Question

An electromagnetic wave travelling in the x-direction has frequency of 2×10 14 Hz and electric field amplitude of 27 Vm −1. From the options given below, which one describes the magnetic field for this wave?

Answer 1

The magnetic field for this wave is

$B(x, t)=(9*10^{-8}T )i sin[2\pi (1.5*10^{-8}x-2*10^{14}t )]$.

Electromagnetic Wave:

The oscillations between an electric field and a magnetic field produce waves known as electromagnetic waves or EM waves. In other words, magnetic and electric fields oscillate to form electromagnetic (EM) waves.

Given,

         Frequency of the wave;  $f=$ $2*10^{14} Hz$

         The amplitude of the Electric Field;  $E_{0} = 27 N/C$

We know that Maxwell's equation is given by,

                                      $\frac{E_{0} }{B_0} =c$        .....................(1)

Where,

         $E_{0} =$ The amplitude of the electric field

         $B_{0}=$ The amplitude of the magnetic field

            $c=$ The velocity of light = $3*10^{8} N/m^{2}$

By solving equation (1), we get

                                     $B_{0} =\frac{27}{3*10^{8} }$

                                      $B_{0}=9*10^{-8}$

We know that,

         Angular frequency; ω $= 2\pi f$

The equation of the magnetic field can be given by

                  $B_{0}sin(kx-\omega t) =B_{0}sin(2\pi (\frac{x}{\lambda}-ft ))$

                  $B_{0}sin(kx-\omega t)=9*10^{-8} sin(2\pi (\frac{x}{\lambda}-2*10^{14} t))$.

Hence,

         The magnetic field for this wave is

                        $B(x, t)=(9*10^{-8}T )i sin[2\pi (1.5*10^{-8}x-2*10^{14}t )]$.

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