Current over distance

 This is a follow up on the last post.

I mentioned in the last post that current I and the length of wire segment over which that current was flowing was all that determined how strong the field strength would be from an antenna.

The actual term that was used by Edmund A. Laporte in his chapter on ANTENNAS in the Radio Engineering Handbook (edited by Henney) was Meter-Amperes. I tend to think in Feet-Amperes but both terms are good. He defines this to be the integral of i dl where i is the rms current in a length of the antenna dl. This means if you take the integral of the current over a length of a halfwave dipole you get the result in Meter- Amperes. The area under the curve represents the summation of the total current over that halfwave length of howevermany meters the antenna is long. Of the current sum was say 10 amps and the length was 20 meters then the result would be 200 meter- Amperes.  In the Radio Engineers Handbook there are charts that shows what the field strength will be in millivolts/ meter at one mile from the antenna for any frequency if we know the Meter-Amperes of the antenna. It appears to be a simple linear function. Double the Meter-Amperes and you double the field strength in millivolts per meter or microvolts per meter. 

Note that the Meter-Amperes is a summation of many equal segments of the antenna with each segment having a slightly different current.segments near the center have a maximum current value and segments moving away from the center have decreasing values of current. Think of it this way. We have a wire 10 meters long. Draw the sinusoid current distribution along the wire and break the 10 meter length of wire up into a number of equal segments each one meter long. If the maximum current near the center is 5 amps we will have two segments near the center that have 5 meter-amperes each, then on each side of center we may have a total of 4 segments ( two on each side) with 3 Meter-Amperes each then on the two ends we may have 4 segments ( two on each side) with 1 meter-ampere each.

By adding these ten segments we get 5 + 5 + 3+3+3+3 +1+1+1+1 = 26 meter-Amperes total. This is a crude estimation, we would have to use integral calculus or maybe break the wire down into 100 sections to get a precise number. However,  it demonstrates that the two center section contribute more Meter- Amperes than any four other sections. It also shows that the ends, where the current is low does not contribute very much at all. 

I made a quick sketch using better numbers. This should help visualize the way the area is broken down into several 1 meter segments with the area above each segment being slightly different depending on the current distribution. The radiation from each segment being proportional to the area associated with each segment. That area being representative of meter-Amperes. At the bottom I included the meter- ampere definition that was in the chapter authored by Laport.







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