Irodov 3.274: Where H Has Sources and B Circulates – Amit & Deepali Sinha Foundation

Problem Irodov 3.274

A magnetic material of permeability $\mu$ has a flat surface facing vacuum. Just outside the surface the magnetic induction is $\vect{B}$ , of magnitude $B$ , pointing out of the material at an angle $\theta$ to the surface normal $\vect{n}$ (the normal points from the material into the vacuum). Find:

the flux of $\vect{H}$ through a sphere $S$ of radius $R$ whose centre lies on the surface;
the circulation of $\vect{B}$ round a square path $\Gamma$ of side $l$ standing across the surface in the plane of $\vect{n}$ and $\vect{B}$ , with its top side in the vacuum and its bottom side in the material, traversed along the top side in the direction of the tangential part of $\vect{B}$ and back along the bottom side.

The magnetic (shaded, μ) below the plane surface, vacuum above. B in the vacuum makes the angle θ with the normal n. The sphere S of radius R is centred on the surface; the square path Γ of side l straddles it, traversed clockwise. — The magnetic (shaded, $\mu$ ) below the plane surface, vacuum above. $\vect{B}$ in the vacuum makes the angle $\theta$ with the normal $\vect{n}$ . The sphere $S$ of radius $R$ is centred on the surface; the square path $\Gamma$ of side $l$ straddles it, traversed clockwise.

Given

$B$ in vacuum at the surface
angle $\theta$ to $\vect{n}$
permeability $\mu$
sphere radius $R$
square side $l$
field uniform on each side over $S$ and $\Gamma$ (assumed)
no free current on the surface (assumed)

Find

(a)

\oint_S \vect{H}\cdot d\vect{S}

; (b)

\oint_\Gamma \vect{B}\cdot d\vect{r}

Solution

Step 1The field inside the magnetic

Two conditions carry the field across the surface. The normal component of $\vect{B}$ is continuous, because the flux of $\vect{B}$ out of a flat pillbox straddling the surface is zero. The tangential component of $\vect{H}$ is continuous, because no free current crosses a thin rectangle straddling it. With $\vect{H} = \vect{B}/\mu_0$ in the vacuum and $\vect{H}' = \vect{B}'/(\mu\mu_0)$ inside, the induction $\vect{B}'$ just inside has

\begin{aligned} B'_n &= B_n = B\cos\theta\\ B'_t &= \mu\mu_0 H_t = \mu B\sin\theta \end{aligned} \tag{1}

Drawn for μ = 2: B' in the magnetic and B in the vacuum. The normal components are equal (the dashed projections reach equal heights); the tangential component inside is μ times the one outside. — Drawn for $\mu = 2$ : $\vect{B}'$ in the magnetic and $\vect{B}$ in the vacuum. The normal components are equal (the dashed projections reach equal heights); the tangential component inside is $\mu$ times the one outside.

Step 2(a) The flux of B through each half of the sphere

Close each hemisphere with the flat disc of radius $R$ that the surface cuts out of the sphere. The flux of $\vect{B}$ out of any closed surface is zero, so the outward flux through the upper (vacuum) hemisphere equals the flux entering through the disc from below, $\pi R^2 B_n$ . For the lower hemisphere the same disc is where the flux leaves, so its own outward flux is the negative of that. By (1) both use the same $B_n$ :

\Phi_B^{\mathrm{vac}} = \pi R^2 B\cos\theta, \qquad \Phi_B^{\mathrm{mag}} = -\pi R^2 B\cos\theta \tag{2}

Step 3(a) The flux of H

On each side $\vect{H}$ is $\vect{B}$ divided by that side’s $\mu\mu_0$ , so the outward flux of $\vect{H}$ through $S$ is the sum of the two terms of (2), each divided by its own factor:

\oint_S \vect{H}\cdot d\vect{S} = \frac{\pi R^2 B\cos\theta}{\mu_0} - \frac{\pi R^2 B\cos\theta}{\mu\mu_0}

Taking out the common factor,

\oint_S \vect{H}\cdot d\vect{S} = \frac{\pi R^2 B\cos\theta\,(\mu - 1)}{\mu\mu_0}

(3)

Step 4(b) The circulation of B round the square

Take the four sides in turn, in the sense of the path. Along the top side, in the vacuum and in the direction of $B$ ’s tangential part, $\vect{B}\cdot d\vect{r}$ adds $B_t l$ . Along the bottom side, inside the magnetic and against the tangential part there, it adds $-B'_t l$ . The two vertical sides run along the normal, one up and one down; with $\vect{B}$ uniform on each side and its normal component the same in both media, their contributions cancel. With (1):

\oint_\Gamma \vect{B}\cdot d\vect{r} = B\sin\theta\; l - \mu B\sin\theta\; l

\oint_\Gamma \vect{B}\cdot d\vect{r} = (1 - \mu)\,Bl\sin\theta

(4)

For $\mu > 1$ it is negative: the circulation runs against the sense of the path.

Check

With $\mu = 1$ there is no material and both results vanish, as they must: $\vect{H} = \vect{B}/\mu_0$ everywhere has zero flux through a closed surface, and a field with no current through $\Gamma$ has zero circulation. At $\theta = 90^\circ$ the field grazes the surface and no flux crosses it, so (3) vanishes; at $\theta = 0$ the field has no tangential part and (4) vanishes.

Remark

Both results are the magnetization showing through. The flux of $\vect{H}$ is not zero because $\nabla\cdot\vect{H} = -\nabla\cdot\vect{J}$ : the magnetization $\vect{J}$ stops at the surface, which therefore carries the effective sources of $\vect{H}$ . The circulation of $\vect{B}$ is not zero because $\Gamma$ encloses molecular current: the surface carries the current $\vect{J}\times\vect{n}$ per unit length, and (4) equals $\mu_0$ times the part of it threading the square, which runs against the sense of $\Gamma$ .