|Class 12 Physics Electromagnetic Waves||Nature of EM waves|
Nature of EM waves
magnetic field --à z-axis.
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.
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
=(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:
= (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
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?
Using Eq. B0 = (E0/c)the magnitude of B is
= (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
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.
(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
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.
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?