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Question #43111

Difficulty: 8/10 | 🟢 Free

The Question

An electromagnetic wave propagates along the +z+z-direction in a linear, isotropic, but non-homogeneous dielectric medium. The medium has a constant electric permittivity ϵ=ϵ0\epsilon = \epsilon_0 and a spatially varying magnetic susceptibility given by χm(z)=αz\chi_m(z) = \alpha z, where α>0\alpha > 0 is a constant with units of m−1\text{m}^{-1}. The electric field of the wave is linearly polarized along the xx-axis and is given by E⃗(z,t)=E(z)e−iωti^\vec{E}(z,t) = E(z) e^{-i\omega t} \hat{i}, where ω\omega is the angular frequency. If cc is the speed of light in vacuum and k0=ω/ck_0 = \omega/c, which of the following differential equations correctly describes the spatial amplitude E(z)E(z)?

Options

  • A.

    d2Edz2−α1+αzdEdz+k02(1+αz)E=0\frac{d^2 E}{dz^2} - \frac{\alpha}{1 + \alpha z} \frac{d E}{dz} + k_0^2 (1 + \alpha z) E = 0

  • B.

    d2Edz2+α1+αzdEdz+k02(1+αz)E=0\frac{d^2 E}{dz^2} + \frac{\alpha}{1 + \alpha z} \frac{d E}{dz} + k_0^2 (1 + \alpha z) E = 0

  • C.

    d2Edz2+k02(1+αz)E=0\frac{d^2 E}{dz^2} + k_0^2 (1 + \alpha z) E = 0

  • D.

    d2Edz2−α1+αzdEdz+k02E=0\frac{d^2 E}{dz^2} - \frac{\alpha}{1 + \alpha z} \frac{d E}{dz} + k_0^2 E = 0

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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

Medium Dependency|Magnetic Permeability and Susceptibility|Vector Calculus (Curl/Divergence)

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to derive the wave equation from Maxwell's equations in a non-homogeneous medium.

A common trap is blindly applying the standard homogeneous wave equation ∇2E⃗−μϵ∂2E⃗∂t2=0\nabla^2 \vec{E} - \mu\epsilon \frac{\partial^2 \vec{E}}{\partial t^2} = 0 by simply substituting the spatially varying permeability μ(z)\mu(z), which leads to the incorrect Option C.

Students must recognize that taking the curl of Faraday's law involves the term ∇×(μH⃗)\nabla \times (\mu \vec{H}), which generates an additional gradient term ∇μ×H⃗\nabla \mu \times \vec{H} due to the spatial variation of the magnetic susceptibility.

This requires rigorous application of vector calculus identities alongside electromagnetic constitutive relations.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Waves - Class 12
ConceptMedium Dependency
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree