Irodov 3.277: One Field Shared by Two Media – Amit & Deepali Sinha Foundation

Problem Irodov 3.277

A very long straight wire carrying a current $I$ lies in the flat boundary between two non-conducting magnetic media, of permeabilities $\mu_1$ and $\mu_2$ , each filling a half-space. Find the magnitude of the magnetic induction everywhere as a function of the distance $r$ from the wire, using the fact that the lines of $\vect{B}$ are circles centred on the wire.

The wire seen end-on, current I (drawn out of the page), lying in the plane boundary between the medium of permeability μ_1 (above) and that of μ_2 (below). — The wire seen end-on, current $I$ (drawn out of the page), lying in the plane boundary between the medium of permeability $\mu_1$ (above) and that of $\mu_2$ (below).

Given

current $I$
permeabilities $\mu_1$ , $\mu_2$
wire in the boundary plane
$\vect{B}$ lines are circles about the wire

Find

B(r)

throughout space

Solution

Step 1B is the same on both halves of a circle

Take the field line of radius $r$ . It crosses the boundary plane at right angles, so where it does, $\vect{B}$ is normal to the boundary, and the normal component of $\vect{B}$ is continuous: $B$ has the same value just above and just below. Inside either medium $B$ cannot change along the line either, because a field directed along concentric circles has zero divergence only if its magnitude does not vary round them. So on the whole circle

B_1(r) = B_2(r) = B(r)

while $H$ differs between the halves:

H_1 = \frac{B}{\mu_1\mu_0}, \qquad H_2 = \frac{B}{\mu_2\mu_0} \tag{1}

Step 2The circulation of H round the circle

The media are non-conducting, so the only free current through the circle is the wire’s $I$ . Half the circle lies in each medium, and $\vect{H}$ is along it with the constant magnitude (1) on each half:

\oint \vect{H}\cdot d\vect{l} = H_1\,\pi r + H_2\,\pi r = I

Substituting (1),

\frac{B\,\pi r}{\mu_0}\left(\frac{1}{\mu_1} + \frac{1}{\mu_2}\right) = I

and solving for $B$ ,

B = \frac{\mu_0\,\mu_1\mu_2}{\mu_1 + \mu_2}\,\frac{I}{\pi r}

(2)

The lines circle the wire by the right-hand rule about $I$ .

The B lines are circles crossing the boundary unbroken; B is the same above and below at a given r, while H_1 and H_2 differ in the ratio μ_2/μ_1 (drawn for μ_1 = 2, μ_2 = 5). — The $\vect{B}$ lines are circles crossing the boundary unbroken; $\vect{B}$ is the same above and below at a given $r$ , while $\vect{H}_1$ and $\vect{H}_2$ differ in the ratio $\mu_2/\mu_1$ (drawn for $\mu_1 = 2$ , $\mu_2 = 5$ ).

Check

With $\mu_1 = \mu_2 = \mu$ , (2) gives $\mu\mu_0I/2\pi r$ , the field of a wire in a uniform medium. With $\mu_1 = 1$ and $\mu_2 \to \infty$ it gives $\mu_0I/\pi r$ , twice the vacuum value: $H_2$ then vanishes, so the whole circulation $I$ is taken by the vacuum half, $H_1\,\pi r = I$ , which is the same result.

Remark

Everywhere the field is that of the same wire in vacuum multiplied by $2\mu_1\mu_2/(\mu_1 + \mu_2)$ , the harmonic mean of the two permeabilities. The two half circles share one $B$ and their contributions to the circulation of $\vect{H}$ add, the way two equal lengths of different materials in series share one current and add their voltages. The extra field comes from the molecular current the media carry along the wire’s surface. A round loop lying in such a boundary has its whole field scaled by the same factor.