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Electromagnetic Waves
Physics · Class 12 · Part 1NCERT pp. 201–2122 sets × 25 questions
Key facts and formulas for Electromagnetic Waves, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| Displacement current | i_d = ε₀ dΦ_E/dt; in a charging capacitor i_d = i_c = C dV/dt |
| Ampere–Maxwell law | ∮B·dl = μ₀(i_c + i_d) = μ₀i_c + μ₀ε₀ dΦ_E/dt |
| Speed and wave relation | c = νλ = 1/√(μ₀ε₀) = 3 × 10⁸ m/s; k = 2π/λ, ω = 2πν |
| Fields in a wave | E₀/B₀ = c (vacuum); E₀ = √2 E_rms; E, B and travel are mutually perpendicular |
| In a medium | v = c/n, n = √(ε_r μ_r); frequency stays same, λ_medium = λ/n |
| Energy and intensity | u_E = ½ε₀E² = u_B = B²/2μ₀; I = ½ε₀cE₀² |
| Momentum and pressure | p = U/c; pressure I/c (absorber), 2I/c (reflector) |
| Photon energy | E (eV) = 1240 / λ (nm) |
| Sources and spectrum | Only accelerated charges radiate; order of wavelength: radio > microwave > IR > visible > UV > X > γ |
| Radar | Round trip: d = ct/2 |
Three sample questions
Q1. The displacement current between the plates of a charging capacitor arises from
- a flow of free charges across the gap
- a time-varying electric field
- a time-varying magnetic field
- a steady electric field
Hint and solution
Hint: Displacement current is not made of moving charges. Ask what changes inside the gap while the capacitor charges.
- i_d = ε₀ dΦ_E/dt — Displacement current is defined by the rate of change of electric flux through the surface.
- Gap: no charge crosses, but E changes — While charging, the charge on the plates grows, so the electric field in the gap changes with time.
- Needs time-varying E field — Moving free charges are conduction current, and steady fields give zero. So the source is a time-varying electric field. Option B.
Answer: B. Displacement current is ε₀ dΦ_E/dt. It needs a changing electric flux, not moving charges.
Q2. A 10 µF capacitor is charged so that the potential difference across it rises at 5 V/s. The displacement current between its plates is
- 2 µA
- 0.5 µA
- 500 µA
- 50 µA
Hint and solution
Hint: Displacement current in the gap equals the conduction current charging the capacitor, which is C times dV/dt.
- i_d = i = dq/dt = C dV/dt — Displacement current in the gap equals the charging current, and q equals C V.
- i = 10 µF × 5 V/s — Substitute the capacitance and the rate of rise of voltage.
- 10 × 5 = 50 µA — Microfarad times volt per second gives microampere. So the displacement current is 50 microampere. Option D.
Answer: D. i_d = i = C dV/dt = 10 µF × 5 V/s = 50 µA.
Q3. In a circuit with a charging capacitor the conduction current in the wires is 2 mA. The displacement current in the gap is
- 2 mA
- zero
- 1 mA
- 4 mA
Hint and solution
Hint: Maxwell made the total current continuous. What must the gap carry so the circuit has no break?
- i_c + i_d continuous — Maxwell added displacement current so that total current has no break in a circuit.
- Wires: i_c = 2 mA, gap: i_c = 0 — Inside the gap no charge flows, so only displacement current can carry the current there.
- i_d = i_c = 2 mA — The two must be equal for continuity. So the displacement current is 2 mA. Option A.
Answer: A. The total current is continuous, so i_d in the gap equals i_c in the wires.
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.