Irodov 3.230: Field Grows Inside, Then Levels Off – Amit & Deepali Sinha Foundation

Problem Irodov 3.230

An infinite plate of thickness $2d$ carries a uniform current density $j$ directed parallel to its faces. The magnetic permeability is $1$ inside and outside the plate. Find the magnetic induction as a function of the distance $x$ from the plate’s median plane.

The plate edge on, thickness 2d, with the uniform current density j out of the page; x is measured from the median plane (dashed). — The plate edge on, thickness $2d$ , with the uniform current density $\vect j$ out of the page; $x$ is measured from the median plane (dashed).

Given

plate thickness $2d$
uniform $j$ parallel to the faces
$\mu = 1$ everywhere

Find

B(x)

x

from the median plane

Solution

Step 1What symmetry leaves of the field

Take $j$ along $z$ (out of the page) and $x$ across the plate. Like a single current sheet, the plate is a stack of thin sheets, each giving a field parallel to the faces and perpendicular to $\vect j$ . So $\vect B$ is along $y$ , depends only on $x$ , and, since the plate looks the same from both sides when turned over about the $z$ axis, it is odd in $x$ :

B_y(-x) = -B_y(x), \qquad B_y(0) = 0 \tag{1}

Step 2Ampère’s law on a rectangle about the median plane

The Amperian rectangle (dashed) of height l with its sides at -x and +x; B points up on the right side and down on the left. — The Amperian rectangle (dashed) of height $l$ with its sides at $-x$ and $+x$ ; $\vect B$ points up on the right side and down on the left.

Take a rectangle in the $xy$ plane with its vertical sides of length $l$ at $-x$ and $+x$ , counterclockwise about $\vect j$ . The horizontal sides are perpendicular to $\vect B$ and contribute nothing; by (1) the two vertical sides contribute equally, so the circulation is $2B_y(x)\,l$ . For $x \le d$ the rectangle encloses the current $j\cdot 2x\,l$ :

2B_y(x)\,l = \mu_0\,j\cdot 2x\,l

For $x \ge d$ it encloses the whole thickness, $j\cdot 2d\,l$ :

2B_y(x)\,l = \mu_0\,j\cdot 2d\,l

B = \begin{cases} \mu_0 j |x|, & |x| \le d\\ \mu_0 j d, & |x| \ge d \end{cases}

(2)

Here $B$ is the magnitude on either side; with signs, $B_y = \mu_0 j x$ inside and $B_y = \pm\mu_0 j d$ outside. The field is parallel to the faces and perpendicular to $\vect j$ , along $\vect j\times\vect n$ with $\vect n$ the normal pointing away from the median plane: opposite on the two sides of it.

Top: the plate edge on, with B at four places, growing inside the plate and equal outside it. Bottom: B_y(x), linear through zero inside the plate and constant outside. — Top: the plate edge on, with $\vect B$ at four places, growing inside the plate and equal outside it. Bottom: $B_y(x)$ , linear through zero inside the plate and constant outside.

Check

The two forms agree at the face: at $x = d$ the inside form gives $\mu_0 j d$ , the outside value, so $B$ is continuous, as it must be with no surface current. Shrinking the plate to a sheet ( $d \to 0$ with $2jd = i$ fixed) gives $\mu_0 j d = \mu_0 i/2$ outside, the field of an infinite current sheet. Units: $\mu_0 j d$ is $(\text{T}\cdot\text{m/A})(\text{A/m}^2)(\text{m}) = \text{T}$ .

Remark

The slope of $B_y$ inside the plate is the current density itself, $dB_y/dx = \mu_0 j$ , which is the one component of $\nabla\times\vect B = \mu_0\vect j$ that survives here; outside, where $j = 0$ , the slope vanishes and the field stays flat, however far away.