Find the vector potential above and below the plane surface current in Ex.5.8. Refe...

Find the vector potential above and below the plane surface current in Ex.5.8.

Reference: Ex.5.8.

Find the magnetic field of an infinite uniform surface current K = K ˆx, flowing over the xy plane

(Fig.5.33)

Solution:

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,

(one Bl comes from the top segment and the other from the bottom), so B =(μ0/2)K, or, more precisely,

Notice that the field is independent of the distance from the plane, just like the electric field of a uniform surface charge