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A long straight cable of length l is placed symmetrically along z-axis and has radius a(ltltl). The cable consists of a thin wire and co-axial conductingtube. An alternating current I(t)=I_0sin (2pivt) flows down the central thin wire and return along the co-axial conducting tube. The induced elelctric field at a distance s form the wire inside the cable isvecE(s,t)=mu_0I_0 v cos (2pivt)log_e(s/a)hatk (i) Calculate the displacement current density inside the cable. (ii) Integrate the displacement current density across the cross-section of the cable to find the total displacement current I^d. (iii) Compare the conduction current I_0 with the displacement current I_0^d. |
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Answer» Solution :(i) Given, the induced electric field at a distance r from the wire INSIDE the cable is `vecE(s,t)=mu_0I_0 v cos (2pivt)log_e(s/a)hatk` DISPLACEMENT current density, `vecJ_D=in_0 (vec(dE))/(dt)=in_0 d/(dt)[mu_0I_0 v cos (2pivt)log_e(s/a)hatk]` `=in_0 mu_0 I_0 v d/(dt)[cos 2pivt]log_e (s/a)hatk=1/(c^2) I_0 v^2 2pi[-sin 2pivt]loe_e (s/a)hatk` `=(v^2)/(c^2)2piI_0 sin 2pivt log_e (a/s)hatk[ :' log_e(s/a)=-log_e(a/s)]` `=1/(lambda^2) 2pi I_0 log_e(a/s) sin 2pivthatk=(2piI_0)/(lambda^2) log_e a/s sin 2pivt hatk` (ii) `I_D=int J_D s DS d theta=int_(s=0)^a int_0^(2pi) d theta=int_(s=0)^a J_Dsdsxx2pi` `=int_0^a[(2pi)/(lambda^2)I_0log_e(a/s)s ds sin 2pivt]xx2pi=((2pi)/lambda)^2I_0 int_(s=0)^alog_e(a/s)s ds sin 2pivt` `((2pi)/lambda)^2I_0 int_(s=0)^alog_e(a/s) 1/2d(s^2).sin 2pivt=(a^2)/2((2pi)/lambda)^2I_0 sin 2pivt int_(s=0)^a log_e(a/s). d(s/a)^2` ` (a^2)/4((2pi)/lambda)^2I_0 sin 2pivt int_(s=0)^a log_e(a/s). d(s/a)^2=-(a^2)/4 ((2pi)/lambda)^2I_0 sin 2pivt int_(s=0)^a log_e(a/s). d(s/a)^2` `=-(a^2)/4 ((2pi)/lambda)^2I_0 sin 2pivtxx(-1)[ :' int_(s=0)^a log_e(a/s). d(s/a)^2=-1]` `I_D=(a^2)/4 ((2pi)/lambda)^2I_0 sin 2pivt=((2pia)/(2lambda))^2I_0sin 2pivt` (iii) The diplacementcurrent, `I_D=((2pia)/(2lambda))^2I_0sin 2pivt=I_(0D)sin 2pivt` where `I_(0D)=((2pia)/(2lambda))^2I_0=((pia)/(lambda))^2I_0 :. (I_0D)/(I_0)=((API)/(lambda))^2` |
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