Calculating ratio energy radiated in a loop and comment on result

In summary, designers of electrical circuits use the equation E=2.6AI0f2/r to calculate the maximum amplitude of the radiated electric field produced by an alternating current. By applying this equation to two different states, where the current and frequency are different, it can be seen that the energy radiated is proportional to the square of the amplitude. State 2, with a current of 0.01A and a frequency of 500MHz, has 5 times more energy radiated than State 1, with a current of 1A and a frequency of 10MHz. This result also shows that higher frequencies and currents result in a higher amount of energy being radiated.
  • #1
craig.16
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0

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 I0 flowing in a loop of area A cm2 to be given by

E=[itex]\frac{2.6AI0f2}{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{2.6AI0f2}{r}[/itex][itex]\mu[/itex]Vm-1

The Attempt at a Solution


For state 1 (where I0=1A and f=10MHz):
E=[itex]\frac{2.6A(1)(10*106)2}{r}[/itex]
=[itex]\frac{2.6*1014A}{r}[/itex]

For state 2 (where I0=0.01A and f=500MHz):
E=[itex]\frac{2.6A(0.01)(5*108)2}{r}[/itex]
=[itex]\frac{6.5*1015A}{r}[/itex]

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

=[itex]\frac{\sqrt{6.5*1015}}{\sqrt{2.6*1014}}[/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?
 
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  • #2
Doesn't matter, worked out what I did wrong.
 

1. What is the formula for calculating the ratio of energy radiated in a loop?

The formula for calculating the ratio of energy radiated in a loop is R = E/E_total, where R is the ratio, E is the energy radiated in the loop, and E_total is the total energy in the loop.

2. How is the energy radiated in a loop measured?

The energy radiated in a loop can be measured using instruments such as a calorimeter or a radiation detector. These instruments can detect and measure the amount of energy being radiated from the loop.

3. What factors affect the ratio of energy radiated in a loop?

The ratio of energy radiated in a loop can be affected by various factors such as the material of the loop, the temperature of the loop, and the distance between the loop and the radiation source.

4. How does the result of the ratio of energy radiated in a loop impact the efficiency of the loop?

The result of the ratio of energy radiated in a loop can impact the efficiency of the loop. A higher ratio indicates that a larger percentage of the energy in the loop is being radiated, which can result in a more efficient loop.

5. What are some practical applications of calculating the ratio of energy radiated in a loop?

Calculating the ratio of energy radiated in a loop can be useful in designing and optimizing various devices that rely on radiation, such as solar panels and infrared heaters. It can also be used in studying and understanding the properties of different materials and their ability to radiate energy.

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