TRANSIENT ELECTROMAGNETIC RESPONSE OF A CONDUCTING INFINITE CYLINDER EMBEDDED IN A CONDUCTING MEDIUM SHRI KRISHNA SINGH* RESUMEN Sc obticne en la forma de integralcs definidas la rcspuesta clectromagnetica transitoria de un cilindro conductor permeable, incrustado en un espacio conductor infinito. La fuenk cs un cable infinito aislado que yace fuera del cilindro y que llcva una corrientc de Heaviside. las corrientes de dcsplazamiento no han sido tomadas en consideracion. Tambien se proporciona una expresi6n de la funci6n de respuesta transitoria bajo la aproximacion quasiestatica. ABSTRACT Transient electromagn~tic response of a conducting permeabk cylinder embedded in a conducting infinite space is obtained in the form of definite integrals. The source is an infinite insulated cable which lies outside the cylinder and carries a Heaviside current. The displacement currents have been neglected. Expression for transient response function under quasi-static approximation is also given. * Institute of Geophysics, U.N.A.M. 7
8 GEOFISICA INTERNACIONAL INTRODUCTION Recently it has been pointed out by several authors (e. g. Ward, 1971; Singh, 1972) that the effect of finite conductivity of the host rock must be considered in electromagnetic (e.m.) exploration of conductive massive sulphide ore bodies. This is particularly true for time-domain e.m. methods where, theoretically, all the freguencies are present. The author (Singh, 1972) has considered the transient response of a conducting, permeable ·sphere embedded in a conducting infinite space under an arbitrarily oriented magnetic dipole exitation. ln this paper the transient response of a conducting, permeable and homogeneous infinite cylinder embedded in a conducting infinite space is considered. The source is an infinite insulated cable which lies parallel to the axis of the cylinder and carries. a Heaviside current. The contributions from displacement-current have been neglected. The transient response, under quasi-static approximation is also considered. We shall obtain the time-domain solution by taking the inverse transform of the known frequency-domain solution which is given by Wait (1952). Frequency-domain solutions have also been given by Kertz ( I960) for a homogeneous cylinder under a uniform field, and by Negi (1962) and Negi et al. (1972) for inhomogeneous cylinders under uniform field and line source, respectively. These results have been specialized to the quasi-static case. Verma ( 1972) has given quasi-static time-domain response of a homogeneous cylinder under uniform field of step and ramp type excitation. An infinitely long current carrying cable in a conductive medium is a difficult source to realize in practice but the result of this paper may be useful (a) in evaluation of the validity of quasi-static approximation, (b) in construction of solution of a more realistic situation in which the source is on top of a conducting half space containing a cylindrical ore body, and (c) in marine exploration. FORMULATION AND SOLUTION Let an isotropic. homogeneous, infinite cylinder of radius a and of electrical properties a 2 µ 2 and E 2 be embedded in an inffoite space of electrical propertier a 1 µ 1 and E 1 . An infinitely long insulated cable, carrying current, lies at point S(r <f> ) parallel to the axis of the o, 0
GEOFISICA INTERNACIONAL 9 cylinder (Fig. I). The point S may be outside the cylinder (r0 > a) or inside the cylinder (r0 < a), Here we shall consider the former case only giving the secondary magnetic components in the outer medium The latter case can be treated in exactly a similar fashion. The expressions for the primary field components are given in Appendix A. A. FRECUENCY-OOMAIN Assuming a harmonically time-varying current I e iwt in the cable, the secondary magnetic field components in the outer medium, in MKS system of units, can be written as (see Wait, 1952 for details): I eiwt .. [ W (w)=-- r 2 1r r n=l I e iwt H:(w)=- --- .,, 2 1r r where, n=O sin n ( q, - ef)0 ) [ h5 (z)) (bd )n nr cos n (q, -- q, 0 ) (bd)n [ h~ <1> (z) I Nn(Z) h5 (z) = 2n (bdt K (Cdz) --- nr n Dn (z) Kn(Cbz) s Nn(Z) hn!J> (z) = - <\ (bd)n Cbz K 0 (Cdz) K~ (Cbz) Nn (z) Dn (z) -[ Dn(Z) In (Cz) I~ (z) -- CK I~ (~z) In (z) J Kn (Cz) I~ (z) ·- CK K~ (Cz) In (z) (l) (2) (3) ( 4) (5)
10 GEOFISICA INTERNACIONAL z2 = x2 a2 = (io µ - € 2) 2 u2 2 2W 2 µ2 W a r b=- a for n = 0 for n ~ 1 I (z) and K (z) are modified Bessel functions. n n (6) (7) (8) (9) (10) (11) If the conduction currents are much greater than the displacement currents, i.e. €. w ~ a. U = 1, 2), as is very often the case in exploration ge6physics, them the expressions for C and z given in equations (6) and (7) simplify to: (12)
