How to Use This Section

CBSE Board-pattern questions on Electromagnetic Waves by mark value, with examiner-rewarded model answers. The Board regulars: displacement current (definition + the capacitor argument), the Ampere-Maxwell law, properties of EM waves, the E-B-propagation geometry, B0=E0/cB_0 = E_0/c numericals, and spectrum bands with production/detection/uses — this chapter is the highest marks-per-revision-hour in the book.

1-Mark Questions (Definitions & Direct)

Q1. Define displacement current. Answer: The current equivalent of a time-varying electric flux, id=ε0dΦE/dti_d = \varepsilon_0\,d\Phi_E/dt, which produces a magnetic field exactly as a conduction current does.

Q2. Write the Ampere-Maxwell law. Answer: Bdl=μ0ic+μ0ε0dΦEdt\oint\vec{B}\cdot d\vec{l} = \mu_0 i_c + \mu_0\varepsilon_0\dfrac{d\Phi_E}{dt} — circulation of B is sourced by conduction plus displacement current.

Q3. What is the ratio E0/B0E_0/B_0 in an electromagnetic wave? Answer: E0/B0=cE_0/B_0 = c, the speed of light in vacuum (3×1083\times10^8 m/s).

Q4. Name the EM radiation used to kill germs in water purifiers. Answer: Ultraviolet rays.

Q5. Which part of the EM spectrum is absorbed by the ozone layer? Answer: Ultraviolet radiation from the sun (ozone layer at about 40-50 km altitude).

Q6. Name the EM waves produced by radioactive nuclei and state one use. Answer: Gamma rays; used in medicine to destroy cancer cells.

2-Mark Questions (Short Answer)

Q7. Why was Ampere's circuital law modified? State the inconsistency. Answer: For a charging capacitor, a flat surface spanning an Amperian loop is pierced by the wire (giving B(2πr)=μ0iB(2\pi r) = \mu_0 i) while a pot-shaped surface through the gap encloses no conduction current (giving B = 0) — the same point gets two different fields. Maxwell added the displacement current ε0dΦE/dt\varepsilon_0\,d\Phi_E/dt, which flows exactly where the wire does not, so every surface now yields the same total current.

Q8. List four properties of electromagnetic waves. Answer: (i) Transverse — EB\vec E \perp \vec B \perp propagation (E^×B^\hat E\times\hat B along k\vec k); (ii) travel at c=1/μ0ε0c = 1/\sqrt{\mu_0\varepsilon_0} in vacuum, needing no medium; (iii) E and B oscillate in phase with B0=E0/cB_0 = E_0/c; (iv) carry energy (and momentum) from place to place.

Q9. An EM wave travels along +z with E along x. Give the direction of B and the expressions for both fields. Answer: B is along y (i^×j^=k^\hat i\times\hat j = \hat k). Ex=E0sin(kzωt)E_x = E_0\sin(kz-\omega t), By=B0sin(kzωt)B_y = B_0\sin(kz-\omega t) with B0=E0/cB_0 = E_0/c.

Q10. Why are microwaves suitable for radar, and what tunes a microwave oven to food? Answer: Their short (cm) wavelengths form narrow, well-directed beams — ideal for locating aircraft and timing vehicles. Ovens select the frequency to match the resonant frequency of water molecules, transferring wave energy efficiently into molecular kinetic energy.

Q11. Distinguish X-rays from gamma rays in origin, given that their wavelength ranges overlap. Answer: X-rays arise from atomic processes — typically high-energy electrons decelerating at a metal target; gamma rays arise from nuclear processes — nuclear reactions and radioactive decay. Production, not wavelength, is the classifier.

3-Mark Questions (Application & Numericals)

Q12. A parallel-plate capacitor is charged by a 0.5 A current. Find (i) the displacement current between the plates, (ii) the rate of change of electric flux. Answer: (i) id=i=0.5i_d = i = 0.5 A (continuity). (ii) dΦE/dt=id/ε0=0.5/8.85×1012=5.6×1010d\Phi_E/dt = i_d/\varepsilon_0 = 0.5/8.85\times10^{-12} = 5.6\times10^{10} V m s1^{-1}.

Q13. The electric field of a plane EM wave is E=120sin(kzωt)i^E = 120\sin(kz - \omega t)\,\hat{i} V/m at 50 MHz. Find B0B_0, k and ω\omega, and write B\vec{B}. Answer: B0=E0/c=4×107B_0 = E_0/c = 4\times10^{-7} T; ω=2πν=3.14×108\omega = 2\pi\nu = 3.14\times10^8 rad/s; k=ω/c=1.05k = \omega/c = 1.05 rad/m; B=4×107sin(1.05z3.14×108t)j^\vec{B} = 4\times10^{-7}\sin(1.05z - 3.14\times10^8 t)\,\hat{j} T.

