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Question #42978

Difficulty: 8/10 | 🟢 Free

The Question

A plane electromagnetic wave with electric field E⃗inc=E0cos⁡(ωt−kx)j^\vec{E}_{\text{inc}} = E_0 \cos(\omega t - kx) \hat{j} is normally incident on a thin, infinite conducting grid located at x=0x=0. The grid has zero electrical resistance but possesses a constant surface self-inductance LsL_s (in Henry). The electric field at the grid induces a surface current density K⃗(t)=K(t)j^\vec{K}(t) = K(t) \hat{j} (in A/m). The dynamics of this current are governed by E(0,t)=LsdKdtE(0,t) = L_s \frac{dK}{dt}, where E(0,t)E(0,t) is the net electric field at x=0x=0.

The induced current generates secondary electromagnetic waves propagating away from the grid in both directions. Assuming the surrounding medium is vacuum (permeability μ0\mu_0, speed of light cc), the reflected electric field for x<0x < 0 is given by E⃗ref=Ercos⁡(ωt+kx+θ)j^\vec{E}_{\text{ref}} = E_r \cos(\omega t + kx + \theta) \hat{j}, where Er>0E_r > 0.

For low frequencies such that ωLs≪μ0c\omega L_s \ll \mu_0 c, what is the approximate phase shift θ\theta of the reflected wave?

Options

  • A.

    π−2ωLsμ0c\pi - \frac{2\omega L_s}{\mu_0 c}

  • B.

    π+2ωLsμ0c\pi + \frac{2\omega L_s}{\mu_0 c}

  • C.

    π−ωLsμ0c\pi - \frac{\omega L_s}{\mu_0 c}

  • D.

    2ωLsμ0c\frac{2\omega L_s}{\mu_0 c}

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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|Boundary Condition Reflection|Phase Difference vs Path Difference|Approximation (Binomial)

Why This Question Is Tricky

Common Mistakes

This question tests the transition of a boundary from transparent to opaque by coupling electromagnetic wave propagation with the transient dynamics of induced currents.

Students commonly misapply the ideal conductor boundary condition, assuming an instantaneous short circuit that yields a rigid phase shift of exactly π\pi.

They fail to recognize that the finite surface inductance requires time for the back-EMF to establish (Lenz's Law), creating a complex, time-dependent reflection coefficient.

By ignoring the differential equation governing the surface current or miscalculating the bidirectional radiation (missing the factor of 2), students fall into the trap of selecting incorrect phase shifts.

The binomial approximation elegantly reveals how the inductive lag manifests as a phase deviation from the ideal π\pi reflection.

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