0 Masteringphysics HW 32)-Google Chrome i Secure https:f LMH0I32Exercise 32.17 m

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0 Masteringphysics HW 32)-Google Chrome i Secure https:f LMH0I32Exercise 32.17 myct/item View?assignment ProhlemID=86000578 Resources previous 9 of 28 next What is the visible light intensity at the surface of the bulb? Express your answer using two significant figures. Exercise 32.17 We can reasonably model a 60-W incandescent ightbulb as a sphere 6.5 cm in diameter. Typically, only about 5 % of the energy goes to visible light: he rest goes largely to nonvisible infrared radlation. Submit My Answere Glve Up Part B What is the amplitude of the electric field at this surface, for a sinusoidal wave with this intensity? Express your answer using two significant figures. Emax = Submit My Anawers ieUp Part C What is the amplitudc of the nagnetic ficld at this surface, for a sinusoidal wave with this intensity? Express your answer usin g two significant figures. Ask me anything 3:06 PM 1020/20177

Explanation / Answer

a)

We have the power (energy per unit time) of the bulb. The intensity is power per unit area. The area of the bulb with diameter d is A = 4r^2 = d^2 . Assuming the bulb is radiating uniformly in all directions and that only 5% of the energy goes in to light.

I = P/A = (0.05*60)/( *0.065^2) = 226 W/m2 = 230W/m2 (using 2sf)

b)

The intensity at a point is the average value of the Poynting vector at that point and the average value of the Poynting vector is related to E and B trough

I = Sav = EmaxBmax/2µ0 = ½ c0Emax^2

Therefore

Emax = sqrt(2I/0c)

= sqrt[(2*226)/(8.854*10^-12*3*10^8)] = 412.53 V/m = 410 V/m (using 2sf)

c)

Bmax = Emax*c*0*µ0 = Emax/c = 412.53/(3*10^8) = 1.4*10^-6 T

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