The U.S. Food and Drug Administration limits the radiation leakage of microwave

Question

The U.S. Food and Drug Administration limits the radiation leakage of microwave ovens to no more than 5.0 mW/cm2 at a distance of 2.00 in. A typical cell phone, which also transmits microwaves, has a peak output power of about 2.0 W.
(a) Approximating the cell phone as a point source, calculate the radiation intensity of a cell phone at a distance of 2.00 in.
1 .
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. mW/cm2

How does the answer compare with the maximum allowable microwave oven leakage? (Enter the ratio of cellphone intensity to microwave intensity.)
Icell phone
Imicrowave
= 2 .

(b) The distance from your ear to your brain is about 2 in. What would the radiation intensity in your brain be if you used a Bluetooth headset, keeping the phone in your pocket, 1.0 m away from your brain? Most headsets are so-called Class 2 devices with a maximum output power of 2.5 mW. (Include both the radiation from the phone and the headset.)
3 .

Explanation / Answer

The microwave oven intensitry I = 5.0 mW/cm^2 = 50 W/m^2 the distance d = 2.0in the cell phone power P2 = 2.0W intensity I = 5.0 mW/cm2 = 50.0 W/m2 distance r = 2.00 in = 2(0.0254 m) Power P = I (4pr^2) = (50.0 W/m2)(4p(2(0.0254 m))^2) = 1.6214639 W The ratio of cellphone power / microwave power is = ( 2.0 W)/(1.6214639 W) = 1.2334533 --------------------------------------------------------------------------------------- (b) The maximum output power of headsets P2 = 2.5*10^-3W The distance from your ear to your brain is about d = 2 in = 2 * 2.54 cm = 0.0508 m The radiation intensity in the brain when we used a Bluetoothheadset is I = (P/d2) Here,P = 2.5 mW at distance r = 1.0m I = (2.0 W/4p(0.0508)2) = 61.7 W/m^2 for headset I = (2.5*10^-3) / 4p(1)^2 =1.99*10^-4 W/cm2 therefore the total intensity I = 61.7 + 0.000199 = 61.7 W/m^2

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