Question

An isotropic point source emits light at wavelength 500 nm, at the rate of 185 W. A light detector is positioned 380 m from the source. What is the maximum rate ∂B/∂t at which the magnetic component of the light changes with time at the detector's location?

Answer 1

Answer:

$\frac{dB}{dt} = 3.49 *10^{6} \ \ T/s$

Explanation:

Given that

An isotropic point source emits light at a wavelength $\lambda$ = 500 nm

Power = 185 W

Radius = 380 m

Let's first calculate the The intensity  of the wave , which is = $\frac{Power }{Area}$

= $\frac{Power}{4 \pi r ^2}$

=  $\frac{185 \ W}{ 4 \pi (380)^2}$

= $1.0195*10^{-4} \ W/m^2$

Now;

The amplitude of the magnetic field is calculated afterwards by using poynting vector

i.e

$I = (\frac{c}{2 \mu_0 })B_{max^2}$

$B_{max^2} = (\frac{2 \mu_0 I}{ c})$

$B_{max^2} = (\frac{2 *4 \pi *10^{-7}*1.0195*10^{-4}}{ 3*10^8})$

$B_{max^2} = 8.5409*10^{-19}$

$B_{max} = \sqrt {8.5409*10^{-19}}$

$B_{max} = 9.242*10^{-10}$

The magnetic field wave equation can now be expressed as;

$B = B_{max} sin (kx - \omega t)$

Taking the differentiation

$\frac{dB}{dt}= - \omega B_{max} \ cos ( kx - \omega t)$

The maximum value ;

$\frac{dB}{dt} = \omega B _{max}$

where ;

$\omega = 2 \pi f\\\omega = \frac{2 \pi c}{\lambda}$

then

$\frac{dB}{dt} = \frac{2 \pi c}{\lambda} B _{max}$

$\frac{dB}{dt} = \frac{2 \pi 3*10^8*9.242*10^{-10}}{500*10^{-9}}$

$\frac{dB}{dt} = 3484751.917$

$\frac{dB}{dt} = 3.49 *10^{6} \ \ T/s$