By P. H Roberts
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For, if the net current is Jy , we have (either by integrating (26) or by using Ampere's circuital relation) bo(d) - bo(O) = j1Jy. Now by (1), (3) and continuity of B. we have B. = bo(d) everywhere in z ;;. d and B. ;;; O. When J y 'i' 0, bo(d) and bo(O) cannot both be zero, and the field energy per unit xy-area is infinite; the state (25) could therefore not be set up in a finite time. (O) = Bid) = 0 for all time; see (27) and (28). 2) It is easily seen from (10) that the nth fourier component Bxit) must satisfy dB xn ---;u = - I] (nn)2 d Bxn , (30) Le.
Suppose that at time t = the magnetic field in the slab is ° ° B o = (bo(z), 0, 0), (t ' 0; ° (t = 0; ° ~ ~ z d); (25) and the current density is jo = (o,tbo(Z),o). ~ Z ~ d). (26) We mustt suppose that the net current flowing in the layer is zero, and that (t = 0). (27) bo(O) = bo(d) = 0, We enquire how the fields and currents decay in time in the absence of any applied potential differences. x Insulator Fig. 2 The fields By, B z' Ex and E z will vanish everywhere for all times, and, in the insulator, B x must also vanish for all time,t so that Bx(O, t) = Bid, t) = 0.
3(a), we findt
An introduction to magnetohydrodynamics by P. H Roberts