JEE Main & Advanced

Electromagnetic Waves

Electromagnetic Waves for JEE Main & Advanced

1
Module 1

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):

EquationStatement
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.

Worked Examples
1

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.

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

Displacement current arises due to:

Q2.

Maxwell's equations predict the existence of:

Q3.

The speed of EM waves in vacuum equals:

Q4.

The Gauss law for magnetism implies:

Q5.

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:

QuantityFormula
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:

  1. Transverse (not longitudinal)
  2. Travel through vacuum (don't need medium)
  3. Speed in vacuum = $c$ (universal constant)
  4. Energy is shared equally between $\vec{E}$ and $\vec{B}$
  5. Carry momentum: $p = U/c$
  6. Exert radiation pressure on absorbing/reflecting surface
  7. Can be polarized (since transverse)
  8. 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: $\langle S\rangle = \dfrac{E_0^2}{2\mu_0 c} = \dfrac{1}{2}c\epsilon_0 E_0^2$.

Worked Examples
1

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.

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

EM waves are:

Q2.

In an EM wave, $\vec{E}, \vec{B}, \vec{c}$:

Q3.

Ratio $E_0/B_0$ in EM wave in vacuum:

Q4.

Poynting vector represents:

Q5.

EM waves transport:

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Module 2

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):

RegionWavelengthProductionDetectionApplications
Radio waves$> 0.1$ mLC oscillatorsAntennasAM/FM radio, TV
Microwaves$10^{-3}$–$0.1$ mKlystron, magnetronCrystal/point detectorsRadar, microwave oven
Infrared (IR)$700$ nm – $10^{-3}$ mHot bodies, moleculesThermopile, photodetectorHeating, remote control, night vision
Visible$400$–$700$ nmAtoms (electron transitions)Eye, photographic plateVision, photography
Ultraviolet (UV)$1$–$400$ nmGas discharges, SunFluorescent screenSterilisation, vitamin D
X-rays$0.01$–$1$ nmInner-shell electron transitionsPhotographic filmMedical imaging, crystallography
Gamma ($\gamma$)$< 0.01$ nmNuclear decayGeiger counter, scintillatorCancer 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
Worked Examples
1

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.

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

Wavelength of visible light lies approximately between:

Q2.

X-rays are produced by:

Q3.

Microwave ovens use frequency around:

Q4.

Which has the highest frequency?

Q5.

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:

CasePressure 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).

Worked Examples
1

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.

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

The momentum of EM wave of energy $U$ in vacuum:

Q2.

Radiation pressure on perfectly absorbing surface (intensity $I$):

Q3.

Intensity of EM wave is proportional to:

Q4.

EM wave from a point source intensity at distance $r$:

Q5.

The solar constant at Earth's surface is approximately:

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