Suppose that the electric field amplitude of an electromagnetic wave is Eo = 120 N/C and that its frequency is v = 50.0 MHz. (a) Determine, Bo, ω, k, and λ. (b) Find expressions for E and B.

Electric field amplitude, Eo = 120 N/C

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

Speed of light, c = 3 x 10⁸m/s

Ans (a).

Magnitude of magnetic field strength is given as:

B₀ =E₀ /c

= 120/(3 x 10⁸)

= 4 x 10⁻⁷ T= 400 nT

Angular frequency of source is given as:

ω =2πν = 2π x 50 x 10⁶ = 3.14 x 10⁸ rad/s

Propagation constant is given as:

k = ω/c   = (3.14 x 10⁸ )/(3 x 10⁸) =6.0 m

Ans (b).

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

Equation of electric field vector is given as:

E = E₀ sin (kx-ωt)j 

= 120 sin [ 1.05 x -3.14 x 10⁸ t] j

And, Magnetic field vector is given as :

B = B₀ sin (kx-ωt)k

= (4 x 10 sin⁻⁷ )[ 1.05 x -3.14 x 10⁸ t] k