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Asked in NEET 2025 · Displacement current equals conduction current
Idea: use the Ampere-Maxwell law, ∮⃗B· d⃗l=μ₀(I+I_d), with the displacement current I_d=ε₀(dΦ_E)/(dt).
The surface charge density grows at a steady rate, so the field between the plates grows steadily and I_d is constant.
Inside the plate region, a circle of radius r<R encloses only the fraction (r²)/(R²) of it, giving B=(μ₀I_dr)/(2π R²), rising with r.
Outside, r>R, the whole displacement current is enclosed and B=(μ₀I_d)/(2π r), falling as 1/r.
So B is non-zero everywhere and peaks at r=R, the imaginary cylinder joining the plate rims.
The pattern is the same as for a solid current-carrying wire, with I_d playing the part of the conduction current.
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