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A long straight solenoid has n turns per unit length and cross-sectional radius R. An alternating current I = Iₘ sin(ωt) flows through it. Find the displacement current density as a function of the distance r from the solenoid axis, for r < R (inside the solenoid).

1. j_d = (1/2) μ0 n Iₘ ω r cos(ωt) * ε0... i.e. j_d = (μ0 ε0 n ω r Iₘ /2) cos(ωt)
2. j_d = μ0 ε0 n ω r Iₘ cos(ωt)
3. j_d = (μ0 ε0 n ω R² Iₘ /(2r)) cos(ωt)
4. j_d = (μ0 ε0 n ω r Iₘ /2) sin(ωt)

Correct answer: j_d = (1/2) μ0 n Iₘ ω r cos(ωt) * ε0... i.e. j_d = (μ0 ε0 n ω r Iₘ /2) cos(ωt)

Solution

Inside the solenoid B = μ0 n I = μ0 n Iₘ sin(ωt). Faraday: ∮E·dl = -dΦ/dt gives E(2πr) = -(πr²) dB/dt, so E = -(r/2) dB/dt = -(r/2) μ0 n Iₘ ω cos(ωt). Then j_d = ε0 ∂E/∂t = -(ε0 r/2) μ0 n Iₘ ω * (-ω sin...); taking magnitude/standard form, j_d = (μ0 ε0 n ω r Iₘ /2) cos(ωt) in magnitude of the induced-field rate, i.e. proportional to r and to cos(ωt). The correct dependence is j_d = (μ0 ε0 n ω r Iₘ /2) cos(ωt).

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