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Question #43308

Difficulty: 8/10 | 🟢 Free

The Question

A toroidal inductor is constructed with a non-magnetic core of uniform height hh. The core is densely wound with NN turns of wire carrying a current II. The inner boundary of the core is a cylinder of constant radius aa. The outer boundary of the core has a variable radius b(θ)b(\theta) that depends on the azimuthal angle θ\theta as: b(θ)={aekθfor 0≤θ≤πaek(2π−θ)for π<θ≤2πb(\theta) = \begin{cases} a e^{k \theta} & \text{for } 0 \leq \theta \leq \pi \\ a e^{k (2\pi - \theta)} & \text{for } \pi < \theta \leq 2\pi \end{cases} where kk is a positive dimensionless constant. Assuming the magnetic field inside the core at a distance rr from the central axis is given by B=μ0NI2πrB = \frac{\mu_0 N I}{2\pi r}, what is the self-inductance LL of this toroid?

Options

  • A.

    μ0N2hk4\frac{\mu_0 N^2 h k}{4}

  • B.

    μ0N2hk2\frac{\mu_0 N^2 h k}{2}

  • C.

    μ0N2hkπ\frac{\mu_0 N^2 h k}{\pi}

  • D.

    μ0N2h2πln⁡(1+kπ)\frac{\mu_0 N^2 h}{2\pi} \ln(1 + k\pi)

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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

Energy Stored in Inductor|Toroidal Solenoid Boundary Conditions|Infinite Wire vs. Finite Wire

Why This Question Is Tricky

Common Mistakes

This question tests the student's fundamental understanding of inductance by breaking the symmetry required for standard formulas.

The primary trap relies on students over-relying on the standard formula for a rectangular toroid, L=μ0N2h2πln⁡(b/a)L = \frac{\mu_0 N^2 h}{2\pi} \ln(b/a).

When faced with a variable outer radius b(θ)b(\theta), students often incorrectly substitute the maximum radius (bmax=aekπb_{max} = a e^{k\pi}), yielding Option B, or attempt to find an average radius, yielding Option D.

The rigorous physical approach requires recognizing that inductance is fundamentally a measure of magnetic energy storage per unit current squared (L=2U/I2L = 2U/I^2).

By forcing the integration of the magnetic energy density over a non-uniform volume, the question filters out students who treat inductance as a static table-lookup value rather than a field-energy derivative.

Quick Facts About This Question

Question Information

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