Irodov 3.278: A Loop Lying on a Magnet – Amit & Deepali Sinha Foundation

Problem Irodov 3.278

A round loop carrying a current lies in the flat boundary between vacuum and a magnetic material of permeability $\mu$ filling the other half-space. At some point on the loop’s axis the induction would be $\vect{B}_0$ if the material were removed. Find the induction $\vect{B}$ at that point with the material in place, and then extend the result to every point of the field.

The loop in a section through its axis: its two crossings of the page carry the current I, drawn out on the right and in on the left. Vacuum above, the magnetic (shaded, μ) below; P is a point on the axis. — The loop in a section through its axis: its two crossings of the page carry the current $I$ , drawn out on the right and in on the left. Vacuum above, the magnetic (shaded, $\mu$ ) below; $P$ is a point on the axis.

Given

loop in the boundary plane
permeability $\mu$ on one side, vacuum on the other
$\vect{B}_0$ at the point without the magnetic

Find

\vect{B}

at that point on the axis, then at every point

Solution

Step 1What the loop’s own field does at its plane

Let $\vect{B}_0$ now denote the loop’s field in vacuum at every point. A plane current distribution is symmetric under reflection in its own plane, and since $\vect{B}$ is an axial vector its parts parallel to the mirror change sign under the reflection while its normal part does not. At a point of the plane itself, which the reflection leaves in place, the parallel parts must therefore vanish:

\vect{B}_0 \perp \text{the boundary plane at every point of it (off the wire)} \tag{1}

Step 2A trial field

Try a field of the same shape as the vacuum field, scaled by a constant $k$ , the same in both half-spaces, with $\vect{H}$ obtained from each side’s permeability:

\vect{B} = k\,\vect{B}_0, \qquad \vect{H}_{\mathrm{vac}} = \frac{k\,\vect{B}_0}{\mu_0}, \qquad \vect{H}_{\mathrm{mag}} = \frac{k\,\vect{B}_0}{\mu\mu_0} \tag{2}

It satisfies every condition but one automatically. Its divergence is zero, as that of $\vect{B}_0$ is. Off the wire $\vect{H}$ has zero curl in each medium, as $\vect{B}_0$ has. At the boundary, by (1), $\vect{B}$ is purely normal and the same on both sides, so its normal component is continuous, and $\vect{H}$ has no tangential component on either side, so that is continuous too.

Step 3The current fixes the scale

The remaining condition is that the circulation of $\vect{H}$ round the wire equals the loop’s current $I$ . Take a small circle round the wire, half in each medium. For the vacuum field $\vect{B}_0$ each half contributes $\mu_0 I/2$ to the circulation of $\vect{B}_0$ (by the mirror symmetry), so with (2)

\oint \vect{H}\cdot d\vect{l} = \frac{k}{\mu_0}\cdot\frac{\mu_0 I}{2} + \frac{k}{\mu\mu_0}\cdot\frac{\mu_0 I}{2} = I

Cancelling $I/2$ ,

k\left(1 + \frac{1}{\mu}\right) = 2

and solving for $k$ ,

k = \frac{2\mu}{1 + \mu}

The trial field now satisfies all the field equations and boundary conditions, and these have only one solution, so it is the field. At the point on the axis, and in fact at every point of the field, on either side,

\vect{B} = \frac{2\mu}{1 + \mu}\,\vect{B}_0

(3)

The field strength, however, differs between the sides: by (2) it is $\vect{H} = 2\mu\vect{B}_0/(1 + \mu)\mu_0$ in the vacuum and $2\vect{B}_0/(1 + \mu)\mu_0$ in the magnetic.

The field lines have exactly the vacuum loop's shape and cross the boundary normally. On the axis B has the same length at mirror points above and below, while H_mag is 1/μ of H_vac (drawn for μ = 4). — The field lines have exactly the vacuum loop’s shape and cross the boundary normally. On the axis $\vect{B}$ has the same length at mirror points above and below, while $\vect{H}_{\mathrm{mag}}$ is $1/\mu$ of $\vect{H}_{\mathrm{vac}}$ (drawn for $\mu = 4$ ).

Check

With $\mu = 1$ the factor is $1$ and $\vect{B} = \vect{B}_0$ , as it must be with no magnetic. With $\mu \to \infty$ the factor tends to $2$ : a very permeable half-space acts as a mirror that adds an image of the loop with the same current, and an image of a loop lying on the mirror coincides with the loop itself, doubling the field. The same factor appears for a straight wire in a boundary between $\mu_1 = 1$ and $\mu_2 = \mu$ , where $B = (\mu_0 I/\pi r)\,\mu/(1 + \mu)$ is the vacuum value $\mu_0 I/2\pi r$ times $2\mu/(1 + \mu)$ .

Remark

The trick works because the boundary lies along the loop’s own symmetry plane, where the vacuum field already meets the boundary at right angles; the magnetic then changes only the field’s size, never its shape. Move the loop off the boundary and that is lost: the lines refract at the surface and the field needs an image loop of strength $(\mu - 1)/(\mu + 1)$ times the current on the vacuum side.