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Question #43076

Difficulty: 8/10 | 🟢 Free

The Question

A long straight wire carries an alternating current I(t)=I0sin⁡(ωt)I(t) = I_0 \sin(\omega t), where I0I_0 is the current amplitude in Amperes and ω\omega is the angular frequency in rad/s. A rigid rectangular conducting loop of side lengths aa and bb (in meters) is placed in the same plane as the wire. The side of length bb is parallel to the wire and is at a distance dd (in meters) from it. The loop has a constant self-inductance LL (in Henries). Due to high-frequency skin effects, the effective resistance of the loop is strictly proportional to the angular frequency, given by R=αωR = \alpha \omega, where α\alpha is a positive constant in Henries. Let M=μ0b2πln⁡(1+ad)M = \frac{\mu_0 b}{2\pi} \ln\left(1 + \frac{a}{d}\right) be the mutual inductance (in Henries) between the wire and the loop. What is the total electrical energy EE dissipated as heat in the loop during exactly one complete cycle of the alternating current?

Options

  • A.

    παM2I02α2+L2 J\frac{\pi \alpha M^2 I_0^2}{\alpha^2 + L^2} \text{ J}

  • B.

    πM2I02α J\frac{\pi M^2 I_0^2}{\alpha} \text{ J}

  • C.

    πLM2I02α2+L2 J\frac{\pi L M^2 I_0^2}{\alpha^2 + L^2} \text{ J}

  • D.

    2παM2I02α2+L2 J\frac{2\pi \alpha M^2 I_0^2}{\alpha^2 + L^2} \text{ J}

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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

Transformer Flux Linkage|Dynamic Resistance (AC Resistance)|Magnetic Flux Scalar Sign

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to integrate Faraday's Law with frequency-dependent AC circuit analysis, specifically targeting the misconception that higher driving frequencies always result in higher energy dissipation per cycle.

Students often blindly apply P=Erms2/RP = \mathcal{E}_{rms}^2 / R while ignoring the inductive reactance XLX_L, leading to the incorrect assumption that inductance doesn't limit the current (Option B).

Alternatively, they might confuse active power with reactive power, using sin⁡ϕ\sin \phi instead of the power factor cos⁡ϕ\cos \phi (Option C).

The core trap lies in realizing that because both the dynamic resistance (R=αωR = \alpha \omega) and the inductive reactance (XL=ωLX_L = \omega L) scale linearly with ω\omega, the impedance ZZ scales with ω\omega, making the induced current amplitude and the energy dissipated per cycle completely independent of the driving frequency.

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Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptTransformer Flux Linkage
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree