2012 Semifinal Exam Part A 9 b.We can apply the results of the previous section to obtain the enclosed charge density Aenc as a function of radius: (0 r<a a<r<2a 入end 2a <r 3a 入 3<r<4a 0 r>4a Defining =0-) we conclude that a charge density-A;exists on the inner surface of the dielectric,a charge density Ai on the outer surface,and no charge on the interior. As with the case of a very long solenoid,we expect the magnetic field to be entirely parallel to the cylinder axis,and to go to zero for large r.Consider an Amperian loop of length l extending along a radius,the inner side of which is at radius r and the outer side of which is at a very large radius.We have on this loop B dl=Holenct Letting B now be the magnetic field at radius r, IB Holencl B=Holend For r>3a,Iend=0,since the charge on the hollow cylinders is not moving.For 2a<r<3a, the loop now encloses the outer surface of the dielectric.In timea chargel passes through the loop,so the current due to the outer surface is lou =lw 2T and thus this is Ienc for 2a <r<3a.For r<2a,the loop now encloses both surfaces of the dielectric;the inner surface contributes a current that exactly cancels the outer one,so again Tencl =0.Putting this together, 0 r<2a B= 2a<r<3a r>3a or,using our expression forλ， 0 r<2a ) L0@入 2 2a <r 3a r>3a Copyright C2012 American Association of Physics Teachers2012 Semifinal Exam Part A 9 b. We can apply the results of the previous section to obtain the enclosed charge density λencl as a function of radius: λencl =    0 r < a λ a < r < 2a λ κ 2a < r < 3a λ 3 < r < 4a 0 r > 4a Defining λi =  1 − 1 κ  λ we conclude that a charge density −λi exists on the inner surface of the dielectric, a charge density λi on the outer surface, and no charge on the interior. As with the case of a very long solenoid, we expect the magnetic field to be entirely parallel to the cylinder axis, and to go to zero for large r. Consider an Amperian loop of length l extending along a radius, the inner side of which is at radius r and the outer side of which is at a very large radius. We have on this loop I B dl = µ0Iencl Letting B now be the magnetic field at radius r, lB = µ0Iencl B = µ0Iencl l For r > 3a, Iencl = 0, since the charge on the hollow cylinders is not moving. For 2a < r < 3a, the loop now encloses the outer surface of the dielectric. In time 2π ω a charge λi l passes through the loop, so the current due to the outer surface is Iout = λi lω 2π and thus this is Iencl for 2a < r < 3a. For r < 2a, the loop now encloses both surfaces of the dielectric; the inner surface contributes a current that exactly cancels the outer one, so again Iencl = 0. Putting this together, B =    0 r < 2a µ0ω 2π λi 2a < r < 3a 0 r > 3a or, using our expression for λi , B =    0 r < 2a ￾ 1 − 1 κ
µ0ωλ 2π 2a < r < 3a 0 r > 3a Copyright c 2012 American Association of Physics Teachers