Class 12 Physics Electromagnetic Waves Properties of EM waves

Properties of EM waves

  1. Velocity of EM waves in free space or vacuum is a fundamental constant.
  • Experimentally it was found that the velocity of EM wave is same as speed of light(c=3x108m/s).
  • The value of c is fundamental constant.
  • Therefore , c=(1/√μ0ε0)
  1. No material medium is necessary for EM waves. But they can propagate with ina medium as well.
  • EM waves require time varying electric and magnetic fields to propagate.
  • If the medium is present then velocity v = (1/√με)
    • Whereμ =permeability of the medium and ε=permittivity of the medium.
  • For example: -Spectacles. When light falls on glass of the spectacle, light rays pass through glass.i.e. Light waves propagate through medium which is glass here.

 

  1. EM waves carry energy and momentum.
  • Total energy stored per unit volume in EM wave ET= E2ε0(partly carried by electric field and partly by magnetic field).
  • As EM waves carry energy and momentum, it becomes an important property for its practical purposes.
  • EM waves are used for communication purposes, voice communication in mobile phones, telecommunication used in radio.

 

  1. EM waves exert pressure. As they carry energy and momentum, they exert pressure.
  • The pressure exerted by EM waves is known as Radiation pressure.
  • For example: -
    • The sunlight which we get from sun is in the form of visible light rays.
    • These light rays are also part of EM waves.
    • If we keep our palm in sun,after some time, palm becomes warm and starts sweating.
    • This happens because sunlight is getting transferred in the form of EM waves and these EM waves carry energy.
    • Suppose total energy transferred to the hand =E.
    • Momentum = (E/c) as c is extremely high, therefore momentum is very small.
    • As momentum is very less, pressure experiencedis also very less.
    • This is the reason due to which the pressure exerted by the sun is not experienced by the hand.

 

Problem:- In a plane electromagnetic wave, the electric field oscillates sinusoidal at a frequency of 2.0 × 1010 Hz and amplitude 48 V m−1.

(a) What is the wavelength of the wave?

(b) What is the amplitude of the oscillating magnetic field?

(c) Show that the average energy density of the E field equals the average energy density

of the B field. [c = 3 × 108 m s−1.]

Answer:- 

Frequency of the electromagnetic wave, ν = 2.0 × 1010 Hz

Electric field amplitude, E0 = 48 Vm−1

Speed of light, c = 3 × 108 m/s

(a) Wavelength of a wave is given as:

λ=(c/ν)

= (3 × 108)/ (2.0 × 1010)

=0.015m

(b) Magnetic field strength is given as:

B0 = (E0/c)

= (48)/ (3 × 108)

=1.6x10-7 T

(c) Energy density of the electric field is given as:

UE = (1/2) (E2ε0)

And, energy density of the magnetic field is given as:

UB =B2(1/2μ0)

Where,

ε0 = Permittivity of free space

μ0 = Permeability of free space

We have the relation connecting E and B as:

E = cB ... (1)

Where,

c= (1/√μ0ε0)   … (2)

Putting equation (2) in equation (1), we get

E= (1/√μ0ε0)   B

Squaring both sides, we get

E2=1/ (μ0ε0) B2

ε0E2 = (B20)

(1/2)ε0E2 = (1/2)μ 0B2

=>UE=UB

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