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Sida. S44. Compute the vector magnetic potential within the outer conductor for the coaxial line whose vector magnetic potential is shown in Fig, 8.20 if the outer radius of the outer conductor is Ta. Select the proper zero reference and sketch the results on the figure: We do this by first finding B within the outer conductor and then “uncurling” the result. to find A. With redirected current / in the outer conductor, the current density is ff 7 Iau = - SSS 8s = mt alTa}? — m(baje “Ob Tine Sinee current J flows in both concuctors, but in opposite directions, Ampere's cirenital law inside the outer conductor gives: an gt | 49a" — pr = = Se trpH, = i- I ba ry me da! de a= np Dat Now, with B= ppjH, we note that Vx A will have a & component only, and from the direction and syimetry of the current, we expect A to be 2-directed, and to vary only with ». Therefore VxA= oA = oH cl and so F Az __ tol [dat o! dp lap 240? Then by direct integration, _ f tal (49) tal @ iat , A: -/ i8ap dp + | aera? de+C= 98 In p| + C As per Fig. 8.20, we establish a zero reference at o = 5e, enabling the evaluation of the integration constant: C- ~at = [25 = $8 In(Sa)] A, = col (4 25) + 98ln (*)| Wh/m A plot of this continues the plot of Fig. 8.20, in which the curve goes negative at = 5a, and then approaches a minimum of —.09ped/i at = Ta, at which point the slape becomes zero. Finally, By expanding Eq.(58), Sec. 8.7 in cartesian coordinates, show that (49) is correct. Eq. (58) can be rewritten as VAHV(V-A)-V XTRA We begin with aa, dA, , GA; Mie Sar ea Gz Then the « component of (WA) is WOW AQ). = 4 PA, i. Pa, A, ar? Geils 
