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Question #42849

Difficulty: 8/10 | 🟢 Free

The Question

A plane electromagnetic wave is normally incident from vacuum (z<0z < 0) onto a non-uniform non-conducting medium (z>0z > 0). The incident electric field is given by E⃗i=E0cos⁡(kz−ωt)y^\vec{E}_i = E_0 \cos(kz - \omega t) \hat{y}. The medium in the region z>0z > 0 has a uniform permittivity ϵ=ϵ0\epsilon = \epsilon_0 but a spatially varying magnetic permeability μ(x)=μ0(1+αx)2\mu(x) = \mu_0 (1 + \alpha x)^2, where α>0\alpha > 0 is a constant. Determine the reflected magnetic field B⃗r\vec{B}_r at the interface z=0z = 0, at the position x=2αx = \frac{2}{\alpha} and time t=0t = 0.

Options

  • A.

    E02cx^\frac{E_0}{2c} \hat{x}

  • B.

    −E02cx^-\frac{E_0}{2c} \hat{x}

  • C.

    E03cx^\frac{E_0}{3c} \hat{x}

  • D.

    −E0cx^-\frac{E_0}{c} \hat{x}

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

Boundary Condition Reflection|Superposition of Magnetic Fields|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of electromagnetic boundary conditions in media with non-uniform magnetic permeability.

A common trap is for students to incorrectly assume the continuity of the tangential magnetic field B⃗\vec{B} (B1t=B2tB_{1t} = B_{2t}) instead of the auxiliary field H⃗\vec{H} (H1t=H2tH_{1t} = H_{2t}).

By treating μ\mu as a scalar constant that cancels out, or ignoring its role in the local wave impedance Z(x)=μ(x)/ϵ0Z(x) = \sqrt{\mu(x)/\epsilon_0}, students will miscalculate the reflection coefficient.

Applying B1t=B2tB_{1t} = B_{2t} yields a reflected electric field of opposite sign (Er=−E0/2E_r = -E_0/2), leading directly to the trap option −E02cx^-\frac{E_0}{2c} \hat{x}.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWaves - Class 11 Part 2
ConceptBoundary Condition Reflection
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree