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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.

User Mary Chang
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1 Answer

3 votes

Answer:


(dB)/(dt) = 3.03 * 10^6 T/s

Step-by-step explanation:

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


P = 170 W

now we know that


P = (N)/(t) ((hc)/(\lambda))

now we know that energy density is given as


u = (B^2)/(2\mu_0) + (\epsilon_0 E^2)/(2)

now we have


E = B c


u = (B^2)/(2\mu_0)

intensity is defined as


I = (P)/(A)

now we have


(I)/(c) = u = (B^2)/(2\mu_0)[/tex]

now we have


(dB)/(dt) = \omega B


(dB)/(dt) = (2\pi c B)/(\lambda)


(dB)/(dt) = (2\pi c √(2\mu_0 I))/(\lambda\sqrt c)

here we have


I = (P)/(4\pi r^2)


I = (170)/(4\pi (410)^2)


I = 8.05 * 10^(-5)

now we have


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


(dB)/(dt) = 3.03 * 10^6 T/s

User Rudresh
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