Question

An isotropic point source emits light at wavelength 510 nm, at the rate of 170 W. A light detector is positioned 410 m from the source. What is the maximum rate dB/dt at which the magnetic component of the light changes with time at the detector's location? The speed of light is c = 3 × 108 m/s, and μ0 = 4π × 10-7 H/m.

Answer 1

Answer:

$\frac{dB}{dt} = 3.03 \times 10^6 T/s$

Explanation:

As we know that the power emitted by the source is given as

$P = 170 W$

now we know that

$P = \frac{N}{t} (\frac{hc}{\lambda})$

now we know that energy density is given as

$u = \frac{B^2}{2\mu_0} + \frac{\epsilon_0 E^2}{2}$

now we have

$E = B c$

$u = \frac{B^2}{2\mu_0}$

intensity is defined as

$I = \frac{P}{A}$

now we have

$\frac{I}{c} = u = \frac{B^2}{2\mu_0}$[/tex]

now we have

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

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

$\frac{dB}{dt} = \frac{2\pi c \sqrt{2\mu_0 I}}{\lambda\sqrt c}$

here we have

$I = \frac{P}{4\pi r^2}$

$I = \frac{170}{4\pi (410)^2}$

$I = 8.05 \times 10^{-5}$

now we have

$\frac{dB}{dt} = \frac{2\pi\sqrt{2\mu_0 c (8.05 \times 10^{-5})}}{(510 nm)}$

$\frac{dB}{dt} = 3.03 \times 10^6 T/s$