect J$ : it exactly undoes the magnetization’s contribution to $\vect B$ .

Step 2The field strength from the circulation of H

Take a closed contour that runs along the axis of the magnet, through its full length $l$ , and returns outside the winding. Outside there is no field, $\vect H = 0$ , so only the inner run counts. The contour links all $N$ turns, so the circulation of $\vect H$ equals $NI$ ; with $H$ uniform along the axis,

\oint \vect H\cdot d\vect l = Hl = NI \tag{2}

The magnet in the state with no outside field: the field strength H (open head) along the axis and the magnetization J (tinted arrows) against it, so that B = 0. — The magnet in the state with no outside field: the field strength $\vect H$ (open head) along the axis and the magnetization $\vect J$ (tinted arrows) against it, so that $\vect B = 0$ .

Step 3Reading off the coercive force

By (1) the material is at $B = 0$ while the field strength in it has the magnitude $H$ of (2), and the field strength at which a magnetized material’s induction falls to zero is its coercive force. Numerically,

\frac{NI}{l} = \frac{300 \times 3.0}{0.15}\text{ A/m}

H_c = \frac{NI}{l} \approx 6.0\text{ kA/m}

(3)

Check

Dimensions: ampere-turns per metre is the unit of $H$ . Order of magnitude: $6$ kA/m is the coercive force of a hardened steel magnet; soft iron has under $0.1$ kA/m (about fifteen ampere-turns over this length would demagnetize it), and modern rare-earth magnets reach about $1000$ kA/m.

Remark

The vanishing outside field is a null measurement: a compass or a field probe beside the magnet only has to read zero, and the coercive force follows from a current and a turn count, with no field measured inside the material. The field strength in the magnet points against the magnetization, as it must, and the current that silences the magnet sits exactly at the point of the hysteresis loop where $B$ crosses zero; a larger current would carry it on toward reversed magnetization.