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## Homework Statement

Designers of electrical circuits often take the maximum amplitude of the radiated electric field (at a large distance r) produced by an alternating current I

_{0}flowing in a loop of area A cm

^{2}to be given by

E=[itex]\frac{2.6AI

_{0}f

^{2}}{r}[/itex][itex]\mu[/itex]Vm

^{-1}

where f is the frequency in MHz. Calculate the ratio energy radiated when the loop carries a current of 1A at 10MHz to that for 0.01A at 500MHz. Comment on your result. Remember that for a wave the energy carried is proportional to the square of the amplitude.

## Homework Equations

E=[itex]\frac{2.6AI

_{0}f

^{2}}{r}[/itex][itex]\mu[/itex]Vm

^{-1}

## The Attempt at a Solution

For state 1 (where I

_{0}=1A and f=10MHz):

E=[itex]\frac{2.6A(1)(10*10

^{6})

^{2}}{r}[/itex]

=[itex]\frac{2.6*10

^{14}A}{r}[/itex]

For state 2 (where I

_{0}=0.01A and f=500MHz):

E=[itex]\frac{2.6A(0.01)(5*10

^{8})

^{2}}{r}[/itex]

=[itex]\frac{6.5*10

^{15}A}{r}[/itex]

As the energy carried is proportional to the square of the amplitude this gives:

=[itex]\frac{\sqrt{6.5*10

^{15}}}{\sqrt{2.6*10

^{14}}}[/itex]

=5

Showing that state 2 has 5 times more energy radiated than state 1. Also the higher the frequency and current, the higher the amount of energy radiated. Is there anything else that I've missed out since this seems like state of the obvious stuff to point out as a result?