A silver wire has resistivity ? = 1.60 × 10-8 O·m and a cross-sectional area of
ID: 2016349 • Letter: A
Question
A silver wire has resistivity ? = 1.60 × 10-8 O·m and a cross-sectional area of 4.90 mm2. The current in the wire is uniform and changing at the rate of 2150 A/s when the current is 140 A. (a) What is the magnitude of the (uniform) electric field in the wire when the current in the wire is 140 A? (b) What is the displacement current in the wire at that time? (c) What is the ratio of the magnitude of the magnetic field due to the displacement current to that due to the current at a distance r from the wire?Explanation / Answer
(a) Electric field E = J = I / A ...............(1) Here = resistivity = 1.6*10-8 .m, current I = 140 A Crosectional area A = 4.9*10-6 m2 plug all values in equation (1) we get , E = 0.457 V /m (b) displacement current Id = 0 * (d /dt) = 0 A * (dE /dt) Id = 0 A (d (I / A) /dt) = 0 (dI /dt) .......................(2) Here 0 = 8.85*10-12 N .m2 /C2 , = 1.6*10-8 .m , dI /dt =2150 A/s plug all values in (2) we get Id = 0.3*10-15 A = 3.0*10-16 A (c) magnetic field B due to I = 0 I /2r B due to Id = 0 Id / 2r B due to Id / B due to I = Id / I = 3.0*10-16 A / 140 A = 2.1*10-18 Here = resistivity = 1.6*10-8 .m, current I = 140 A Crosectional area A = 4.9*10-6 m2 plug all values in equation (1) we get , E = 0.457 V /m (b) displacement current Id = 0 * (d /dt) = 0 A * (dE /dt) Id = 0 A (d (I / A) /dt) = 0 (dI /dt) .......................(2) Here 0 = 8.85*10-12 N .m2 /C2 , = 1.6*10-8 .m , dI /dt =2150 A/s plug all values in (2) we get Id = 0.3*10-15 A = 3.0*10-16 A (c) magnetic field B due to I = 0 I /2r B due to Id = 0 Id / 2r B due to Id / B due to I = Id / I = 3.0*10-16 A / 140 A = 2.1*10-18Related Questions
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