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Question #94420

Difficulty: 5/10 | 🟢 Free

The Question

A parallel plate capacitor consists of two circular plates of radius $R$ (in m) separated by a small distance $d$ (in m). The space between the plates is filled with a non-uniform dielectric material whose permittivity varies with the radial distance $r$ (in m) from the central axis as $\epsilon(r) = \epsilon_0 \frac{R}{r}$ for $0 < r \le R$ , where $\epsilon_0$ is the vacuum permittivity. A time-varying potential difference $V(t) = V_0 \sin(\omega t)$ is applied across the plates, where $V_0$ is the voltage amplitude (in V) and $\omega$ is the angular frequency (in rad/s). Assume quasi-static conditions and neglect edge effects. What is the amplitude of the induced magnetic field at a radial distance $r = R/2$ from the central axis? (Let $\mu_0$ be the vacuum permeability)

Options

A.
$\frac{\mu_0 \epsilon_0 R V_0 \omega}{d}$
B.
$\frac{\mu_0 \epsilon_0 R V_0 \omega}{2d}$
C.
$\frac{\mu_0 \epsilon_0 R V_0 \omega}{4d}$
D.
$\frac{2 \mu_0 \epsilon_0 R V_0 \omega}{d}$

Check Your Answer

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

Displacement Current Source|Non-Uniform Dielectric Filling|Vector Integration

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

Students typically memorize the magnetic field inside a parallel plate capacitor as $B \propto r$ , derived from the assumption of a uniform displacement current density $J_d$ .

In this problem, the non-uniform dielectric $\epsilon(r) \propto 1/r$ makes $J_d$ spatially variant.

A common trap is to either calculate the total displacement current and assume it is uniformly distributed, or to erroneously apply the standard formula $B = \frac{\mu_0 \epsilon r}{2} \frac{\partial E}{\partial t}$ using the local permittivity at $r = R/2$ .

Both incorrect methods yield an amplitude of $\frac{\mu_0 \epsilon_0 R V_0 \omega}{2d}$ (Option B).

The correct approach requires recognizing that $D = \epsilon(r)E$ and performing a rigorous surface integral of $\frac{\partial D}{\partial t}$ to find the enclosed displacement current, revealing that the magnetic field is actually uniform inside this specific capacitor.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electromagnetic Waves - Class 12
Concept	Displacement Current Source
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free