Q14. Write Maxwell's four equations and state what each expresses. Answer: EdA=Q/ε0\oint\vec E\cdot d\vec A = Q/\varepsilon_0 (charges source E); BdA=0\oint\vec B\cdot d\vec A = 0 (no monopoles); Edl=dΦB/dt\oint\vec E\cdot d\vec l = -d\Phi_B/dt (changing B makes E); Bdl=μ0ic+μ0ε0dΦE/dt\oint\vec B\cdot d\vec l = \mu_0 i_c + \mu_0\varepsilon_0\,d\Phi_E/dt (currents and changing E make B).

Q15. Name the EM radiation for each: (i) maintaining the earth's warmth via the greenhouse effect, (ii) treating cancer, (iii) aircraft navigation, (iv) photographing in fog/night from satellites (heat signatures), (v) sterilising water, (vi) broadcasting. Answer: (i) Infrared; (ii) gamma rays (also X-rays for certain cancers); (iii) microwaves (radar); (iv) infrared; (v) ultraviolet; (vi) radio waves.

5-Mark Questions (Long Answer)

Q16. (a) Explain, with the charging-capacitor argument, the need for displacement current and state the Ampere-Maxwell law. (b) Show that the displacement current between the plates equals the conduction current in the leads. (c) A 100 pF capacitor is connected to V=230sin(300t)V = 230\sin(300t) V; find the peak displacement current. Answer:

  1. (a) Two surfaces with the same boundary give different enclosed conduction currents (wire-pierced disc vs gap-passing pot surface) — Ampere's law contradicts itself. Maxwell's repair: a changing electric flux acts as the displacement current id=ε0dΦE/dti_d = \varepsilon_0\,d\Phi_E/dt, giving Bdl=μ0ic+μ0ε0dΦE/dt\oint\vec B\cdot d\vec l = \mu_0 i_c + \mu_0\varepsilon_0\,d\Phi_E/dt.
  2. (b) Between plates of area A, E=Q/(Aε0)E = Q/(A\varepsilon_0), so ΦE=Q/ε0\Phi_E = Q/\varepsilon_0 and id=ε0dΦE/dt=dQ/dt=ii_d = \varepsilon_0\,d\Phi_E/dt = dQ/dt = i — exactly the charging current.
  3. (c) idmax=CV0ω=100×1012×230×300=6.9×106i_d^{max} = CV_0\omega = 100\times10^{-12}\times230\times300 = 6.9\times10^{-6} A =6.9 μ= 6.9\ \muA.

Q17. (a) Describe the nature of electromagnetic waves (geometry, speed, phase, medium). (b) Derive nothing but state the meaning of c=1/μ0ε0c = 1/\sqrt{\mu_0\varepsilon_0} and v=1/μεv = 1/\sqrt{\mu\varepsilon}. (c) The magnetic field of a wave is By=2×107sin(0.5×103x+1.5×1011t)B_y = 2\times10^{-7}\sin(0.5\times10^3x + 1.5\times10^{11}t) T; find the wavelength, frequency and the electric field. Answer:

  1. (a) Transverse waves: E, B and propagation mutually perpendicular, E^×B^\hat E\times\hat B along propagation; E and B in phase with B0=E0/cB_0 = E_0/c; self-sustaining field oscillations needing no medium; speed in vacuum the same for all wavelengths.
  2. (b) The vacuum speed is fixed by the electric and magnetic constants (3×1083\times10^8 m/s — light's measured speed, identifying light as an EM wave); in a medium, μ\mu and ε\varepsilon replace μ0,ε0\mu_0, \varepsilon_0, so the speed (and hence refractive index n=μrεrn = \sqrt{\mu_r\varepsilon_r}) depends on the medium.
  3. (c) λ=2π/k=2π/500=1.26\lambda = 2\pi/k = 2\pi/500 = 1.26 cm; ν=ω/2π=23.9\nu = \omega/2\pi = 23.9 GHz; E0=cB0=60E_0 = cB_0 = 60 V/m, directed along z (so that E^×B^\hat E\times\hat B points along x-x, the propagation direction): Ez=60sin(0.5×103x+1.5×1011t)E_z = 60\sin(0.5\times10^3x + 1.5\times10^{11}t) V/m.