Irodov 3.229: A Current Sheet Halves Its Jump – Amit & Deepali Sinha Foundation

Problem Irodov 3.229

Find the magnitude and the direction of the magnetic induction $\vect B$ produced by

an infinite plane carrying a surface current whose linear density $\vect i$ (current per unit width) is the same vector at every point of the plane;
two parallel infinite planes, one carrying the uniform linear current density $\vect i$ and the other the uniform density $-\vect i$ .

(a) The plane seen edge on; its current density i runs into the page. — (a) The plane seen edge on; its current density $\vect i$ runs into the page.

(b) Two parallel planes edge on, with current densities i (into the page) and - i (out of it). — (b) Two parallel planes edge on, with current densities $\vect i$ (into the page) and $-\vect i$ (out of it).

Given

(a) one plane, uniform $\vect i$
(b) two parallel planes, $\vect i$ and $-\vect i$

Find

\vect B

, magnitude and direction, everywhere

Solution

Step 1What symmetry allows for one plane

Let $\vect n$ be the unit normal pointing from the plane toward the field point. The plane is unchanged by any shift along itself, so $\vect B$ depends only on the distance from it. The field of each current element is perpendicular to the element, so $\vect B$ has no component along $\vect i$ . A normal component is ruled out too: rotating the plane by $180^\circ$ about an axis along $\vect i$ leaves the currents as they were but swaps the two sides, so the normal component would have to point away from the plane on both sides or toward it on both, and then the flux of $\vect B$ out of a box enclosing a piece of the plane would not vanish. What is left is a field parallel to the plane and perpendicular to $\vect i$ , opposite on the two sides:

\vect B = B\,\frac{\vect i\times\vect n}{i} \tag{1}

with $B$ the same at every distance, as the next step shows.

Step 2Plane (a): Ampère’s law on a rectangle

(a) The field on both sides of the plane, along i· n: parallel to the plane, opposite on the two sides. The dashed rectangle, the Amperian contour, straddles the plane and is traversed counterclockwise. — (a) The field on both sides of the plane, along $\vect i\times\vect n$ : parallel to the plane, opposite on the two sides. The dashed rectangle, the Amperian contour, straddles the plane and is traversed counterclockwise.

Take a rectangle in the plane perpendicular to $\vect i$ , with two sides of length $l$ parallel to the current plane, one on each side of it. The other two sides cross the plane perpendicularly and contribute nothing by (1). Go round it counterclockwise, as the figure marks, with $\vect i$ into the page. On each side of length $l$ the field then runs against the direction of travel (to the right above the plane while the path goes left, to the left below it while the path goes right), so the circulation is $-2Bl$ . The positive normal of a counterclockwise contour points out of the page, against $\vect i$ , so the enclosed current counts as $-i\,l$ :

-B\,l - B\,l = -\mu_0\,i\,l

\vect B = \frac{\mu_0}{2}\,\vect i\times\vect n

(2)

The magnitude is $B = \mu_0 i/2$ at every distance from the plane. The direction is parallel to the plane and perpendicular to $\vect i$ , and it reverses across the plane: seen with $\vect i$ running into the page, $\vect B$ points to the right above the plane and to the left below it.

Step 3Planes (b): superposition

(b) Between the planes the two fields point the same way and add; outside they point opposite ways and cancel.

Each plane gives (2). Let $\vect n_0$ be the unit normal pointing from the plane carrying $\vect i$ toward the plane carrying $-\vect i$ . Between the planes the field point is on the $\vect n_0$ side of the first plane and on the $-\vect n_0$ side of the second:

\vect B = \frac{\mu_0}{2}\,\vect i\times\vect n_0 + \frac{\mu_0}{2}\,(-\vect i)\times(-\vect n_0) = \mu_0\,\vect i\times\vect n_0

Outside, beyond the second plane, the point is on the $\vect n_0$ side of both, and beyond the first on the $-\vect n_0$ side of both:

\vect B = \pm\frac{\mu_0}{2}\left(\vect i\times\vect n_0 + (-\vect i)\times\vect n_0\right) = 0

\begin{gathered}\vect B = \mu_0\,\vect i\times\vect n_0 \ \text{between the planes}\\ \vect B = 0 \ \text{outside}\end{gathered}

(3)

Between the planes the magnitude is $B = \mu_0 i$ , and $\vect B$ is parallel to the planes and perpendicular to the currents.

Check

The jump of the tangential field across a current sheet must equal $\mu_0$ times the sheet’s current density. For (a), the field goes from $\mu_0 i/2$ one way to $\mu_0 i/2$ the other, a jump of $\mu_0 i$ . In (b), across each plane the field goes from $0$ outside to $\mu_0 i$ between, again a jump of $\mu_0 i$ . Units: $\mu_0 i$ is $(\text{T}\cdot\text{m/A})(\text{A/m}) = \text{T}$ .

Remark

Pair (b) is the magnetic counterpart of a parallel-plate capacitor: two sheets of opposite source strength confine the whole field to the gap and leave none outside. A long solenoid cut along its axis shows the same pattern: the winding crosses the cut at the top and at the bottom with opposite currents, and its inside field $\mu_0 n I$ is exactly $\mu_0 i$ , with $i = nI$ the current per unit length of the winding.