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Question #43253

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor with circular plates of radius R=5 cmR = 5 \text{ cm} is filled with a lossy dielectric material. The material has a relative permittivity (dielectric constant) K=9K = 9 and a uniform electrical conductivity σ=3×10−3 S/m\sigma = \sqrt{3} \times 10^{-3} \text{ S/m}. An alternating electric field E(t)=E0sin⁡(ωt)E(t) = E_0 \sin(\omega t) is established inside the dielectric, where E0=5×104 V/mE_0 = 5 \times 10^4 \text{ V/m} and ω=4π×106 rad/s\omega = 4\pi \times 10^6 \text{ rad/s}. Neglecting edge effects, what is the amplitude of the induced magnetic field BmaxB_{max} at a radial distance r=2 cmr = 2 \text{ cm} from the central axis of the capacitor? (Take 14πϵ0=9×109 N m2/C2\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2 and μ0=4π×10−7 T m/A\mu_0 = 4\pi \times 10^{-7} \text{ T m/A})

Options

  • A.

    0.4π \muT0.4\pi \text{ \mu T}

  • B.

    0.2π \muT0.2\pi \text{ \mu T}

  • C.

    0.23π \muT0.2\sqrt{3}\pi \text{ \mu T}

  • D.

    0.2(1+3)π \muT0.2(1+\sqrt{3})\pi \text{ \mu T}

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

Ampere’s Law Displacement Current|Maxwell's Correction to Ampere's Law|Vector Calculus

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of the Ampere-Maxwell law in a non-ideal (lossy) dielectric medium.

A common misconception is that displacement current only exists in a vacuum or that conduction current is zero in any dielectric.

Students often calculate only the displacement current (Option B) or only the conduction current (Option C).

Furthermore, even if they recognize both exist, they may incorrectly add their amplitudes algebraically (Option D) instead of recognizing the 90-degree phase difference between the conduction current (in phase with the electric field) and the displacement current (proportional to the time derivative of the electric field).

The correct approach requires phasor addition of the current densities.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMoving Charges and Magnetism - Class 12
ConceptAmpere’s Law Displacement Current
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree