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Question #42760

Difficulty: 8/10 | 🟢 Free

The Question

A homogeneous cylindrical poorly conducting rod of length LL (in m\text{m}), cross-sectional area AA (in m2\text{m}^2), and constant permittivity ϵ\epsilon (in F⋅m−1\text{F}\cdot\text{m}^{-1}) has an initial uniform electrical conductivity σ0\sigma_0 (in Ω−1⋅m−1\Omega^{-1}\cdot\text{m}^{-1}). Due to uniform Joule heating from a prior transient current, the momentum relaxation time of the charge carriers decreases, resulting in a new uniform steady-state conductivity σ=σ0/3\sigma = \sigma_0 / \sqrt{3}. Immediately after this heating phase, an AC source is connected across the rod, driving a total uniform current I(t)=I0cos⁡(ωt)I(t) = I_0 \cos(\omega t) (in A\text{A}) through it, where I0I_0 is the current amplitude and tt is time. If the angular frequency of the AC source is ω=σ0/(3ϵ)\omega = \sigma_0 / (\sqrt{3}\epsilon) (in rad⋅s−1\text{rad}\cdot\text{s}^{-1}), which of the following expressions gives the amplitude of the potential difference V0V_0 (in V\text{V}) across the rod?

Options

  • A.

    32I0LAσ0 V\sqrt{\frac{3}{2}} \frac{I_0 L}{A \sigma_0} \text{ V}

  • B.

    3I0LAσ0 V\sqrt{3} \frac{I_0 L}{A \sigma_0} \text{ V}

  • C.

    32I0LAσ0 V\frac{\sqrt{3}}{2} \frac{I_0 L}{A \sigma_0} \text{ V}

  • D.

    I0LAσ0 V\frac{I_0 L}{A \sigma_0} \text{ V}

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

Conductor Interior Potential|Ohm's Law Validity|Relaxation Time Dependency|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to model a conductor under high-frequency AC conditions where the quasi-static assumption (∇⋅J⃗c=0\nabla \cdot \vec{J}_c = 0) fails.

Students commonly fall into two traps: (1) ignoring the displacement current (ϵ∂E⃗/∂t\epsilon \partial \vec{E}/\partial t) which arises from the transient continuity equation, and (2) neglecting the temperature dependence of the momentum relaxation time, which alters the macroscopic conductivity σ\sigma due to Joule heating.

By assuming a purely resistive DC-like behavior (V=IRV = IR), a student will incorrectly choose option B.

If they account for the displacement current but fail to update the conductivity from its initial state, they will choose option C.

The correct approach requires synthesizing the modified conductivity with the complex admittance of the leaky dielectric.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectrostatic Potential and Capacitance - Class 12
ConceptConductor Interior Potential
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree