Class 12 Physics Electromagnetic Waves Nature of EM waves

Nature of EM waves

  • EM waves are transverse waves.
  • The transverse waves are those in which direction of disturbance or displacement in the medium is perpendicular to that of the propagation of wave.
  • The particles of the medium are moving in a direction perpendicular to the direction of propagation of wave.

 

  • In case of EM waves the propagation of wave takes place along x-axis, electric and magnetic fields are perpendicular to the wave propagation.
  • This means wave propagation ---à x-axis , electric field ------> y-axis,

 magnetic field --à z-axis.

  • Because of this EM waves are transverse waves in nature.
  • Electric field of EM wave is represented as:
    • Ey = E0 sin(kx–ωt)
      • Where Ey= electric field along y-axis and x=direction of propagation of wave.
      • Wave number k=(2π/λ)
    • Magnetic field of EM wave is represented as:
      • Bz=B0sin(kx-ωt)
        • Where BZ = electric field along z-axis and x=direction of propagation of wave.

Problem:-

Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and thatits frequency is ν = 50.0 MHz (a) Determine, B0, ω, k, and λ. (b) Find expressions for E andB.

Answer:-

Electric field amplitude, E0 = 120 N/C

Frequency of source, ν = 50.0 MHz = 50 × 106 Hz

Speed of light, c = 3 × 108 m/s

  • Magnitude of magnetic field strength is given as:

B0=(E0/c)

=(120)/(3x108)=4 x10-7 T=400nT

Angular frequency of source is given as:

ω = 2πν = 2π ×50 × 106

= 3.14 × 108rad/s

Propagation constant is given as:

k=(ω/c)

= (3.14 × 108)/ (3x108) =1.05rad/m

Wavelength of wave is given as:

λ=(c/v) =(3x108)/(50x106) = 6.0m

(b) Suppose the wave is propagating in the positive x direction. Then, the electric field vectorwill be in the positive y direction and the magnetic field vector will be in the positive zdirection. This is because all three vectors are mutually perpendicular.

Equation of electric field vector is given as:

E=E0 sin (kx –ωt)ˆj

=120 sin [1.05 x - 3.14x108t]ˆj

And, magnetic field vector is given as:

B=B0 sin (kx –ωt)ˆk

=4 x10-7) sin1.05 x - 3.14x108t]ˆk

 

Problem:-

A plane electromagnetic wave of frequency25 MHz travels in free space along the x-direction. At a particularpoint in space and time, E = 6.3 ˆj V/m. What is B at this point?

Answer:-

 Using Eq. B0 = (E0/c)the magnitude of B is

B= (E/c)

= (6.3 V)/ (3x108m/s)

=2.1 x10-8 T

To find the direction, we note that E is along y-direction and thewave propagates along x-axis. Therefore, B should be in a directionperpendicular to both x- and y-axes. Using vector algebra, E × B shouldbe along x-direction. Since, (+ ˆj) × (+ ˆk) = ˆi, B is along the z-direction.

Thus, B = 2.1 × 10–8ˆkT

Problem:-

The magnetic field in a plane electromagnetic wave isgiven by

By = 2 × 10–7 sin (0.5×103x+1.5×1011t) T.(a) What is the wavelength and frequency of the wave?

(b) Write an expression for the electric field.

Answer:-

(a) Comparing the given equation with

By = B0 sin [2 π ((x/ λ) + (t/T))]

We get, λ = (2 π)/ (0.5x103) m =1.26cm and

(1/T) = ν = (1.5 x1011)/ (2 π)

= 23.9 GHz

 (b) E0 = B0c = 2×10–7 T × 3 × 108 m/s = 6 × 101 V/m

The electric field component is perpendicular to the direction ofpropagation and the direction of magnetic field. Therefore, theelectric field component along the z-axis is obtained as

Ez = 60 sin (0.5 × 103x + 1.5 × 1011 t) V/m

Problem:-

Light with an energy flux of 18 W/cm2 falls on a non-reflectingsurface at normal incidence. If the surface has an area of20 cm2, find the average force exerted on the surface during a 30minute time span.

Answer:-

The total energy falling on the surface is

U = (18 W/cm2) × (20 cm2) × (30 × 60)

= 6.48 × 105J

Therefore, the total momentum delivered (for complete absorption) is

p = (U/c) = (6.48 x 105J)/ (3x108m/s)

= 2.16 × 10–3 kg m/s

The average force exerted on the surface is

F = (p/t) = (2.16x10-3)/ (0.18x104)

=1.2 x10-6 N

How will your result be modified if the surface is a perfect reflector?

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