76. We take the current (i= 50 A) to flow in the +xdirection, and the electron to be at a point Pwhich is r= 0.050 m above the wire (where “up” is the +ydirection). Thus, the field produced by the current points in the +zdirection at P. Then, combining Eq. 29-4 with Eq. 28-2, we obtain GGFeirve=−×µ02πbgej±.k (a) The electron is moving down: Gvv±j (where v= 1.0 ×107m/s is the speed) so ( )
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