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Question #42645

Difficulty: 8/10 | 🟢 Free

The Question

A plane electromagnetic wave propagates in a vacuum. At $x=0$ , the electric field is $\mathbf{E}(0,t) = \alpha q(t) \hat{j}$ and the magnetic field is $\mathbf{B}(0,t) = \beta \frac{di}{dt} \hat{k}$ , where $q(t)$ is the instantaneous charge on a capacitor in an ideal LC circuit, $i(t) = \frac{dq}{dt}$ is the current, and $\alpha, \beta$ are positive constants. The capacitor charge varies as $q(t) = q_0 \cos(\omega t)$ . Let $E_0 = \alpha q_0$ and $S_0 = \frac{\alpha \beta q_0^2 \omega^2}{2\mu_0}$ . Which of the following correctly describes the electric field $\mathbf{E}(x,t)$ and the time-averaged Poynting vector $\langle \mathbf{S} \rangle$ of the wave?

Options

A.
$\mathbf{E}(x,t) = E_0 \cos(\omega t - kx) \hat{j} \text{ and } \langle \mathbf{S} \rangle = S_0 \hat{i}$
B.
$\mathbf{E}(x,t) = E_0 \cos(\omega t + kx) \hat{j} \text{ and } \langle \mathbf{S} \rangle = -S_0 \hat{i}$
C.
$\mathbf{E}(x,t) = E_0 \sin(\omega t - kx) \hat{j} \text{ and } \langle \mathbf{S} \rangle = S_0 \hat{i}$
D.
$\mathbf{E}(x,t) = E_0 \sin(\omega t + kx) \hat{j} \text{ and } \langle \mathbf{S} \rangle = -S_0 \hat{i}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Wave Equation Sign Convention|LC Circuit Oscillations|Differential Equations

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to link the kinematics of wave propagation (sign convention in the wave equation) with the electrodynamics of time-varying fields and LC circuit oscillations.

A common trap is assuming that standard wave equations always take the form $f(\omega t - kx)$ (propagating in $+x$ ) without verifying the instantaneous direction of the Poynting vector $\mathbf{S} \propto \mathbf{E} \times \mathbf{B}$ .

By tying the boundary fields to $q(t)$ and $d^2q/dt^2$ of an LC circuit, the problem introduces a $\pi$ phase shift (negative sign) in the magnetic field.

Students who fail to account for this second derivative sign will incorrectly calculate $\mathbf{S}$ in the $+x$ direction and choose the wrong wave equation sign.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Waves - Class 11 Part 2
Concept	Wave Equation Sign Convention
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free