00:01
So here we have the intensity of radiation at a distance r from a point source, and that point source is radiating a total power p.
00:07
This would be the intensity for part a.
00:10
This would be the intensity equaling the power divided by the effective area, essentially.
00:16
This would be equalling rather the power divided by 4 pi r squared.
00:23
And so at a distance here we have r is equaling 2 .0 inches.
00:30
From a cell phone radiating a total power of p equaling 2 .0 watts, or better yet, we can say 2 .0 times 10 to the third millawatts.
00:48
We can say that the intensity then i is equaling 2 .0 times 10 to the third milliwatts, and this would be divided by 4 pi, multiplied by 2 .0 inches, multiplied by 2 .54 centimeters for every 1 inch.
01:12
And this would be squared.
01:15
And this is giving us 6 .2 milliwats for every centimeter squared.
01:23
So this would be our final answer for part a.
01:26
And from here you can see that the intensity, this intensity i, is 24 % higher than the maximum allowed leakage from a microwave.
02:03
Again, at this distance of 2 inches.
02:08
So continuing on for part b, we have a bluetooth headset...