Electromagnetic Waves
Electromagnetic Waves for JEE Main & Advanced
Maxwell's Equations and EM Wave Theory
Displacement Current and Maxwell's EquationsTopic 1
Maxwell's Modification: Ampere's law $\oint \vec{B}\cdot d\vec{l} = \mu_0 I$ is inconsistent during capacitor charging — no conduction current flows in the gap. Maxwell introduced displacement current to fix this.
Displacement Current: Current associated with changing electric flux: $$I_d = \epsilon_0\frac{d\Phi_E}{dt}$$ For parallel plate capacitor: $\Phi_E = EA = (Q/\epsilon_0 A)A = Q/\epsilon_0$, so $I_d = dQ/dt$ = conduction current entering capacitor.
Modified Ampere-Maxwell Law: $$\oint \vec{B}\cdot d\vec{l} = \mu_0(I_c + I_d) = \mu_0\left(I_c + \epsilon_0\frac{d\Phi_E}{dt}\right)$$
Maxwell's Four Equations (in integral form):
| Equation | Statement |
|---|---|
| Gauss's law (electric) | $\oint \vec{E}\cdot d\vec{A} = Q_{\text{enc}}/\epsilon_0$ |
| Gauss's law (magnetic) | $\oint \vec{B}\cdot d\vec{A} = 0$ (no magnetic monopoles) |
| Faraday's law | $\oint \vec{E}\cdot d\vec{l} = -d\Phi_B/dt$ |
| Ampere-Maxwell | $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_c + \mu_0\epsilon_0\,d\Phi_E/dt$ |
These predict the existence of EM waves (Maxwell 1864; verified by Hertz 1887).
Speed of EM Waves in Vacuum: $$c = \frac{1}{\sqrt{\mu_0\epsilon_0}} = 3 \times 10^8 \text{ m/s}$$
In medium with permittivity $\epsilon = \epsilon_0\epsilon_r$ and permeability $\mu = \mu_0\mu_r$: $v = c/\sqrt{\epsilon_r\mu_r} = c/n$ where $n$ = refractive index.
A parallel plate capacitor is being charged at rate $0.5$ A. Find displacement current between plates.
Show solution
By continuity of current, displacement current equals conduction current: $$I_d = \frac{dQ}{dt} = 0.5 \text{ A}$$
Final Answer: $I_d = 0.5$ A.
Find speed of EM waves in a medium with $\epsilon_r = 4$, $\mu_r = 1$.
Show solution
$$v = \frac{c}{\sqrt{\epsilon_r\mu_r}} = \frac{3 \times 10^8}{\sqrt{4}} = 1.5 \times 10^8 \text{ m/s}$$
Final Answer: $v = 1.5 \times 10^8$ m/s.
Displacement current arises due to:
Maxwell's equations predict the existence of:
The speed of EM waves in vacuum equals:
The Gauss law for magnetism implies:
Speed of EM waves in a medium with $n = 1.5$:
Properties and Propagation of EM WavesTopic 2
Electromagnetic Wave: Transverse wave consisting of mutually perpendicular oscillating $\vec{E}$ and $\vec{B}$ fields, both perpendicular to direction of propagation. Carries energy and momentum.
Wave equations (in vacuum): $$\nabla^2 E = \mu_0\epsilon_0\frac{\partial^2 E}{\partial t^2}, \quad \nabla^2 B = \mu_0\epsilon_0\frac{\partial^2 B}{\partial t^2}$$
Plane Wave Solution (propagating along $+x$): $$E_y = E_0\sin(\omega t - kx), \quad B_z = B_0\sin(\omega t - kx)$$ $\vec{E}, \vec{B}, \vec{c}$ form right-handed triad: $\hat{c} = \hat{E}\times\hat{B}$.
