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# Suppose that the electric field amplitude of an electromagnetic wave is E 0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B0,w, k, and l. (b) Find expressions for E and B.

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.

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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$$\times \pi \times 50$$\times$$10^{6}$

=3.14$\times$10rad s-1

Propagation constant(k)

$\frac{2\pi }{\lambda }$

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

=$\frac{\omega }{c}$

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

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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