Irodov 3.273: Normal B, Tangential H Carry Across – Amit & Deepali Sinha Foundation

Problem Irodov 3.273

A uniform isotropic magnetic material of permeability $\mu$ has a flat surface facing vacuum. Just outside the surface, in the vacuum, the magnetic induction has magnitude $B$ and the vector $\vect B$ makes an angle $\alpha$ with the normal to the surface. What is the magnitude $B'$ of the magnetic induction inside the material, right next to the surface?

The flat surface of the magnetic material (shaded, permeability μ) with vacuum above. The induction B in the vacuum meets the surface at the angle α from the normal (dashed). — The flat surface of the magnetic material (shaded, permeability $\mu$ ) with vacuum above. The induction $\vect B$ in the vacuum meets the surface at the angle $\alpha$ from the normal (dashed).

Given

$B$ in vacuum at the surface
$\alpha$ from the normal
permeability $\mu$

Find

B'

inside the magnetic, at the surface

Solution

Step 1What crosses the surface unchanged

Two field equations fix the field on the far side of the surface. A thin pillbox straddling it has zero net flux of $\vect B$ , so the normal component of $\vect B$ is the same on both sides. A thin rectangular loop straddling it encloses no conduction current, so the circulation of $\vect H$ round it vanishes and the tangential component of $\vect H$ is the same on both sides.

B'_n = B_n, \qquad H'_t = H_t \tag{1}

The vectors on both sides of the surface, drawn for μ = 3. Left: B bends away from the normal, its normal component unchanged (dashed projections at equal heights). Right: H keeps its tangential component (equal offsets along the surface). — The vectors on both sides of the surface, drawn for $\mu = 3$ . Left: $\vect B$ bends away from the normal, its normal component unchanged (dashed projections at equal heights). Right: $\vect H$ keeps its tangential component (equal offsets along the surface).

Step 2The two components inside

Resolving $\vect B$ along the normal and along the surface gives its components in vacuum. The normal one carries straight across by (1):

B'_n = B_n = B\cos\alpha \tag{2}

For the tangential one, write each side’s $H$ through its own $B$ : $\vect H = \vect B/\mu_0$ in vacuum and $\vect H' = \vect B'/(\mu\mu_0)$ in the magnetic. Then the second condition of (1) reads

\begin{aligned} \frac{B'_t}{\mu\mu_0} &= \frac{B_t}{\mu_0} = \frac{B\sin\alpha}{\mu_0}\\[2pt] B'_t &= \mu B\sin\alpha \end{aligned} \tag{3}

Step 3The magnitude

The two components (2) and (3) are perpendicular, so

B'^2 = B'^2_n + B'^2_t = B^2\cos^2\alpha + \mu^2B^2\sin^2\alpha

and taking the root,

B' = B\sqrt{\mu^2\sin^2\alpha + \cos^2\alpha}

(4)

Check

Three special cases. With $\mu = 1$ the material is vacuum and the formula gives $B' = B$ , as it must. At normal incidence, $\alpha = 0$ , only the normal component exists and it is continuous, so $B' = B$ for any $\mu$ . At grazing incidence, $\alpha = 90^\circ$ , only the tangential $H$ exists and it is continuous, so $B' = \mu\mu_0 H = \mu B$ , which the formula also gives.

Remark

Dividing (3) by (2) gives the refraction law $\tan\alpha' = \mu\tan\alpha$ for the angle $\alpha'$ the inner field makes with the normal. For iron, with $\mu$ in the thousands, the field inside runs almost along the surface for any $\alpha$ not very close to zero; read backwards, field lines leave an iron surface almost exactly along its normal, which is why the shape of a pole piece fixes the shape of the field in the air beside it.