Class 12 Physics Chapter 8: Electromagnetic Waves worksheet with answer key

Answer: (c) 0.35 A

i_d = ε₀ dΦ_E/dt = ε₀ A dE/dt = 8.85 × 10⁻¹² × 0.020 × 2.0 × 10¹² ≈ 0.35 A. 0.18 A halves it; 35 A slips two powers of ten; no charge crosses the gap, but a changing field carries a displacement current.

Answer: (a) bombard a metal target with high-energy electrons

X-rays are commonly generated by bombarding a metal target with high-energy electrons, as in an X-ray tube. Hot bodies give infrared and visible light; a semiconductor diode (an LED) gives infrared or light; radioactive nuclei emit gamma rays.

Answer: (d) ε₀ dΦ_E/dt

i_d = ε₀ dΦ_E/dt, the rate of change of electric flux times ε₀. μ₀ε₀ dΦ_E/dt is the term as it appears in the Ampere–Maxwell law, μ₀ times the current; the magnetic flux Φ_B belongs to Faraday's law.

Answer: (c) it matches the resonant frequency of water molecules, which take up the energy as kinetic energy

The oven's frequency is selected to match the resonant frequency of water molecules, so energy goes efficiently into their kinetic energy and any food containing water heats up. Microwaves are far below visible frequencies and do not ionise molecules.

Answer: (b) absorbs ultraviolet radiation

Tanning is the skin's response to ultraviolet: exposure induces more melanin. Ordinary glass absorbs ultraviolet, so the radiation that would tan her does not reach her; visible light clearly passes, since the room is sunlit.

∮B·dl = μ₀ i_c + μ₀ε₀ dΦ_E/dt. i_c is the conduction current, the current of flowing charges, through a surface bounded by the loop; ε₀ dΦ_E/dt is the displacement current, ε₀ times the rate of change of electric flux through the same surface.

• ∮B·dl = μ₀ i_c + μ₀ε₀ dΦ_E/dt (1)
• i_c conduction current of moving charges; ε₀ dΦ_E/dt displacement current from the changing electric flux (1)

μ₀ε₀ = 4π × 10⁻⁷ × 8.85 × 10⁻¹² = 1.11 × 10⁻¹⁷ s² m⁻²; 1/√(1.11 × 10⁻¹⁷) ≈ 3.0 × 10⁸ m s⁻¹. This is the speed c of electromagnetic waves in vacuum given by Maxwell's equations, equal to the measured speed of light: light is an electromagnetic wave.

• 1/√(μ₀ε₀) ≈ 3.0 × 10⁸ m s⁻¹ (1)
• equals the speed of light: light is an electromagnetic wave (1)

A magnetron (a klystron, another special vacuum tube, is also used for microwaves). The oven's frequency is chosen to match the resonant frequency of water molecules, so the energy of the waves goes efficiently into the kinetic energy of the water molecules, raising the temperature of any food containing water.

• magnetron (or klystron) (1)
• frequency matches the resonant frequency of water molecules; energy goes into their kinetic energy (1)

(a) 2 cm lies between 0.1 m and 1 mm: microwaves, detected by point contact diodes. (b) 5 μm lies between 1 mm and 700 nm: infrared, detected by thermopiles, a bolometer or infrared photographic film. (c) 100 nm lies between 400 nm and 1 nm: ultraviolet, detected by photocells or photographic film.

• (a) microwave; point contact diode (1)
• (b) infrared; thermopile or bolometer (1)
• (c) ultraviolet; photocell or photographic film (1)

Take a loop of radius r round the wire at a point P near the capacitor. For a flat surface bounded by the loop and pierced by the wire, the current through it is i(t), so B(2πr) = μ₀i(t): there is a field at P. For a pot-shaped surface with the same rim whose bottom lies between the plates, no conduction current passes through it, so the law gives B(2πr) = 0. The same loop gives two answers. Maxwell added the displacement current ε₀ dΦ_E/dt: through the pot-shaped surface the electric flux between the plates changes, and ε₀ dΦ_E/dt equals i, so both surfaces carry the same total current and give the same B.

• flat surface: B(2πr) = μ₀i(t) (1)
• pot-shaped surface between the plates: no conduction current, B = 0, a contradiction (1)
• displacement current ε₀ dΦ_E/dt = i through the second surface restores consistency (1)

Take a circular loop of radius r coaxial with the plates. No conduction current crosses it; the displacement current through it is the fraction r²/R² of the total i, since E is uniform over the plate area. So B(2πr) = μ₀ i r²/R², B = μ₀ i r/(2πR²). B is zero on the axis and rises linearly with r to μ₀i/(2πR) at the rim, equal to the field just outside the plates.

• displacement current through the loop = i r²/R² (1)
• B = μ₀ i r/(2πR²) (1)
• zero on the axis, linear rise to μ₀i/(2πR) at the rim (1)