Irodov 3.231: The Plane Shields Its Underside Completely

Problem Irodov 3.231

A very long straight wire meets an infinite conducting plane at right angles, at the point $O$ , and lies entirely on one side of the plane. A direct current $I$ flows along the wire toward $O$ and from there spreads radially outward over the whole plane. Find the magnetic induction at every point of space.

The wire and the plane, side on: the current I comes down the wire to O and spreads outward over the plane (the horizontal line). — The wire and the plane, side on: the current $I$ comes down the wire to $O$ and spreads outward over the plane (the horizontal line).

Given

current $I$
long wire ending at $O$ , perpendicular to the plane
radial spreading over the plane

Find

\vect B

everywhere

Solution

Step 1The field circles the wire’s axis

Let the axis be the line of the wire, continued through the plane. The currents are symmetric about it, so $B$ depends only on the distance $r$ from the axis and on the height above or below the plane. Every current, in the wire and on the plane, flows in a plane containing the axis. A reflection in any such plane leaves the whole current system unchanged, and $\vect B$ , an axial vector, must then be perpendicular to that plane at its own points.

So $\vect B$ is tangent to circles about the axis, with a magnitude constant round each circle.

Step 2Ampère’s law on a circle above the plane

Take a circle of radius $r$ about the axis, parallel to the plane, on the wire’s side. The flat disc spanning it is crossed by the wire and by nothing else, so it carries the current $I$ :

B\cdot 2\pi r = \mu_0 I

B = \frac{\mu_0 I}{2\pi r}

(1)

For the current running toward the plane, the right-hand rule makes $\vect B$ circulate clockwise seen from the wire’s side.

Step 3Ampère’s law on a circle below the plane

A circle of radius $r$ on the other side of the plane spans a flat disc that no current crosses: the wire ends at $O$ and the plane’s currents stay in the plane.

B\cdot 2\pi r = 0

The wire above the plane with circles of B about it, clockwise seen from above; the radial current on the plane; no field on the other side. — The wire above the plane with circles of $\vect B$ about it, clockwise seen from above; the radial current on the plane; no field on the other side.

B = \begin{cases} \dfrac{\mu_0 I}{2\pi r} & \text{on the wire's side of the plane}\\[6pt] 0 & \text{on the other side} \end{cases}

(2)

Here $r$ is the distance from the axis of the wire. On the wire’s side $\vect B$ is tangent to circles about the axis, in the right-hand sense of the current in the wire: clockwise seen from the wire’s side when the current flows toward the plane.

Check

Across the plane the tangential field must jump by $\mu_0$ times the surface current density. At distance $r$ from $O$ the current $I$ crosses a circle of circumference $2\pi r$ , so the surface density is $I/(2\pi r)$ , and $\mu_0 I/(2\pi r)$ is exactly the jump from the value (1) above the plane to zero below it.

Remark

On its own side the field is exactly that of an infinitely long wire, even though the wire stops at $O$ : the radial currents on the plane supply what the missing half of the wire would have given there, and on the far side they cancel the stub’s field completely. Ampère’s law settles this in two lines, where Biot and Savart would need an integral over the whole plane.