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Question #42722

Difficulty: 8/10 | 🟢 Free

The Question

A long coaxial cable consists of a solid inner cylindrical conductor of radius aa and a thick outer coaxial conducting cylindrical shell of inner radius bb and outer radius cc. Both conductors are made of a non-magnetic metal with resistivity ρ\rho and permeability μ0\mu_0. A high-frequency alternating current I(t)=I0cos⁡(ωt)I(t) = I_0 \cos(\omega t) flows down the inner conductor and returns via the outer shell. The angular frequency ω\omega is sufficiently high such that the skin depth δ=2ρωμ0\delta = \sqrt{\frac{2\rho}{\omega \mu_0}} is much smaller than aa, bb, and (c−b)(c-b). What is the time-averaged power dissipated as heat per unit length of the cable?

Options

  • A.

    I024πρωμ02(a+bab)\frac{I_0^2}{4\pi} \sqrt{\frac{\rho \omega \mu_0}{2}} \left( \frac{a+b}{ab} \right)

  • B.

    I022πρωμ02(a+bab)\frac{I_0^2}{2\pi} \sqrt{\frac{\rho \omega \mu_0}{2}} \left( \frac{a+b}{ab} \right)

  • C.

    I024πρωμ02(a+cac)\frac{I_0^2}{4\pi} \sqrt{\frac{\rho \omega \mu_0}{2}} \left( \frac{a+c}{ac} \right)

  • D.

    I024πρωμ02(1a)\frac{I_0^2}{4\pi} \sqrt{\frac{\rho \omega \mu_0}{2}} \left( \frac{1}{a} \right)

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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

Ohm's Law Validity|Frequency Sensitivity of Z|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the transition from lumped circuit theory to distributed electromagnetic field theory via the high-frequency skin effect.

The primary trap lies in identifying the correct surfaces where the current flows.

Students often incorrectly assume the return current in the outer shell distributes uniformly or flows on the outer surface (radius cc).

However, to minimize the system's inductance (magnetic field energy), the high-frequency return current strongly crowds to the inner surface of the outer shell (radius bb).

Additionally, students may forget the factor of 1/21/2 when calculating time-averaged power from peak current I0I_0, or fail to correctly approximate the effective cross-sectional area 2πrδ2\pi r \delta for a thin cylindrical shell.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterCurrent Electricity - Class 12
ConceptOhm's Law Validity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree