Question

The intensity of the sunlight reaching Earth is 1380 W m−2. Assume this light to be a plane, monochromatic wave. Find the amplitudes of electric and magnetic fields in this wave.

Answer 1

Electric field  $E_0 = 1.0192 \times 10{3}\frac{Newton}{coulomb}$

Magnetic field $B_0 = 3.4\times10^{-6}$

Explanation:

Given that,

Intensity of sunlight at the earth $I = 1380 \frac{W}{m^{2}}$

When light is a plane and electromagnetic wave then amplitudes of electric and magnetic field have to be found.  

Relation b/w field  and intensity of the wave is given by,  

$I = \frac{1}{2}\epsilon_0E^{2}_0 c$

Where,

$c = \textrm{the speed of light in vacuum} = 3 \times10^{8} ms^{-1}$

$\epsilon_0 = \textrm{electric permittivity of free space(vacuum)} = 8.85\times10^{-12} C^{2}N^{-1}m^{-2}$

$E_0 = \textrm{amplitude of electric field}$

$E_0 = \sqrt{\frac{2I}{c\epsilon_0}$  

$= \sqrt{\frac{2\times1380}{3\times 10^{8} \times 8.85\times10^{-12}}$  

Electric field  $E_0 = 1.0192 \times 10{3}\frac{Newton}{coulomb}$

Relation b/w electric field  and magnetic field $B_0$ in free space is given by

$B_0 = \frac{E_0}{c}$  

$B_0 = \frac{1.0192 \times 10^{3}}{3\times 10^{8}}$

Magnetic field $B_0 = 3.4\times10^{-6}$  

Answer 2

Option (a) is Answer

Explanation:

Since,  

• Intensity of sunlight reaching earth = 1380 $Wm^-^2$

• Amplitude of Magnetic Field is --

$E_0 = \sqrt \frac {2I}{\epsilon _0 c}$

= $\sqrt \frac {2 \times 1380}{8.85 \times 10^{-12} \times 3 \times 10^8}$

= $1.02 \times 103 NC^-^1$

• So, $B_0 = \frac {E_0}{c}$  

= $\frac {1.02 \times 10^3}{3 \times 10^8T}$

= $3.4 \times 10^-^6 T$