Question

The oscillating electric field in a plane electromagnetic wave is given by
`{50\sin (\omega t-kx) V/m}`, and the frequency of electric field is 2 × 10⁷ Hz. Find: (a) Find
`{\omega}`,
`{\lambda}`, k and write the expression for the electric field (b) Find
`{B_(0)}` and write the expression for magnetic field (c) Predict the Direction of propagation of electromagnetic wave.​

Answer 1

Here, we are given with:

`{:\implies \quad \longrightarrow \begin{cases}\sf E_(0)= 50\: V/m\\ \sf \\u = 2* 10^(7)Hz\end{cases}}`

(a) So now, we can thus obtain

`{:\implies \quad \sf \omega =2\pi \\u}`

`{:\implies \quad \sf \omega =2\pi (2* 10^(7))}`

`{:\implies \quad \boxed{\bf{\omega = 4\pi * 10^(7)\: s^(-1)}}}`

Now, finding
`\lambda`

`{:\implies \quad \sf \lambda =(c)/(\omega)=(3* 10^(8))/(4\pi * 10^(7))}`

`{:\implies \quad \boxed{\bf{\lambda \approx 2.39\:m}}}`

Now, finding k

`{:\implies \quad \sf k=(2\pi)/(\lambda)=(200\pi)/(239)}`

`{:\implies \quad \boxed{\bf{k\approx 2.63\:m^(-1)}}}`

Thus, expression for the electric field is:

`{:\implies \quad \boxed{\bf{E=50\sin \bigg(4\pi * 10^(7)t-(263x)/(100)\bigg)}}}`

(b) Now, here

`{:\implies \quad \sf B_(0)=(E_(0))/(c)=(50)/(3* 10^(8))}`

`{:\implies \quad \boxed{\bf{B_(0)\approx 16.67×10^(-7)\:T}}}`

Thus, expression for the magnetic field:

`{:\implies \quad \boxed{\bf{B_(y)=16.67* 10^(-7)\sin \bigg(4\pi * 10^(7)t-(263x)/(100)\bigg)\:T}}}`

(c) The electromagnetic wave propagates along Z-axis