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Question #43255

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor consists of two circular plates of radius aa separated by a distance dd. The space between the plates is completely filled with a lossy dielectric material of uniform permittivity ϵ\epsilon and uniform conductivity σ\sigma. An alternating voltage V(t)=V0sin⁡(ωt)V(t) = V_0 \sin(\omega t) is applied across the plates. At a specific angular frequency ω0\omega_0, the amplitude of the conduction current density in the dielectric equals the amplitude of the displacement current density. What is the amplitude of the magnetic field B0B_0 at a radial distance rr (r<ar < a) from the central axis of the plates when the applied voltage has an angular frequency ω=3ω0\omega = \sqrt{3}\omega_0?

Options

  • A.

    μ0rσV0d\frac{\mu_0 r \sigma V_0}{d}

  • B.

    3μ0rσV02d\frac{\sqrt{3} \mu_0 r \sigma V_0}{2 d}

  • C.

    (1+3)μ0rσV02d\frac{(1+\sqrt{3}) \mu_0 r \sigma V_0}{2 d}

  • D.

    μ0rσV02d\frac{\mu_0 r \sigma V_0}{\sqrt{2} d}

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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

Ampere’s Law Displacement Current|Capacitive Reactance Dependency|Frequency Sensitivity of Z

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Ampere-Maxwell Law in a lossy dielectric medium, bridging microscopic field theory with macroscopic AC circuit behavior.

A common trap is treating the capacitor as purely reactive (ignoring conduction current entirely, leading to Option B) or algebraically adding the amplitudes of conduction and displacement currents (leading to Option C).

Students must recognize that conduction current (in phase with the electric field) and displacement current (proportional to the time derivative of the electric field) are orthogonal in phase space.

Consequently, their amplitudes must be added in quadrature (phasor addition) to find the total enclosed effective current driving the induced magnetic field.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMoving Charges and Magnetism - Class 12
ConceptAmpere’s Law Displacement Current
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree