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Question #43319

Difficulty: 8/10 | 🟢 Free

The Question

An ideal capacitor of capacitance CC, initially charged to a charge Q0Q_0, is connected at t=0t=0 across an ideal solenoid to form a closed LC circuit. The solenoid has NN turns, length ll, and cross-sectional area AA. It is completely filled with a non-linear magnetic core material. The magnetic field BB and the magnetic intensity HH inside the core are related by the equation H=αB+βB3H = \alpha B + \beta B^3, where α\alpha and β\beta are positive constants. Assuming no resistance in the circuit and no magnetic flux leakage, what is the maximum magnitude of the magnetic field BmaxB_{\text{max}} inside the solenoid during the subsequent oscillations?

Options

  • A.

    (α2+2βQ02CAl−αβ)1/2\left( \frac{\sqrt{\alpha^2 + \frac{2 \beta Q_0^2}{C A l}} - \alpha}{\beta} \right)^{1/2}

  • B.

    (α2+4βQ02CAl−α2β)1/2\left( \frac{\sqrt{\alpha^2 + \frac{4 \beta Q_0^2}{C A l}} - \alpha}{2\beta} \right)^{1/2}

  • C.

    (α2+6βQ02CAl−α3β)1/2\left( \frac{\sqrt{\alpha^2 + \frac{6 \beta Q_0^2}{C A l}} - \alpha}{3\beta} \right)^{1/2}

  • D.

    (α2+8βQ02CAl−α4β)1/2\left( \frac{\sqrt{\alpha^2 + \frac{8 \beta Q_0^2}{C A l}} - \alpha}{4\beta} \right)^{1/2}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

LC Circuit Oscillations|Energy Density Locality|Magnetic Field Line Behavior in Matter

Why This Question Is Tricky

Common Mistakes

This question tests the student's fundamental understanding of magnetic energy density in non-linear media.

A common trap is to blindly apply the linear formula u=12BHu = \frac{1}{2} B H (or U=12Li2U = \frac{1}{2} L i^2), which assumes a constant permeability μ\mu.

In a non-linear core where H(B)H(B) is a polynomial, the true energy density must be derived from first principles using the work done by the source: u=∫HdBu = \int H dB.

Students who incorrectly use u=12BHu = \frac{1}{2} B H will obtain an incorrect coefficient for the non-linear term, leading them to Option B.

Those who confuse energy with co-energy (u=∫BdHu = \int B dH) will select Option C.

The problem forces a return to the fundamental electrodynamic definition of work rather than relying on memorized formulas for linear oscillators.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Induction - Class 12
ConceptLC Circuit Oscillations
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree