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Question #43294

Difficulty: 8/10 | 🟢 Free

The Question

A rigid wire frame consists of two identical square loops, Loop 1 (left) and Loop 2 (right), each of side length LL, sharing a common vertical branch. The common branch contains an ideal parallel-plate capacitor of capacitance CC. The left vertical branch of Loop 1 has a resistance RR, and the right vertical branch of Loop 2 has a resistance RR. All other wire segments have negligible resistance. The frame lies in the x−yx-y plane. A spatially uniform magnetic field B⃗1=4αtk^\vec{B}_1 = 4\alpha t \hat{k} is confined to the region of Loop 1, and B⃗2=αtk^\vec{B}_2 = \alpha t \hat{k} is confined to the region of Loop 2, where α>0\alpha > 0 is a constant and t≥0t \ge 0 is time. At t=0t=0, the capacitor is uncharged. Assuming the self-inductance of the loops is negligible, what is the magnitude of the displacement current IdI_d through the capacitor as a function of time tt?

Options

  • A.

    3αL2Re−2t/RC\frac{3\alpha L^2}{R} e^{-2t/RC}

  • B.

    5αL2Re−2t/RC\frac{5\alpha L^2}{R} e^{-2t/RC}

  • C.

    3αL2Re−t/RC\frac{3\alpha L^2}{R} e^{-t/RC}

  • D.

    5αL22Re−t/RC\frac{5\alpha L^2}{2R} e^{-t/RC}

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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

Magnetic Flux Scalar Sign|Kirchhoff’s Loop Rule|Maxwell's Correction to Ampere's Law|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to rigorously apply Faraday's Law and Kirchhoff's Laws in a multi-mesh circuit with time-varying magnetic fields.

The primary trap lies in the sign convention of the induced EMFs when transitioning from loop equations to nodal analysis.

Because both magnetic fields point in the same direction (+k^+\hat{k}), students often incorrectly assume the induced EMFs will simply add or subtract algebraically without tracing the actual mesh current directions.

A clockwise current in the right loop corresponds to an EMF pointing downwards in the right branch, which acts with opposite polarity to the left loop's EMF relative to the top node.

Failing to account for this spatial boundary condition leads to an incorrect effective driving voltage (e.g., 5αL25\alpha L^2 instead of 3αL23\alpha L^2) and an incorrect displacement current.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Induction - Class 12
ConceptMagnetic Flux Scalar Sign
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree