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ExamsJEE AdvancedPhysics

In a material with conductivity sigma = 5.0 S/m and relative permittivity epsilon_r = 1, the electric field is E = 250 sin(10¹⁰ * t) V/m. Determine the conduction current density Jc, the displacement current density Jd, and the frequency f at which their amplitudes are equal.

  1. Jc = 1250 sin(10¹⁰ t) A/m²; Jd = 22.1 cos(10¹⁰ t) A/m²; f = 90 GHz
  2. Jc = 22.1 cos(10¹⁰ t) A/m²; Jd = 1250 sin(10¹⁰ t) A/m²; f = 90 GHz
  3. Jc = 1250 sin(10¹⁰ t) A/m²; Jd = 22.1 sin(10¹⁰ t) A/m²; f = 50 GHz
  4. Jc = 1250 cos(10¹⁰ t) A/m²; Jd = 22.1 cos(10¹⁰ t) A/m²; f = 50 GHz

Correct answer: Jc = 1250 sin(10¹⁰ t) A/m²; Jd = 22.1 cos(10¹⁰ t) A/m²; f = 90 GHz

Solution

Jc = 5 * 250 sin(10¹⁰ t) = 1250 sin(10¹⁰ t) A/m². Jd = epsilon₀ * dE/dt = 8.854e-12 * 250 * 10¹⁰ * cos(10¹⁰ t) = 22.1 cos(10¹⁰ t) A/m². For equal amplitudes: sigma = omega*epsilon₀ -> f = sigma/(2*pi*epsilon₀) = 5/(2*pi*8.854e-12) approx 90 GHz.

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