Chapter 30211P30.73Note that the current Iexists in the conductor with acurrent density JIA=, whereAaaaa=−−LNMOQP=π2222442.Therefore, JIa=22.To find the field at either point P1or P2, find Bswhichwould exist if the conductor were solid, using Ampère’slaw. Next, find B1and B2that wouldbe due to theconductors of radius a2that couldoccupy the void wherethe holes exist. Then use the superposition principle andsubtract the field that would be due to the part of theconductor where the holes exist from the field of thesolid conductor.θθθBs-B1-B2P1rrr2+a2()2a/2a/2Bs−′B 1B 2P2FIG. P30.73(a)At point P1, BJars=µπ022ej, Bra1022=−bgdi, and B2022=+.BB B BrBIrrraIrs=−−=−−−+LNMMOQPP=−−−LNMMMOQPPP=−−LNMOQP1220202114212224µafdirected to the left(b)At point P2, Brs=022and ′= ′=+BB02222.The horizontal components of ′B1and ′B2cancel while their vertical components add.BB BBrrBr
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