and, where, GEOFISICA INTERNACIONAL z2 = iw/32 /32 = a µ a2 2 2 11 (13) (14) Applying asymptotic expansions of modified Bessel functions, it can be easly shown that, under quasi-static approximation (ICzi, ICbzl, !Cdzi ~ I), where, R,(z) ~ [ n~l z l~ (z) -- nK 10 (z) J z l~(z) + nK In (z) (15) which is the quasi-static response function. Note that there is an error in the quasi-static response function given by Wait ( 1952). B. TIME-DOMAIN Given secondary magnetic field components, under harmonic excitation, the transient response can be written as: n=l sin n(</1 - 1>0 ) (bd)" [h~. (t)] (16)
12 with, GEOFISICA INTERNACIONAL. - I H5(t) = --- (J) 2 ·,rr n = 0 h5 (t) = --- nr 27Ti hs (t) = -- nq, 21T i sin n(i/) - <Po) (bd)'1 [hs (t)) n<f> E+i~ f E - i~ f s st d h (s) <P (s) e s nr h5 (s) <P (s) est ds nq, f - i 00 (17) (18) ( 19) where s = iw, E is the real positive constant and rp(s) is the Laplac, transform of the primary input pulse. In this paper we shall assume a Heaviside pulse ( d, (s) = _L ). . s We shall neglect the contribution from displacement currents. Thus the solution would be valid for t ~ E f; U = 1, 2). From equation (13)wenotethat J I Z = S 112 {3 (20) It is easly shown that the integrands of equations ( 18), and ( 19), are double-valued functions of complex s. In order to make these integrands single-valued functions of s, so that we can evaluate the integrals using residue theorem, we introduce a branch cut along negative reals axis as shown in Fig. 2 and require that - 1T < arg s <1r. Now, considering the contour of Fig. 2, we have
GEOFISICA INTERNACIONAL 13 1 = 21ri[ sum of the residues] (21) It is not difficult to show that there are no poles of the integrands in or on the contour (see Singh, 1972 for a proof of a similar problem) and that the integral over the large circular arcs BC and FA vanish in the limit as the radius R ~ oo Thus Jim R ~ oo The integral around ED as 8 ~ 0 gives the static part of the solu· tion from the pole at the origin (which is also a branch point). But this part of the solution is not of interest in exploration and, therefore, we shall ignore it. Considering equation (18) in detail und substituting on DC ,therefore, z = u ei1r/2 and on FE u2 -i1r s=-- e {32 ,therefore, z = u e-i1r/2 we can write lim [ R ~ ~ _2_1r_i_ c st J s e h. (s) -- ds nr S J D -f 21Ti O N0 (iu) -u 21;a2 4n (bd) K (iCdu) ---K (iCbu) e ,_. O Dn (iu) " du (22) u 0
14 GEOFISICA INTERNACIONAL Using the following properties of modified Bessel functions 7f n K [z e±i_!!.] == ±1T ie+i -2-[ - J (z) ±i Y (z)] (23) n 2 2 n n we can write equation (22) in the following form: ~ -- n (bd)" f H(2) (Cdu) Pn(u)_ H(2)(Cbu) e·u2t / il2 ~ n Qn(U) n U (25) 0 where, P (u) = J (Cu) J , (u) - CK J, (Cu) J (u) (26) n n n n n · Q (u) = H< 2> (Cu) J , (u) - CK H<2>, (Cu) J (u} (27) n n n n n Similarily we can show that Jim [ 21T i J hs (s) e' 1 ~ J nr S F n (bd)" J (28) 0
GEOFISICA INTERNACIONAL 15 where bar on Qn (u) represents complex conjugate of Qn (u). Now, h\r (t), ignoring the static part, can be obtained by summing equations (25) and (28) and can be written as: h5 (t) = ·- 2n(bd)" Re nr [ J~ H(2) (Cdu) Pn (u2_ H(2) (Chu) e-u 2 t/132 duu-J (29) n Qn(U) n 0 Following a similar procedure as given above, we can write: h5 (t) = 5 (bd)" Re n(/) n [J CbH<2>(Cdu) Pn(u) n Qn (u) H;'' · (Cbu) e·"' ,1,' du J (30) where, Re in equations (29) and (30) means that the real part of the expression in the brackets must be taken. Equations (29) and (30) give the transient response functions for a Heaviside input pulse when displacement-currents are negligible. These are a function of following dimensionless parameters: t/{32 , K, C, Cb, Cd In numerical integration one must remember that the zeros of Qn (u) lie very close to the line of integration for small values of C. Since most of the contribution to the integrals come from this region, special care must be taken.
16 GEOFISICA INTERNACIONAL QUASI-STATIC TRANSIENT RESPONSE FUNCTION Quasi-static transient response function, R (t), for a Heaviside input pulse is given, from equation (15 ), by: n E + i~ 1 R (t) = --- n 2 7T i J R (s) e81 ~ n S (3 1) where Rn (s) is obtained from equation (15) by noting the relation given in equation (20). We shall consider non-permeable (K = 1) and permeable (K =I= 1) cylinder cases separately: Non-Permeable Case (K = 1 ): It is easily shown that Rn ( s) is a single-valued function of s and has simple poles at s = 0 (where the residue is zero) and at the zeros of In-I (z) = 0 (32) which are all imaginary. If z = iy 1 . is a zero, then z =-iY. 1 . is also a zero. By applying residue th~orJm and recurrence relatiorii of modified Bessel functions and noting the relation given in equation (24) we can show that ~ R (t) = 4n n [ j= I e-Y 2 n - I ,jt/{3 2 2 Yn -1,i (33) where iy 0 _1 i is a zero of I 0 _1 (z). When n = 1, the result is same as given by Verma (1972).
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