00:01
Have an radio tower which is emitting power p equals to 30 .0 kilowatt equally in all directions.
00:07
So assuming that radio waves that hit the earth are reflected, so for the part a, we have to calculate the magnitude of the pointing vector as at a distance small r equals to 12 .0 kilometer.
00:20
Okay.
00:21
So the radiation emitted by the radio tower will be in the form of hemispair because it will reflect from the earth surface, so area of the hemispair, this will be equals to 2 pi r square.
00:34
Okay, so pointing vector as it can be written as power divided by area and it can be written as e squared by mu not multiplied by c.
00:44
Okay, so power can be calculated from here.
00:49
So pointing vector as can be calculated from here.
00:52
So as this will be equals to power p divided by area and this will be the area.
00:56
So, substituting 2 pi r square, so now we can substitute values.
01:00
So power p it is equals to 30 .0 multiplied by 10 to the power 3 watt divided by 2 pi multiplied by r which is 12 .0 multiplied by 10 to the power 3 meter whole square.
01:12
So from here after solving pointing vector s comes out to be 3 .32 multiplied by 10 to the power minus 5 watt per meter square.
01:21
So this is the answer for the part a.
01:23
Okay, now we can calculate the part b in which we have to calculate the root mean square value of the electric field.
01:30
So we have to calculate here erms.
01:33
So now we know that the power or pointing vector as this is equal to erms square divided by mu not multiplied by c...