Problem 6.16A coaxial cable consists of two very long cylindrical tubes, separated by linear insulating material of magnetic susceptibility Xm- A current I flows down the inner conductor and returns along the outer one; in each case, the current dis- tributes itself uniformly over the surface (Fig. 6.24). Find the magnetic field in the region between the tubes. As a check, calculate the magnetization and the bound currents, and confirm that (together, of course, with the free currents) they generate the correct field. FIGURE 6.24 NOTE: (PLEASE SOLVE BY EXPLAİNİNG İNTERMEDİATE STEPS İN ANSWERS.) (ANSWERED) Problem 6.16 Xml ô. fH«dl = Isame = I, so H =ộ. B = Ho(1+Xm)H =| 4o(1 + Xm); -ộ. M= XmH| 2ns XmI z = 0. K, = Mxân at s = a; 2ma J, = V×M s ds 278 Xml2, at s = b. 2mb Total enclosed current, for an amperian loop between the cylinders: po(1+ Xm)1 ô.r Xm!, -2па — (1 + Хт)1, so I+ 2ла B · dl = µolenc = µo(1+Xm)I → B =

Problem 6.16A coaxial cable consists of two very long cylindrical tubes, separated by linear insulating material of magnetic susceptibility Xm- A current I flows down the inner conductor and returns along the outer one; in each case, the current dis- tributes itself uniformly over the surface (Fig. 6.24). Find the magnetic field in the region between the tubes. As a check, calculate the magnetization and the bound currents, and confirm that (together, of course, with the free currents) they generate the correct field. FIGURE 6.24 NOTE: (PLEASE SOLVE BY EXPLAİNİNG İNTERMEDİATE STEPS İN ANSWERS.) (ANSWERED) Problem 6.16 Xml ô. fH«dl = Isame = I, so H =ộ. B = Ho(1+Xm)H =| 4o(1 + Xm); -ộ. M= XmH| 2ns XmI z = 0. K, = Mxân at s = a; 2ma J, = V×M s ds 278 Xml2, at s = b. 2mb Total enclosed current, for an amperian loop between the cylinders: po(1+ Xm)1 ô.r Xm!, -2па — (1 + Хт)1, so I+ 2ла B · dl = µolenc = µo(1+Xm)I → B =