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ExamsJEE AdvancedPhysics

The magnetic field of a plane electromagnetic wave is given by B = 3 * 10⁻⁸ * sin[200*pi*(y + ct)] i_hat (in Tesla), where c = 3 * 10⁸ m/s. What is the corresponding electric field?

  1. E = -10⁻⁶ * sin[200*pi*(y + ct)] k_hat V/m
  2. E = -9 * sin[200*pi*(y + ct)] k_hat V/m
  3. E = 9 * sin[200*pi*(y + ct)] k_hat V/m
  4. E = 3 * 10⁻⁸ * sin[200*pi*(y + ct)] k_hat V/m

Correct answer: E = -9 * sin[200*pi*(y + ct)] k_hat V/m

Solution

Amplitude of E: E₀ = c * B₀ = 3*10⁸ * 3*10⁻⁸ = 9 V/m. Wave propagation direction: argument (y + ct) means wave travels in -y direction (since at t=0, the wave front is at y=0, and at t>0 the same phase is at y = -ct, moving in -y). Unit vector of propagation: k_hat_prop = -j_hat. Relationship: E cross B must point in direction of propagation. E_hat cross B_hat = k_hat_prop = -j_hat. B is along i_hat. If E is along -k_hat (negative z): (-k_hat) cross (i_hat) = -(k_hat cross i_hat) = -(j_hat) = -j_hat. Yes! This works. So E = -9 * sin[200*pi*(y+ct)] k_hat V/m.

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