8. Suppose that the electric field amplitude of an electromagnetic wave is E_0 = 120 N/C and that its frequency is \nu =50.0\ MHz (a) Determine,B_0 , \omega , \ k , \lambda (b) Find expressions for E and B.

Answers (1)
S Sayak

E0 = 120 NC-1 

\nu =50.0\ MHz

(a)

 Magnetic\ Field \ amplitude(B0) =\frac{E_{0}}{c}

=\frac{120}{3\times 10^{8}}

=400 nT

Angular frequency (\omega)  = 2\pi \nu

=2\times \pi \times 50\times10^{6}

=3.14\times10rad s-1

Propagation constant(k)

\frac{2\pi }{\lambda }

=\frac{2\pi \nu }{\lambda\nu }

=\frac{\omega }{c}

=\frac{3.14\times 10^{8} }{3\times 10^{8}}

=1.05 rad m-1

 Wavelength(\lambda) = \frac{c}{\nu }

=\frac{3\times 10^{8}}{50\times 10^{6}}= 6 m

Assuming the electromagnetic wave propagates in the positive z-direction the Electric field vector will be in the positive x-direction and the magnetic field vector will be in the positive y-direction as they are mutually perpendicular and E_{0}\times B_{0}  gives the direction of propagation of the wave.

\vec{E}=E_{0}sin(kx-\omega t)\hat{i}

= 120sin(1.05x-3.14\times 10^{8} t)NC^{-1}\hat{i}

 

\vec{B}=B_{0}sin(kx-\omega t)\hat{j}

= 400sin(1.05x-3.14\times 10^{8} t)\ nT\hat{j}

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