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Question #43282

Difficulty: 8/10 | 🟢 Free

The Question

An LC oscillator circuit consists of an ideal capacitor of capacitance CC and an inductor of inductance LL. The inductor is a solenoid of length ll, cross-sectional area AA, and NN turns, completely filled with a perfectly laminated magnetic core of permeability μ\mu (such that L=μN2AlL = \frac{\mu N^2 A}{l}). The core material exhibits a temperature-dependent hysteresis B-H loop, transitioning from a soft to a hard magnetic material as the absolute temperature TT increases. The area of the B-H loop (representing the energy dissipated per unit volume per cycle) is given by AH=αTμBmax2A_H = \frac{\alpha T}{\mu} B_{max}^2, where BmaxB_{max} is the peak magnetic field in the core during the cycle and α\alpha is a positive constant with units of K−1K^{-1}. Assuming the damping is weak (αT≪1\alpha T \ll 1), which of the following correctly gives the equivalent series resistance ReqR_{eq} of the inductor due to hysteresis loss, and the time t0t_0 taken for the current amplitude to decrease to 1/e1/e of its initial value?

Options

  • A.

    Req=αTπLC,t0=2πLCαTR_{eq} = \frac{\alpha T}{\pi} \sqrt{\frac{L}{C}}, \quad t_0 = \frac{2\pi\sqrt{LC}}{\alpha T}

  • B.

    Req=αT2πLC,t0=4πLCαTR_{eq} = \frac{\alpha T}{2\pi} \sqrt{\frac{L}{C}}, \quad t_0 = \frac{4\pi\sqrt{LC}}{\alpha T}

  • C.

    Req=αTπLC,t0=πLCαTR_{eq} = \frac{\alpha T}{\pi} \sqrt{\frac{L}{C}}, \quad t_0 = \frac{\pi\sqrt{LC}}{\alpha T}

  • D.

    Req=2αTLC,t0=LCαTR_{eq} = 2\alpha T \sqrt{\frac{L}{C}}, \quad t_0 = \frac{\sqrt{LC}}{\alpha T}

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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

Soft vs Hard Magnetic Materials|LC Circuit Oscillations|Temperature Coefficient (Alpha)

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to map thermodynamic energy dissipation (hysteresis loop area) to an equivalent lumped circuit parameter (ReqR_{eq}).

A primary trap is equating the energy loss per cycle directly to Imax2ReqToscI_{max}^2 R_{eq} T_{osc} instead of using the correct average power relation Pavg=Irms2Req=12Imax2ReqP_{avg} = I_{rms}^2 R_{eq} = \frac{1}{2} I_{max}^2 R_{eq}, which leads to an incorrect factor of 2 in the resistance (Option B).

Another critical trap is confusing the decay rate of the circuit's total energy with the decay rate of the current amplitude.

Energy decays at a rate of 2γ2\gamma, while amplitude decays at γ\gamma.

Students who calculate the energy decay time will incorrectly select Option C.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMagnetism and Matter - Class 12
ConceptSoft vs Hard Magnetic Materials
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree