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A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:

Asked in NEET 2025 · Displacement current equals conduction current

Answer: (1) Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

Step-by-step solution

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.

Why the other options are wrong

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