Example 6-7: Displacement CurrentDensityThe conduction current flowing through a wire with conductivity σ = 2 × 107 S/m and relative permittivity ‘r = 1 is givenby Ic = 2 sin ωt (mA). If ω = 109 rad/s, find the displacementcurrent.Solution: The conduction current Ic = J A = σEA, where Ais the cross section of the wire. Hence,E = IcσA = 2 × 10−3 sin ωt2 × 107A = 1 × 10−10Asin ωt (V/m).Application of Eq. (6.44), with D = ‘E, leads toId = JdA = ‘A∂E∂t = ‘A∂∂t!1 × 10−10Asin ωt”= ‘ω × 10−10 cos ωt = 0.885 × 10−12 cos ωt (A),where we used ω = 109 rad/s and ‘ = ‘0 = 8.85×10−12 F/m.Note that Ic and Id are in phase quadrature (90◦ phase shiftbetween them). Also, Id is about nine orders of magnitudesmaller than Ic, which is why the displacement current usuallyis ignored in good conductors.

image 135 - Example 6-7: Displacement CurrentDensityThe conduction current flowing through a wire with conductivity σ = 2 × 107 S/m and relative permittivity 'r = 1 is givenby Ic = 2 sin ωt (mA). If ω = 109 rad/s, find the displacementcurrent.Solution: The conduction current Ic = J A = σEA, where Ais the cross section of the wire. Hence,E = IcσA = 2 × 10−3 sin ωt2 × 107A = 1 × 10−10Asin ωt (V/m).Application of Eq. (6.44), with D = 'E, leads toId = JdA = 'A∂E∂t = 'A∂∂t!1 × 10−10Asin ωt"= 'ω × 10−10 cos ωt = 0.885 × 10−12 cos ωt (A),where we used ω = 109 rad/s and ' = '0 = 8.85×10−12 F/m.Note that Ic and Id are in phase quadrature (90◦ phase shiftbetween them). Also, Id is about nine orders of magnitudesmaller than Ic, which is why the displacement current usuallyis ignored in good conductors.
This content is for Premium members only.
sign up for premium and access unlimited solutions for a month at just 5$(not renewed automatically) images - Example 6-7: Displacement CurrentDensityThe conduction current flowing through a wire with conductivity σ = 2 × 107 S/m and relative permittivity 'r = 1 is givenby Ic = 2 sin ωt (mA). If ω = 109 rad/s, find the displacementcurrent.Solution: The conduction current Ic = J A = σEA, where Ais the cross section of the wire. Hence,E = IcσA = 2 × 10−3 sin ωt2 × 107A = 1 × 10−10Asin ωt (V/m).Application of Eq. (6.44), with D = 'E, leads toId = JdA = 'A∂E∂t = 'A∂∂t!1 × 10−10Asin ωt"= 'ω × 10−10 cos ωt = 0.885 × 10−12 cos ωt (A),where we used ω = 109 rad/s and ' = '0 = 8.85×10−12 F/m.Note that Ic and Id are in phase quadrature (90◦ phase shiftbetween them). Also, Id is about nine orders of magnitudesmaller than Ic, which is why the displacement current usuallyis ignored in good conductors. already a member please login

8   +   1   =