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neonofarm [45]
3 years ago
10

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?
Physics
1 answer:
Schach [20]3 years ago
7 0

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

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