Exercise 4.1.13: Infinite current sheet

The sheet is invariant under translations in $x$ and $y$, so $H$ depends on $z$ only. Each strip $dy$ of the sheet is a line current along $x$ whose field circles it in the $yz$-plane. By Biot–Savart's law $J_S\times a_R$ has no $x$-component, and the $z$-components of the strips at $+y$ and $-y$ cancel. Hence $H=H_y(z)a_y$, and by the mirror symmetry $z\to -z$ (which reverses the circulation) $H_y(-z)=-H_y(z)$.

Take the rectangle in the $yz$-plane with sides of length $\ell$ parallel to $y$ at $z=\pm h$, traversed so that the enclosed current $K\ell$ (along $+x$) is positive, i.e. counter-clockwise when seen from $+x$: along $+a_y$ at $z<0$ and along $-a_y$ at $z>0$. The $z$-sides contribute nothing ($H\perp dl$):

with $a_n$ the unit normal pointing from the sheet to the field point. The field is uniform and does not depend on the distance, exactly like $E$ of a charged plane. Numbers: $H=500$ A/m, $B=0.63$ mT on each side.

Superpose: between the sheets ($0<z<d$) the two contributions, $-\frac{K}{2}{a}_y$ from the lower sheet and $-\frac{K}{2}{a}_y$ from the upper one (its current is $-K{a}_x$ and the field point is below it), add. Outside they cancel:

With $K=1000$ A/m, $B=1.26$ mT between the plates, zero outside.

Units: A/m. This is the field inside a solenoid seen from close up: the winding is a cylindrical current sheet $K=nI$, the field $nI$ inside and $0$ outside (Home exercise 4.1.7). The tangential component of $H$ jumps by $K$ across a current sheet, $a_n\times (H_1-H_2)=J_S$, which is the magnetic boundary condition of the next section.