= Solution
Choose $+x$ along the <electric current>, $+z$ upward normal to the sheet, and $+y$ to complete a right-handed frame. In the interior region far from its edges, treat the sheet as infinite in its plane. Its <surface current density> is
$$
\mathbf K=\frac Iw\widehat{\mathbf x}.
$$
Reflection and translational <symmetry> make the <magnetic field> parallel to the sheet, perpendicular to its current, and equal in magnitude with opposite direction above and below. Choose an <Ampère's circuital law> rectangle with length $s$ along $y$, straddling the full thickness. The two long sides each contribute $Bs$ with the same sign; on the vertical sides $\mathbf B\cdot d\mathbf l=0$ since $\mathbf B$ is horizontal. The enclosed <electric current> is $Ks=Is/w$. Thus
$$
2Bs=\mu_0\frac Iw s,\qquad\boxed{B=\frac{\mu_0I}{2w}.}
$$
The <right-hand rule> fixes the directions:
$$
\boxed{\mathbf B_{\rm above}=-\frac{\mu_0I}{2w}\widehat{\mathbf y},\qquad\mathbf B_{\rm below}=+\frac{\mu_0I}{2w}\widehat{\mathbf y}.}
$$
The <current density> inside the finite-thickness conductor is $\mathbf J=I/(wt)\,\widehat{\mathbf x}$, so integrating through thickness $t$ gives the same $\mathbf K$. These are exterior fields in the broad-sheet approximation, not exact finite-width edge fields.
Back to article page
