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.
in mW/cm2?

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

(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.)
in mW/cm2?

Explanation / Answer

a. A uniform point source produces a radiation intensity given by

Power Density = P / (4pi R^2)

where P is the total power and R is the distance from the point source to the point where you are measuring the power density. This is nothing but the total power divided by the area of a sphere; I have assumed that the radiation is emitted equally in all directions (not using the dipole "butterfly" pattern here). Now, in your case

R = 2 in = 0.0508 meters

and, thus,

Power Density = 2 / (4pi (0.0508)^2) = 61.67 W/m^2

= 61.76*1000/100^2 = 6.17 mW/cm^2

so your cell phone is just above the FDA regulations 2in away from it.

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