We can reasonably model a 40W incandescent lightbulb as a sphere 6.0cm in diamet

Question

We can reasonably model a 40W incandescent lightbulb as a sphere 6.0cm in diameter. Typically, only about 5.0% of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation.

A) What is the visible light intensity (in W/m^2) at the surface of the bulb?

B) What is the amplitude of the electric field at this surface, for a sinusoidal wave with this intensity?
Emax= _________V/m

C) What is the amplitude of the magnetic field at this surface, for a sinusoidal wave with this intensity?
Bmax = _________

Explanation / Answer

a.)I = P/A = 0.05*40/(pi*(0.06^2)) = 176.84 W/m^2

b.)Emax = sqrt(2I/e0c) = sqrt(2*176.84/8.85x10^-12*3x10^8) = 364.98 V/m

c.)Bmax = Emax/c = 364.98/(3x10^8) = 1.216 x 10^-6 T = 1.216 uT

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