Simple Nature - Light and Matter

circulates around the y axis, so at the point of interest, it’s reallythe x component of the field that we want to compute:∫dB x =dB R cos α∫ kI dl dx=c 2 s 3 (3 sin θ cos θ cos α)∫3kI dl ∞1( xz)=c 2 0 s 3 s 2 dx∫3kIz dl ∞x=c 2 (x 2 + r 2 dx) 5/20kI dl z=c 2 r 3kI dl sin φ=c 2 r 2m / The field of an infinite U.In the more general case, l, the current loop is not planar, the pointof interest is not in the end-planes of the U’s, and the U shapeshave their ends staggered, so the end-piece dl is not the only part ofeach U whose current is not canceled. Without going into the gorydetails, the correct general result is as follows:dB =kI dl × rc 2 r 3 ,n / The geometry of the Biot-Savart law. The small arrowsshow the result of the Biot-Savartlaw at various positions relativeto the current segment dl. TheBiot-Savart law involves a crossproduct, and the right-handrule for this cross product isdemonstrated for one case.which is known as the Biot-Savart law. (It rhymes with “leo bazaar.”Both t’s are silent.) The distances dl and r are now defined asvectors, dl and r, which point, respectively, in the direction of thecurrent in the end-piece and the direction from the end-piece tothe point of interest. The new equation looks different, but it isconsistent with the old one. The vector cross product dl × r has amagnitude r dl sin φ, which cancels one of r’s in the denominatorand makes the dl × r/r 3 into a vector with magnitude dl sin φ/r 2 .The field at the center of a circular loop example 11Previously we had to do quite a bit of work (examples 9 and 10),to calculate the field at the center of a circular loop of current ofradius a. It’s much easier now. Dividing the loop into many shortsegments, each dl is perpendicular to the r vector that goes fromit to the center of the circle, and every r vector has magnitude a.Therefore every cross product dl × r has the same magnitude,a dl, as well as the same direction along the axis perpendicular670 Chapter 11 Electromagnetism