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Question #42994

Difficulty: 8/10 | 🟢 Free

The Question

A non-linear inductor with a ferromagnetic core operates in a regime where magnetic saturation begins to occur. The total magnetic flux linkage Φ\Phi (in Wb\text{Wb}) through the inductor varies with the current II (in A\text{A}) according to the relation Φ(I)=2I−14I2\Phi(I) = 2I - \frac{1}{4}I^2 for 0≤I≤4 A0 \le I \le 4 \text{ A}. Calculate the total magnetic energy stored in the inductor when the current is slowly increased from 0 A0 \text{ A} to 4 A4 \text{ A}.

Options

  • A.

    163 J\frac{16}{3} \text{ J}

  • B.

    8 \text{ J}

  • C.

    323 J\frac{32}{3} \text{ J}

  • D.

    16 \text{ J}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Energy Stored in Inductor|Ohm's Law Validity|Faraday’s Law Vector Form

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental thermodynamic derivation of magnetic energy, weaponizing the standard formula U=12LI2U = \frac{1}{2}LI^2.

Students are conditioned to use this formula, which is only valid for linear inductors where permeability (and thus LL) is constant.

Faced with a non-linear flux-current relationship, a common trap is to calculate the static inductance Lstatic=ΦI=2−14IL_{static} = \frac{\Phi}{I} = 2 - \frac{1}{4}I, which at I=4 AI = 4 \text{ A} gives L=1 HL = 1 \text{ H}, leading to the incorrect answer of 8 J8 \text{ J}.

Another trap is integrating ∫Φ dI\int \Phi \, dI, which calculates the magnetic co-energy (323 J\frac{32}{3} \text{ J}) rather than the actual stored energy.

The problem forces students to reject linear assumptions and correctly apply U=∫I dΦU = \int I \, d\Phi to account for magnetic saturation.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Induction - Class 12
ConceptEnergy Stored in Inductor
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree