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Question #43072

Difficulty: 8/10 | 🟢 Free

The Question

An ideal LC circuit consists of an inductor and a parallel-plate capacitor with circular plates of radius RR and separation dd. The circuit oscillates with an angular frequency ω\omega. The current in the circuit is given by i(t)=I0cos⁡(ωt)i(t) = I_0 \cos(\omega t), where tt is time. Neglecting edge effects, what is the magnitude of the Poynting vector at the radial edge (r=Rr=R) of the region between the capacitor plates at time t=π6ωt = \frac{\pi}{6\omega}? (Here ϵ0\epsilon_0 is the vacuum permittivity)

Options

  • A.

    3I028π2ϵ0ωR3\frac{\sqrt{3} I_0^2}{8 \pi^2 \epsilon_0 \omega R^3}

  • B.

    3I028π2ϵ0ωR3\frac{3 I_0^2}{8 \pi^2 \epsilon_0 \omega R^3}

  • C.

    I028π2ϵ0ωR3\frac{I_0^2}{8 \pi^2 \epsilon_0 \omega R^3}

  • D.

    I024π2ϵ0ωR3\frac{I_0^2}{4 \pi^2 \epsilon_0 \omega R^3}

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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

LC Oscillations Energy Cycle|Maxwell's Correction to Ampere's Law|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to transition from lumped-circuit theory to field theory by evaluating the Poynting vector inside a capacitor.

A common misconception is treating the near-field electric and magnetic fields in a reactive circuit as being in phase, analogous to far-field electromagnetic plane waves (where S∝cos⁡2(ωt)S \propto \cos^2(\omega t)).

In reality, the electric field is proportional to the accumulated charge q(t)q(t), while the induced magnetic field (via the Ampere-Maxwell law) is proportional to the displacement current dq/dtdq/dt.

This derivative relationship introduces a strict π/2\pi/2 phase shift between E⃗\vec{E} and B⃗\vec{B}.

Students who incorrectly assume the fields are in phase will calculate the Poynting vector using cos⁡2(ωt)\cos^2(\omega t) or sin⁡2(ωt)\sin^2(\omega t), leading directly to the trap options.

The correct approach requires recognizing this phase quadrature to find the instantaneous power flow.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptLC Oscillations Energy Cycle
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree