Answer:
[tex]\frac{dB}{dt} = 3.49 *10^{6} \ \ T/s[/tex]
Explanation:
Given that
An isotropic point source emits light at a wavelength [tex]\lambda[/tex] = 500 nm
Power = 185 W
Radius = 380 m
Let's first calculate the The intensity of the wave , which is = [tex]\frac{Power }{Area}[/tex]
= [tex]\frac{Power}{4 \pi r ^2}[/tex]
= [tex]\frac{185 \ W}{ 4 \pi (380)^2}[/tex]
= [tex]1.0195*10^{-4} \ W/m^2[/tex]
Now;
The amplitude of the magnetic field is calculated afterwards by using poynting vector
i.e
[tex]I = (\frac{c}{2 \mu_0 })B_{max^2}[/tex]
[tex]B_{max^2} = (\frac{2 \mu_0 I}{ c})[/tex]
[tex]B_{max^2} = (\frac{2 *4 \pi *10^{-7}*1.0195*10^{-4}}{ 3*10^8})[/tex]
[tex]B_{max^2} = 8.5409*10^{-19}[/tex]
[tex]B_{max} = \sqrt {8.5409*10^{-19}}[/tex]
[tex]B_{max} = 9.242*10^{-10}[/tex]
The magnetic field wave equation can now be expressed as;
[tex]B = B_{max} sin (kx - \omega t)[/tex]
Taking the differentiation
[tex]\frac{dB}{dt}= - \omega B_{max} \ cos ( kx - \omega t)[/tex]
The maximum value ;
[tex]\frac{dB}{dt} = \omega B _{max}[/tex]
where ;
[tex]\omega = 2 \pi f\\\omega = \frac{2 \pi c}{\lambda}[/tex]
then
[tex]\frac{dB}{dt} = \frac{2 \pi c}{\lambda} B _{max}[/tex]
[tex]\frac{dB}{dt} = \frac{2 \pi 3*10^8*9.242*10^{-10}}{500*10^{-9}}[/tex]
[tex]\frac{dB}{dt} = 3484751.917[/tex]
[tex]\frac{dB}{dt} = 3.49 *10^{6} \ \ T/s[/tex]
