I would like the equation and coinciding proof that will allow me to obtain the EM produced by power lines?

Question

Regarding the typical electromagnetic radiation (EM) produced by a current flowing through a straight power line: What is the wavelength (or frequency if you prefer) of this radiation. Is there a function with the variables being: 1) the distance you are from the actual wire (since the intensity dissipates at a rate proportional to the distance squared) and 2) if necessary, the voltage and wattage of the current. I would assume that the EM is in the form of radio waves (and even possibly slightly higher frequency microwave radiation), but I would like to prove this. My goal is the proof, in the form of a function.

Answer 1

The frequency will be the same as that of the current in the wire. The reason is that the radiation is a combination of electric and magnetic fields. Mathematically, these fields are calculated from something called the vector potential (the electric field would be the negative time derivative of the vector potential while the magnetic field would be the curl of the vector potential [this last is the definition of the vector potential, actually]). The vector potential itself is gotten from an integral over space of the current density (the current divided by the area it passes through).

Now, the frequency of the current is generally represented as a sine or cosine function of time (e.g. cos[wt] for instance, with w being the angular frequency) multiplied onto the magnitude of the current. As the integral to get the vector potential is only over space, the sine function is a constant as far as the integral is concerned, so it can be pulled out of the integral unaffected. So the vector potential has the exact same frequency as the current.

Now, the vector potential is not the fields, so you need to get the electric and magnetic fields from the vector potential. Now, the magnetic field is the result of taking spatial derivatives (derivatives with respect to x,y, and z for instance), so again, the sine function representing the frequency is a constant with repect to the derivative, so it's unaffected, and thus the magnetic field is unaffected. Now, actually you could stop here, as the electric and magnetic fields have to have the same frequency, but this can also be seen by considering the electric field, which will be (in the absence of free, un-canceled-out charge) the negative time derivative of the vector potential. Now, this does affect the sine function, but not so much that the frequency is changed at all (the derivative of cos[wt] with respect to time for instance is -w*sin[wt], which has the same frequency). So, the frequency of the radiation is exactly the same as the frequency of the current.

To find the intensity, you'd actually have to work this all out in detail, but it would be proportional to the current (in amperes) squared (there would be a lot of other constants too).

Answer 2

It's 60 Hz indeed, just like in the house. It's actually quite simple to inductively couple a loop of wire to a transmition line. You can work it all out from Ampere's law and Faraday's law directly. Don't worry about radiation (that's the D term in the latter). It's strickly a near field effect because 60 Hz is so slow. Of course, you need to know how to use these equations to get that formula you want. Also, the loop would have to be rather large and conspicuous, and it's against the law. You'd be stealing electricity.

Answer 3

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Answer 4

If the power is 60 hz AC, then you'll have 60 hz radiation coming off. This is extremely low frequency. Very little energy per photon.