Suppose that the electric field amplitude of an electromagnetic wave is ${{E}_{0}}=120N{{C}^{-1}}$ and that its frequency is $\nu =50.0MHz$.
(a) Determine $B_0$, ω, k and λ
(b) Find the expression for E and B

Hint: Electromagnetic waves bring energy to their system through their electric and magnetic fields. As the value for amplitude and frequency of wave is given, we shall use the electric field intensity and magnetic field intensity in the electromagnetic wave to determine the values.

Complete step by step answer:
The equation of an electromagnetic wave can be written as
$E={{E}_{0}}\sin \left( kx-\omega t \right)$ on basis of electric field intensity and
$B={{B}_{0}}\sin \left( kx-\omega t \right)$ on the basis of magnetic field intensity.
(a) The values given in the question are ${{E}_{0}}=120N{{C}^{-1}}$ and $\nu =50.0Mhz$
Substituting frequency value in the angular frequency formula,
$\omega =2\pi f$
Where, ‘f’ is the frequency of the electromagnetic wave
$\omega =2\pi \times 50\times {{10}^{6}}$
$\omega =100\pi \times {{10}^{6}}rad{{s}^{-1}}$
We also know that for an electromagnetic wave, $\dfrac{\omega }{k}=c$
Where, ‘c’ is the velocity of light and
              ‘k’ is a constant
$\dfrac{100\pi \times {{10}^{6}}}{k}=3\times {{10}^{8}}$
So, $k=\dfrac{\pi }{3}rad{{m}^{-1}}$
The wavelength is given as,
$\lambda =\dfrac{c}{\nu }$
Let us substitute the known values,
$\lambda =\dfrac{3\times {{10}^{8}}}{50\times {{10}^{6}}}$
$\lambda =6m$
The electric field intensity can also be expressed in terms of magnetic intensity as
${{B}_{0}}=\dfrac{{{E}_{0}}}{c}$
${{B}_{0}}=\dfrac{120}{3\times {{10}^{8}}}=40\times {{10}^{-8}}T$
(b) The standard equation for electric field can be given as
$E={{E}_{0}}\sin \left( kx-\omega t \right)$
On substituting the known values, we get
$E=120\sin \left( \dfrac{\pi }{3}x-100\pi \times {{10}^{6}}t \right)$
The standard equation for magnetic field is given as
$B={{B}_{0}}\sin \left( kx-\omega t \right)$
Let’s substitute the known values,
$E=40\times {{10}^{-8}}\sin \left( \dfrac{\pi }{3}x-100\pi \times {{10}^{6}}t \right)$
The expression for E and B is derived.
Consider the wave is moving along the x-axis, the electric field along the y-axis and the magnetic field along the z-axis. The expressions can be written as
$E=120\sin (1.05x-3.14\times {{10}^{8}}t)\widehat{j}$ and
$B=4\times {{10}^{-7}}\sin (1.05x-3.14\times {{10}^{8}}t)\widehat{k}$

Note: The electromagnetic wave equation can also be written as $E={{E}_{0}}\cos \left( \omega t-kx \right)$. The behaviour is the same for both but the difference between them lies in how a cos and sine wave begins at x=0 and t=0. The sine wave begins from zero and proceeds to build a sinusoidal wave, but the cosine wave starts only from 1.