Key Relations:
| Quantity | Formula |
|---|---|
| Wave speed | $c = \omega/k = E_0/B_0$ |
| $\vec{E}, \vec{B}$ ratio | $E_0/B_0 = c$ in vacuum |
| Wavelength | $\lambda = c/f = 2\pi/k$ |
Properties of EM Waves:
- Transverse (not longitudinal)
- Travel through vacuum (don't need medium)
- Speed in vacuum = $c$ (universal constant)
- Energy is shared equally between $\vec{E}$ and $\vec{B}$
- Carry momentum: $p = U/c$
- Exert radiation pressure on absorbing/reflecting surface
- Can be polarized (since transverse)
- Obey reflection, refraction, interference, diffraction
Energy Density: $u = \dfrac{1}{2}\epsilon_0 E^2 + \dfrac{B^2}{2\mu_0} = \epsilon_0 E^2$ (since electric and magnetic components contribute equally).
Average energy density: $\langle u\rangle = \dfrac{1}{2}\epsilon_0 E_0^2$.
Poynting Vector: $\vec{S} = \dfrac{1}{\mu_0}(\vec{E}\times\vec{B})$ gives instantaneous energy flux (W/m²). Average: $\langle S\rangle = \dfrac{E_0^2}{2\mu_0 c} = \dfrac{1}{2}c\epsilon_0 E_0^2$.
The electric field amplitude in an EM wave is $300$ V/m. Find magnetic field amplitude.
Show solution
$$B_0 = E_0/c = 300/(3 \times 10^8) = 10^{-6} \text{ T} = 1\,\mu\text{T}$$
Final Answer: $B_0 = 1\,\mu$T.
An EM wave has $E_0 = 60$ V/m. Find average energy density.
Show solution
$$\langle u\rangle = \frac{1}{2}\epsilon_0 E_0^2 = \frac{1}{2}(8.85 \times 10^{-12})(60)^2$$ $$= \frac{1}{2}(8.85 \times 10^{-12})(3600) = 1.59 \times 10^{-8} \text{ J/m}^3$$
Final Answer: $\langle u\rangle \approx 1.59 \times 10^{-8}$ J/m³.
EM waves are:
In an EM wave, $\vec{E}, \vec{B}, \vec{c}$:
Ratio $E_0/B_0$ in EM wave in vacuum:
Poynting vector represents:
EM waves transport:
Electromagnetic Spectrum and Applications
EM Spectrum — Production, Detection, UseTopic 1
EM Spectrum spans frequencies from $\sim 10^0$ Hz to $> 10^{22}$ Hz. All regions travel at $c$ in vacuum.
Major Regions (in order of increasing frequency / decreasing wavelength):
| Region | Wavelength | Production | Detection | Applications |
|---|---|---|---|---|
| Radio waves | $> 0.1$ m | LC oscillators | Antennas | AM/FM radio, TV |
| Microwaves | $10^{-3}$–$0.1$ m | Klystron, magnetron | Crystal/point detectors | Radar, microwave oven |
| Infrared (IR) | $700$ nm – $10^{-3}$ m | Hot bodies, molecules | Thermopile, photodetector | Heating, remote control, night vision |
| Visible | $400$–$700$ nm | Atoms (electron transitions) | Eye, photographic plate | Vision, photography |
| Ultraviolet (UV) | $1$–$400$ nm | Gas discharges, Sun | Fluorescent screen | Sterilisation, vitamin D |
| X-rays | $0.01$–$1$ nm | Inner-shell electron transitions | Photographic film | Medical imaging, crystallography |
| Gamma ($\gamma$) | $< 0.01$ nm | Nuclear decay | Geiger counter, scintillator | Cancer therapy, sterilisation |
Visible Light Spectrum (VIBGYOR):
- Violet: $380-450$ nm
- Indigo: $450-485$ nm
- Blue: $485-500$ nm
- Green: $500-565$ nm
- Yellow: $565-590$ nm
- Orange: $590-625$ nm
- Red: $625-700$ nm
Mnemonic: "VIBGYOR" (Violet → Red).
