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Electromagnetic Waves - Electromagnetic Spectrum and Characteristics

Grade 12ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Electromagnetic (EM) waves are produced by accelerated charges and do not require a material medium for propagation.

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The electric field E⃗\vec{E} and magnetic field B⃗\vec{B} vectors are perpendicular to each other and also perpendicular to the direction of wave propagation, making EM waves transverse in nature.

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The speed of EM waves in a vacuum is given by c=1μ0ϵ0≈3×108 m/sc = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \approx 3 \times 10^8 \text{ m/s}, where μ0\mu_0 is the permeability and ϵ0\epsilon_0 is the permittivity of free space.

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The ratio of the amplitudes of the electric and magnetic fields is constant and equal to the speed of light: E0B0=c\frac{E_0}{B_0} = c.

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EM waves carry energy and momentum. The energy density is shared equally between the electric and magnetic fields: uE=uBu_E = u_B.

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The Electromagnetic Spectrum consists of (in order of increasing frequency): Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, and Gamma rays.

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Wavelength (λ\lambda) and frequency (ν\nu) are related by the wave equation c=νλc = \nu \lambda.

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Radiation pressure PP is exerted by EM waves on a surface. For total absorption, P=IcP = \frac{I}{c}, and for total reflection, P=2IcP = \frac{2I}{c}, where II is intensity.

📐Formulae

c=1μ0ϵ0c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}

c=νλc = \nu \lambda

E0B0=c\frac{E_0}{B_0} = c

uavg=12ϵ0E02=B022μ0u_{avg} = \frac{1}{2} \epsilon_0 E_0^2 = \frac{B_0^2}{2 \mu_0}

I=uavgc=12ϵ0E02cI = u_{avg} c = \frac{1}{2} \epsilon_0 E_0^2 c

p=Uc (Momentum transferred for total absorption)p = \frac{U}{c} \text{ (Momentum transferred for total absorption)}

E=hν=hcλE = h\nu = \frac{hc}{\lambda}

💡Examples

Problem 1:

A plane electromagnetic wave travels in vacuum along the z-direction. If the frequency of the wave is 30 MHz30 \text{ MHz} and the amplitude of the electric field is 48 V/m48 \text{ V/m}, calculate the wavelength and the amplitude of the magnetic field.

Solution:

  1. Wavelength: λ=cν=3×108 m/s30×106 Hz=10 m\lambda = \frac{c}{\nu} = \frac{3 \times 10^8 \text{ m/s}}{30 \times 10^6 \text{ Hz}} = 10 \text{ m}.
  2. Magnetic Field Amplitude: B0=E0c=483×108=1.6×10−7 TB_0 = \frac{E_0}{c} = \frac{48}{3 \times 10^8} = 1.6 \times 10^{-7} \text{ T}.

Explanation:

The wavelength is calculated using the relation between speed of light, frequency, and wavelength. The magnetic field strength is derived from the constant ratio E0B0=c\frac{E_0}{B_0} = c.

Problem 2:

Identify the part of the electromagnetic spectrum which: (i) is used in radar systems, (ii) is used to take photographs of internal body parts, and (iii) is used to treat muscular strain.

Solution:

(i) Microwaves (ii) X-rays (iii) Infrared rays

Explanation:

Microwaves are used in radar due to their short wavelengths. X-rays have high penetrating power, making them suitable for medical imaging of bones. Infrared rays produce a heating effect and are used for physical therapy/muscular relief.

Electromagnetic Spectrum and Characteristics Revision - Class 12 Physics ICSE