(A)Find the Total Power Radiated by the Sun Assuming That 1400 W.M-2 Is the Intensity Of

Question 7

(a) Find the total power radiated by the Sun assuming that 1400 W.m-2 is the intensity of the sunlight reaching Earth. The Sun is 1.501011 m away and the radius of Earth is 6.37106 m.

(b) The peak electric field 10.0 m away from a small light bulb is 2.00 V.m-1. Find (i) the peak magnetic field; (ii) average intensity of the light 10.0 m away; (iii) the total power radiated by the light bulb.

(c) A He-Ne laser has a power output of 5.0 mW in a beam of 1.00 mm diameter. Find (i) the intensity of the beam; (ii) the peak electric field in the beam; (iii) the peak magnetic field in the beam.

(d) The lights in the laboratory are 40.0 W fluorescent tubes that are cylindrical tubes about 2m long. Assume that all the light is emitted radially from these tubes at an average wavelength of 550 nm. Estimate (i) the intensity of the electromagnetic waves produced by a single tube 1.00 m away from it; (ii) the magnitude of the electric field in these electromagnetic waves; (iii) the speed of these electromagnetic waves.

(Reference: http://phys.csuchico.edu/kagan/4B/Home_Page.html)

a04/p2/emproblems/em07.doc 6/27/2018 4:40 PM 1 Solution

(a) Since the Sun radiates EM waves uniformly in all directions this intensity will be the same over the surface area of a sphere that is 1.501011m away. Using the definition of power and the definition of intensity d U P P I    . A d t A 4 r 2 Solving for the power, P= I(4 p r2 ) = (1400)(4 p )(1.50 x 10 11 ) 2 W P= 3.96 x 1026 W .

(b) E 2.00 (i) The ratio of the peak fields is, B =o =T = 6.67 nT . o c 3.00 108

(ii) The average intensity is related to the peak fields 2 2 Eo (2.00) -2 - 3 - 2 I = =-7 8 W.m = 5.31 10 W.m . 2moc 2(4 p创 10 )(3.00 10 )

(iii) Using the definition of intensity P P I� � p P =4 p r2 I = 4 (10.0) 2 (0.00531) W 6.67 W . A4p r 2 (c) (i) The intensity is the power per unit area P 5 10-3 S = =W.m-2 = 6.64 10 3 W.m -2 . A p(0.0005)2 (ii) The Poynting Vector is related to the peak field 2 Eo 3 -1 Savg=� E o m2 = o cS avg 2.19 10 V.m . 2moc E (iii) The ratio of the peak fields is the speed of light: B =o = 7.30m T . o c (d) (i) The intensity is the power per unit area. The area in this case is a cylinder of radius r = 1.00 m and length L = 2.00 m.

P 40.0 S = =W.m-2 = 3.18 W.m -2 avg A 2p (1)(2) (ii) The average intensity is the Poynting Vector

2 Eo Savg= � E o2m c S avg 2mo c

-7 8 -1 -1 Eo =(2)(4p 创 10 )(3 10 )(3.18) V.m = 49.0 V.m

(iii) The speed of EM waves is always the speed of light 3.00108 m.s-1 a04/p2/emproblems/em07.doc 6/27/2018 4:40 PM 2