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Electromagnetic Waves - Properties of EM Waves

Grade 12CBSEPhysics

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

🔑Concepts

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Electromagnetic waves are produced by accelerating charges and do not require any material medium for propagation.

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

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The oscillations of E⃗\vec{E} and B⃗\vec{B} are in the same phase. The direction of propagation is given by the vector cross product E⃗×B⃗\vec{E} \times \vec{B}.

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The velocity of EM waves in a vacuum is a fundamental constant, c=1μ0ϵ0≈3×108 m/sc = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \approx 3 \times 10^8 \text{ m/s}.

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In any medium, the velocity is v=1μϵv = \frac{1}{\sqrt{\mu \epsilon}}, where μ\mu and ϵ\epsilon are the permeability and permittivity of the medium respectively.

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EM waves carry energy and momentum. The energy is shared equally between the electric and magnetic field components.

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The ratio of the amplitudes of the electric field (E0E_0) and magnetic field (B0B_0) is equal to the speed of light: c=E0B0c = \frac{E_0}{B_0}.

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Electromagnetic waves exert radiation pressure on the surfaces they strike. For total absorption, momentum delivered is p=Ucp = \frac{U}{c} where UU is energy.

📐Formulae

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

v=cn=1μϵv = \frac{c}{n} = \frac{1}{\sqrt{\mu \epsilon}}

c=E0B0c = \frac{E_0}{B_0}

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 (For complete absorption)p = \frac{U}{c} \text{ (For complete absorption)}

Ez=E0sin⁡(kx−ωt)E_z = E_0 \sin(kx - \omega t)

By=B0sin⁡(kx−ωt)B_y = B_0 \sin(kx - \omega t)

💡Examples

Problem 1:

A plane electromagnetic wave of frequency 25 MHz25 \text{ MHz} travels in free space along the xx-direction. At a particular point in space and time, E⃗=6.3j^ V/m\vec{E} = 6.3 \hat{j} \text{ V/m}. What is B⃗\vec{B} at this point?

Solution:

Given E=6.3 V/mE = 6.3 \text{ V/m} and the wave travels in the xx-direction. We know B0=E0cB_0 = \frac{E_0}{c}. Substituting values: B=6.33×108=2.1×10−8 TB = \frac{6.3}{3 \times 10^8} = 2.1 \times 10^{-8} \text{ T}. Since E⃗\vec{E} is along the yy-axis (j^\hat{j}) and propagation is along the xx-axis (i^\hat{i}), B⃗\vec{B} must be along the zz-axis (k^\hat{k}) because E⃗×B⃗\vec{E} \times \vec{B} must point in the direction of propagation. Therefore, B⃗=2.1×10−8k^ T\vec{B} = 2.1 \times 10^{-8} \hat{k} \text{ T}.

Explanation:

The relationship between field magnitudes is B=E/cB = E/c. The directions are determined by the right-hand rule where the direction of propagation is i^=j^×k^\hat{i} = \hat{j} \times \hat{k}.

Problem 2:

Light with an energy flux of 18 W/cm218 \text{ W/cm}^2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 cm220 \text{ cm}^2, find the average force exerted on the surface during a 3030 minute time span.

Solution:

Total energy U=Flux×Area×TimeU = \text{Flux} \times \text{Area} \times \text{Time}. U=18×20×(30×60)=6.48×105 JU = 18 \times 20 \times (30 \times 60) = 6.48 \times 10^5 \text{ J}. Total momentum delivered p=Uc=6.48×1053×108=2.16×10−3 kg m/sp = \frac{U}{c} = \frac{6.48 \times 10^5}{3 \times 10^8} = 2.16 \times 10^{-3} \text{ kg m/s}. Average force F=pt=2.16×10−31800=1.2×10−6 NF = \frac{p}{t} = \frac{2.16 \times 10^{-3}}{1800} = 1.2 \times 10^{-6} \text{ N}.

Explanation:

Radiation pressure results from the momentum transfer of photons to the surface. For a non-reflecting surface, all momentum is absorbed.