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:
**E**_{y}= E_{0}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:
**B**_{z}=B_{0}sin(kx-ωt)- Where B
_{Z}= electric field along z-axis and x=direction of propagation of wave.

- Where B

__Problem:-__

Suppose that the electric field amplitude of an electromagnetic wave is E_{0} = 120 N/C and thatits frequency is ν = 50.0 MHz (a) Determine, B_{0}, ω, k, and λ. (b) Find expressions for E andB.

__Answer:-__

Electric field amplitude, E_{0} = 120 N/C

Frequency of source, ν = 50.0 MHz = 50 × 10^{6} Hz

Speed of light, c = 3 × 10^{8} m/s

- Magnitude of magnetic field strength is given as:

B_{0}=(E_{0}/c)

=(120)/(3x10^{8})=4 x10^{-7} T=400nT

Angular frequency of source is given as:

ω = 2πν = 2π ×50 × 10^{6}

= 3.14 × 10^{8}rad/s

Propagation constant is given as:

k=(ω/c)

= (3.14 × 10^{8})/ (3x10^{8}) =1.05rad/m

Wavelength of wave is given as:

λ=(c/v) =(3x10^{8})/(50x10^{6}) = 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**=E_{0} sin (kx –ωt)ˆj

=120 sin [1.05 x - 3.14x10^{8}t]ˆj

And, magnetic field vector is given as:

**B**=B_{0} sin (kx –ωt)ˆk

=4 x10^{-7}) sin1.05 x - 3.14x10^{8}t]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. B_{0} = (E_{0}/c)the magnitude of B is

B= (E/c)

= (6.3 V)/ (3x10^{8}m/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^{–8}**ˆk**T

__Problem:-__

The magnetic field in a plane electromagnetic wave isgiven by

B_{y} = 2 × 10^{–7} sin (0.5×10^{3}x+1.5×10^{11}t) 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

B_{y} = B_{0} sin [2 π ((x/ λ) + (t/T))]

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

(1/T) = ν = (1.5 x10^{11})/ (2 π)

= 23.9 GHz

(b) E_{0} = B_{0}c = 2×10^{–7} T × 3 × 10^{8} m/s = 6 × 10^{1} 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

E_{z} = 60 sin (0.5 × 10^{3}x + 1.5 × 10^{11} t) V/m

__Problem:-__

Light with an energy flux of 18 W/cm^{2} falls on a non-reflectingsurface at normal incidence. If the surface has an area of20 cm^{2}, 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/cm^{2}) × (20 cm^{2}) × (30 × 60)

= 6.48 × 10^{5}J

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

p = (U/c) = (6.48 x 10^{5}J)/ (3x10^{8}m/s)

= 2.16 × 10^{–3} kg m/s

The average force exerted on the surface is

F = (p/t) = (2.16x10^{-3})/ (0.18x10^{4})

=1.2 x10^{-6} N

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

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