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Question #42976

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor with circular plates of radius RR is filled completely with a linear dielectric material of relative permittivity ϵr\epsilon_r. The electric field E⃗\vec{E} between the plates is uniform, directed along the central axis, and its magnitude is increasing at a constant rate dEdt=α\frac{dE}{dt} = \alpha. What is the magnitude of the magnetic field BB at a radial distance rr (r<Rr < R) from the central axis inside the dielectric? (Assume edge effects are negligible, and μ0,ϵ0\mu_0, \epsilon_0 are the permeability and permittivity of free space, respectively).

Options

  • A.

    12μ0ϵ0αr\frac{1}{2} \mu_0 \epsilon_0 \alpha r

  • B.

    12μ0ϵ0(ϵr−1)αr\frac{1}{2} \mu_0 \epsilon_0 (\epsilon_r - 1) \alpha r

  • C.

    12μ0ϵ0ϵrαr\frac{1}{2} \mu_0 \epsilon_0 \epsilon_r \alpha r

  • D.

    12μ0ϵ0ϵrαR2r\frac{1}{2} \mu_0 \epsilon_0 \epsilon_r \alpha \frac{R^2}{r}

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

Faraday’s Law Vector Form|Displacement Current Source|Dielectric Polarization Logic|Vector Calculus

Why This Question Is Tricky

Common Mistakes

This question tests a deep understanding of the Ampere-Maxwell law in the presence of a dielectric medium.

A common trap for students is to memorize the displacement current strictly as Id=ϵ0dΦEdtI_d = \epsilon_0 \frac{d\Phi_E}{dt}, which is only valid in a vacuum.

When a dielectric is present, the changing electric field induces a time-varying polarization P⃗\vec{P}, creating a polarization current density J⃗p=∂P⃗∂t\vec{J}_p = \frac{\partial \vec{P}}{\partial t}.

The true source of the magnetic field is the total displacement current density J⃗d=∂D⃗∂t=ϵ0∂E⃗∂t+∂P⃗∂t=ϵ0ϵr∂E⃗∂t\vec{J}_d = \frac{\partial \vec{D}}{\partial t} = \epsilon_0 \frac{\partial \vec{E}}{\partial t} + \frac{\partial \vec{P}}{\partial t} = \epsilon_0 \epsilon_r \frac{\partial \vec{E}}{\partial t}.

Ignoring the polarization term leads to the incorrect vacuum-only answer (Option A), while using only the polarization term leads to Option B.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Induction - Class 12
ConceptFaraday’s Law Vector Form
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree