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Question #42919

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor consists of two circular plates of radius RR separated by a small distance dd. The space between the plates is filled with a leaky dielectric material with constant permittivity ϵ\epsilon. The conductivity of the dielectric varies radially as σ(r)=σ0rR\sigma(r) = \sigma_0 \frac{r}{R}, where rr is the distance from the central axis. At t=0t=0, a constant direct current I0I_0 begins to flow into the positive plate and out of the negative plate from an external circuit. Assuming the electric field magnitude EE between the plates is spatially uniform at any given instant and edge effects are negligible, what is the magnitude of the magnetic field BB at a distance rr (r<Rr < R) from the central axis as a function of time tt? (Assume quasi-static conditions where the induced electric field due to the time-varying magnetic field is negligible, and μ0\mu_0 is the vacuum permeability).

Options

  • A.

    μ0I02πr[r3R3+(r2R2−r3R3)e−2σ03ϵt]\frac{\mu_0 I_0}{2\pi r} \left[ \frac{r^3}{R^3} + \left( \frac{r^2}{R^2} - \frac{r^3}{R^3} \right) e^{-\frac{2 \sigma_0}{3 \epsilon} t} \right]

  • B.

    μ0I02πr[r3R3+(r2R2−r3R3)e−σ0ϵt]\frac{\mu_0 I_0}{2\pi r} \left[ \frac{r^3}{R^3} + \left( \frac{r^2}{R^2} - \frac{r^3}{R^3} \right) e^{-\frac{\sigma_0}{\epsilon} t} \right]

  • C.

    μ0I0r22πR3(1−e−2σ03ϵt)\frac{\mu_0 I_0 r^2}{2\pi R^3} \left( 1 - e^{-\frac{2\sigma_0}{3\epsilon} t} \right)

  • D.

    μ0I0r2πR2\frac{\mu_0 I_0 r}{2\pi R^2}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Flux and Surface Geometry|Maxwell's Correction to Ampere's Law

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of the Ampere-Maxwell Law in a system where both conduction and displacement currents coexist and vary spatially.

A common trap is assuming the conduction current enclosed by an Amperean loop is simply Jc×πr2J_c \times \pi r^2, ignoring the required area integral ∫Jc2πrdr\int J_c 2\pi r dr for a radially varying conductivity.

This leads to an incorrect time constant (Option B).

Another trap is ignoring the displacement current entirely (Option C) or assuming the magnetic field remains constant in time as it would in an ideal, non-leaky capacitor (Option D).

The student must correctly bridge Gauss's Law (to relate EE to total current) and Ampere's Law (to find BB) while handling non-trivial surface geometry integrations.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectric Charges and Fields - Class 12
ConceptFlux and Surface Geometry
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree