Irodov 3.281: Inside a Magnet, H Runs Backward – Amit & Deepali Sinha Foundation

Problem Irodov 3.281

A permanent magnet has the shape of a ring with a narrow gap between its poles. The mean diameter of the ring is $d = 20$ cm, the gap is $b = 2.0$ mm wide, and the magnetic induction in the gap is $B = 40$ mT. Neglecting the spreading of the flux at the edges of the gap, find the magnitude of the magnetic field strength $\vect H$ inside the magnet.

The ring magnet seen from above, with the narrow gap of width b and the mean diameter d. The closed lines are B, through the magnet and across the gap (gap widened for clarity). — The ring magnet seen from above, with the narrow gap of width $b$ and the mean diameter $d$ . The closed lines are $\vect B$ , through the magnet and across the gap (gap widened for clarity).

Given

$d = 20$ cm
$b = 2.0$ mm
$B = 40$ mT in the gap
no spreading at the gap edges

Find

the magnitude

H

inside the magnet

Solution

Step 1The field strength in the gap

With no spreading at the edges, the flux tube keeps the ring’s cross-section across the gap, and the normal component of $\vect B$ is continuous at each face, so the induction is $B$ both in the gap and in the magnet. In the gap, which is vacuum, the field strength is

H_0 = \frac{B}{\mu_0} \tag{1}

In the magnet, however, $\vect H$ is not $\vect B/\mu_0$ : a permanent magnet’s magnetization is not fixed by a linear law, so $H$ there has to come from the circulation.

Step 2The circulation of H is zero

Take the contour along the mean circle of the ring: a run of length $\pi d - b$ in the magnet and $b$ across the gap. No conduction current threads it (there is no winding), so the circulation of $\vect H$ along it is zero. Counting $H_l$ as the component of $\vect H$ along the contour’s sense, taken along $\vect B$ , the constant values in each part give

H_l(\pi d - b) + H_0\,b = 0

Solving for $H_l$ with (1):

H_l = -\frac{bB}{\mu_0(\pi d - b)} \tag{2}

The ring with its field strength: H_0 spans the gap along B, while H in the magnet (a short stub) points against B. The true ratio H/H_0 1/313 is far smaller than drawn. — The ring with its field strength: $\vect H_0$ spans the gap along $\vect B$ , while $\vect H$ in the magnet (a short stub) points against $\vect B$ . The true ratio $H/H_0 \approx 1/313$ is far smaller than drawn.

The minus sign says that inside the magnet $\vect H$ points against $\vect B$ , as the figure shows.

Step 3The magnitude

The magnitude is the modulus of (2). Numerically, with $\pi d - b = 0.6283 - 0.0020 = 0.6263$ m,

\frac{bB}{\mu_0(\pi d - b)} = \frac{(2.0\times10^{-3})(0.040)}{(4\pi\times10^{-7})(0.6263)}\text{ A/m}

H = \frac{bB}{\mu_0(\pi d - b)} \approx 0.10\text{ kA/m}

(3)

Since $b$ is a third of a percent of $\pi d$ , dropping it from the denominator gives $H \approx bB/(\mu_0\pi d)$ , the same $0.10$ kA/m.

Check

Units: $bB/\mu_0$ is a length times a field strength, and dividing by the length $\pi d - b$ leaves A/m. Size: in the magnet $\vect B = \mu_0(\vect H + \vect J)$ with $\vect H$ reversed, so the magnetization exceeds $B/\mu_0 = 31.8$ kA/m by only $0.10$ kA/m, a third of a percent: the reverse $H$ is a small correction, as a narrow gap should make it. As a limit, closing the gap ( $b \to 0$ ) makes $H \to 0$ : a closed ring magnet has no $\vect H$ inside at all, only $\vect B = \mu_0\vect J$ .

Remark

Inside a permanent magnet with a gap, $\vect H$ runs backward against $\vect B$ : it is the demagnetizing field of the magnet’s own poles at the faces of the gap. This is why a magnet is stored with a keeper bridging its poles: closing the gap removes the reverse $\vect H$ , which over time and with knocks would push the material down its hysteresis loop toward demagnetization. The wider the gap, the larger the reverse $H$ , and the lower the magnet’s working point slides down its demagnetization curve, toward $B = 0$ at the coercive force.