Question

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0xx10^10 H_z and amplitude 48V_m^-1 (a) What is the wavelength o f the wave? (b) What is the amplitude of the oscillating magnetic field. (c) Show that the average energy density of the field E equals the average energy density of the field B.[c=3xx10^8 ms^-1].

Answer 1

Given:

Frequency $(f)$   $= 2.0\times10^10_Hz$

Amplitude$(E_0)$  $= 48 V_m^{-1}$

To find:

(a) Wave length of wave

(b) Amplitude of the oscillating magnetic field.

(c) Show that the average energy density of the field E equals the average energy density of the field B.

Solution:

(a) Wavelength is given by

                          $\lambda = \frac{c}{f}$

                                                               [ $c=3\times 10^8m s^{-1}$]

                        $\lambda = \frac{3\times 10^8}{2\times 10^{10}}$

                        $\lambda = 1.5 \times 10^{-2} m$

          Thus, wavelength is   $\lambda = 1.5 \times 10^{-2} m$

(b) Amplitude of the oscillating magnetic field$(B_0) = \frac{E_0}{c}$

                      $B_0= \frac{48}{3\times10^8}$

                     $B_0 = 1.6\times 10^{-7}T$

Thus, Amplitude of the oscillating magnetic field $B_0 = 1.6\times 10^{-7}T$

(c)  Average energy density of the electric field $u_E= \frac{1}{2} \epsilon_0E^2$

                                                                                        $E_{rms} = \frac{\epsilon_0}{\sqrt{2} }$

So,                $u_E= \frac{1}{4} \epsilon_0 (E_0)^2$     ........$(i)$

Average energy density of the magnetic field

                          $u_b= \frac{(B_0)^2}{4\mu_0} ...........(ii)$

Putting value of   $(B_0) = \frac{E_0}{c}$

                         $u_b= \frac{(E_0)^2}{4\mu_0c^2}$

                        $u_b= \frac{(E_0)^2}{4\mu_0\frac{1}{\epsilon_0\mu_0} } ......(ii)$                                      $c= \frac{1}{\sqrt{\epsilon_0\mu_0} }$

After substituting the values from $eq (i) and (ii)$

we get,

                            $u_b = u_E$

Thus, The average energy density of the field E equals the average energy density of the field B.