Question 5.8: Find the magnetic field of an infinite uniform surface curre...

First of all, what is the direction of B? Could it have any x component? No: A glance at the Biot-Savart law (Eq. 5.42) reveals that B is perpendicular to K. Could it have a z component? No again. You could confirm this by noting that any vertical contribution from a filament at +y is canceled by the corresponding filament at −y. But there is a nicer argument: Suppose the field pointed away from the plane. By reversing the direction of the current, I could make it point toward the plane (in the Biot-Savart law, changing the sign of the current switches the sign of the field). But the z component of B cannot possibly depend on the direction of the current in the xy plane. (Think about it!) So B can only have a y component, and a quick check with your right hand should convince you that it points to the left above the plane and to the right below it.

With this in mind, we draw a rectangular Amperian loop as shown in Fig. 5.33, parallel to the yz plane and extending an equal distance above and below the surface. Applying Ampère’s law,

\pmb{B}(\pmb{r})=\frac{\mu _{0}}{4\pi}\int{\frac{\pmb{K}(\acute{\pmb{r}})\times \hat{\eta} }{\eta ^{2}} }d\acute{a}       and       \pmb{B}(\pmb{r})=\frac{\mu _{0}}{4\pi}\int{\frac{\pmb{J}(\acute{\pmb{r}})\times \hat{\eta} }{\eta ^{2}} }d\acute{\tau },            (5.42)

(one Bl comes from the top segment and the other from the bottom), so B =(μ_{0}/2)K, or, more precisely

\pmb{B}=\begin{cases} +(μ_{0}/2)  K  \hat{y}        for      z < 0,    \\ – (μ_{0}/2)     K    \hat{y}        for       z > 0 .\end{cases}                    (5.58)

Notice that the field is independent of the distance from the plane, just like the
electric field of a uniform surface charge (Ex. 2.5).