Suppose that the electric field amplitude of an electromagnetic wave travelling along the x-axis is \( E_0=120\,\mathrm{N\,C^{-1}} \) and its frequency is \( \nu=50\,\mathrm{MHz} \). The expression for the electric field \( E_y \) (along the y-axis) is (\( E_y \) in \( \mathrm{N\,C^{-1}} \), x in m, t in s)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 120\sin\left(1.05\,x-3.14\times10^{8}\,t\right) \)
For a wave travelling along the x-axis with its electric field along y, \( E_y=E_0\sin(kx-\omega t) \). The angular frequency is \( \omega=2\pi\nu \). The magnitude of the wave vector, \( k \), follows from \( \omega=ck \), with \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \).
The data are:
\( E_0=120\,\mathrm{N\,C^{-1}} \)
\( \nu=50\,\mathrm{MHz}=5\times10^{7}\,\mathrm{Hz} \)
The angular frequency is:
\( \begin{aligned} \omega &= 2\pi\nu \\ &= 2\times3.14\times5\times10^{7}\,\mathrm{s^{-1}} \\ &= 3.14\times10^{8}\,\mathrm{rad\,s^{-1}} \end{aligned} \)
The value of \( k \) is:
\( \begin{aligned} k &= \frac{\omega}{c} \\ &= \frac{3.14\times10^{8}\,\mathrm{rad\,s^{-1}}}{3\times10^{8}\,\mathrm{m\,s^{-1}}} \\ &\approx 1.05\,\mathrm{rad\,m^{-1}} \end{aligned} \)
Therefore, the electric field is:
\( E_y=120\sin\left(1.05\,x-3.14\times10^{8}\,t\right) \)