Suppose that the electric field amplitude of an electromagnetic wave is E_{0} = 120 N/C and that its frequency is v = 50.0 MHz.

**(a)** Determine, B_{0}, ω, k, and λ.

**(b)** Find expressions for E and B.

#### Solution

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

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

Speed of light, c = 3 × 10^{8 }m/s

**(a)** Magnitude of magnetic field strength is given as:

`"B"_0 = "E"_0/"c"`

= `120/(3 xx 10^8)`

= 4 × 10^{−7 }T

= 400 nT

Angular frequency of source is given as:

ω = 2πv

= 2π × 50 × 10^{6}

= 3.14 × 10^{8} rad/s

Propagation constant is given as:

`"k" = ω/"c"`

= `(3.14 xx 10^8)/(3 xx 10^8)`

= 1.05 rad/m

Wavelength of wave is given as:

`lambda = "c"/"v"`

= `(3 xx 10^8)/(50 xx 10^6)`

= 6.0 m

**(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:

`vec"E" = "E"_0 sin ("k""x" - ω"t")hat"j"`

= `120 sin [1.05"x" - 3.14 xx 10^8"t"]hat"j"`

And, magnetic field vector is given as:

`vec"B" = "B"_0 sin ("kx" - ω"t")hat"k"`

`vec"B" = (4 xx 10^-7) sin [1.05"x" - 3.14 xx 10^8"t"]hat"k"`