Pradeep Asked a Question
August 5, 2021 11:45 ampts 30 pts
B 235 Amperian loop Sheet of current K B Amperian loop FIGURE 5.32 FIGURE 5.33 Example 5.8. Find the magnetic field of an infinite uniform surface current K = K*, 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 fromn 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 ay 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, B dl= 2Bl = uolene=uoKl, (one B comes from the top segment and the other from the bottom), so B = (Ao/2) K, or, more precisely, B - +(Mo/2) Ky for z< 0, -(4o/2) Ký for z > 0. (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).
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  • Bidyut sinha thankyou
    please see the attached picture.
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