23. (a) We refer to the (very large) wire length as L and seek to compute the ﬂux per meter: Φ B /L . Using the right-hand rule discussed in Chapter 30, we see that the net Feld in the region between the axes of antiparallel currents is the addition of the magnitudes of their individual Felds, as given by Eq. 30-19 and Eq. 30-22. There is an evident reﬂection symmetry in the problem, where the plane of symmetry is midway between the two wires (at what we will call x = `/ 2, where ` =20mm=0 . 020m) ;thenetFe ldatanypo in t0 <x<`/ 2 is the same at its “mirror image” point ` − x . The central axis of one of the wires passes through the origin, and that of the other passes through x = ` . We make use of the symmetry by integrating over 0 <x<`/ 2andthen multiplying by 2: Φ B=2 Z `/ 2 0 BdA=2 Z d/ 2 0 B ( Ldx )+2 Z `/ 2 d/ 2 B ( Ldx ) where d =0 . 0025 m is the diameter of each wire. We will use R = d/ 2, and r instead of x in the following steps. Thus, using the equations from Ch. 30 referred to above, we Fnd Φ B L=2 Z R0 µ µ 0 i 2 πR 2r + µ 0 i 2 π (
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.