Notable points:
- Ozone layer absorbs harmful UV-B and UV-C
- Microwave background radiation ($2.7$ K) — relic of Big Bang
- IR radiation responsible for greenhouse effect
- X-rays were discovered by Roentgen in 1895
- Gamma rays have highest penetration
A FM radio station broadcasts at $100$ MHz. Find wavelength.
Show solution
$$\lambda = c/f = \frac{3 \times 10^8}{100 \times 10^6} = 3 \text{ m}$$
Final Answer: $\lambda = 3$ m.
Find frequency of red light of wavelength $700$ nm.
Show solution
$$f = c/\lambda = \frac{3 \times 10^8}{700 \times 10^{-9}} = 4.286 \times 10^{14} \text{ Hz}$$
Final Answer: $f \approx 4.29 \times 10^{14}$ Hz.
Wavelength of visible light lies approximately between:
X-rays are produced by:
Microwave ovens use frequency around:
Which has the highest frequency?
Infrared radiation has wavelength:
Energy, Momentum, Intensity and Radiation PressureTopic 2
Intensity $I$ = power per unit area carried by EM wave (W/m²): $$I = \langle S\rangle = \frac{1}{2}\epsilon_0 c E_0^2 = \frac{E_0^2}{2\mu_0 c} = \frac{cB_0^2}{2\mu_0}$$
Relation between $E_0, B_0$, and intensity: $$I = \frac{E_{\text{rms}}^2}{\mu_0 c} = c\epsilon_0 E_{\text{rms}}^2$$
Momentum carried by EM wave: $$p = U/c$$ where $U$ is the energy of the wave.
Radiation Pressure:
| Case | Pressure on surface |
|---|---|
| Perfectly absorbing surface | $P_r = I/c$ |
| Perfectly reflecting surface | $P_r = 2I/c$ |
| Surface with reflectivity $R$ | $P_r = (1+R)I/c$ |
Solar Constant (intensity of sunlight at Earth): $\sim 1370$ W/m².
For point source of power $P$, intensity at distance $r$: $I = P/(4\pi r^2)$ (inverse square law).
Power radiated by accelerating charge: $P = q^2a^2/(6\pi\epsilon_0 c^3)$ (Larmor formula).
A laser beam delivers $500$ W on $1$ mm² area. Find intensity, $E_0$, and radiation pressure on perfectly absorbing surface.
Show solution
$$I = P/A = 500/10^{-6} = 5 \times 10^8 \text{ W/m}^2$$ $E_0 = \sqrt{2I/(\epsilon_0 c)} = \sqrt{(2 \times 5 \times 10^8)/(8.85 \times 10^{-12} \times 3 \times 10^8)}$ $= \sqrt{10^9/(2.655 \times 10^{-3})} = \sqrt{3.77 \times 10^{11}} \approx 6.14 \times 10^5$ V/m. $P_r = I/c = 5 \times 10^8/(3 \times 10^8) \approx 1.67$ Pa.
Final Answer: $I = 5 \times 10^8$ W/m²; $E_0 \approx 6.14 \times 10^5$ V/m; $P_r \approx 1.67$ Pa.
Solar constant = $1370$ W/m². Find force exerted on a perfectly reflecting solar sail of $100$ m².
Show solution
$$F = (2I/c) \cdot A = \frac{2 \times 1370 \times 100}{3 \times 10^8}$$ $$= \frac{274000}{3 \times 10^8} \approx 9.13 \times 10^{-4} \text{ N}$$
Final Answer: $F \approx 9.13 \times 10^{-4}$ N $\approx 0.91$ mN.
The momentum of EM wave of energy $U$ in vacuum:
Radiation pressure on perfectly absorbing surface (intensity $I$):
Intensity of EM wave is proportional to:
EM wave from a point source intensity at distance $r$:
The solar constant at Earth's surface is approximately:
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