# How to calculate magnetic/electric field near an overhead power line?

1. Nov 11, 2011

### arroy_0205

The usual overhead power distribution wires are said to have potential of 230V in a certain locality. (These are voltage levels in wires seen along streets electricity pillars and these enter house of consumers.) How does one calculate magnetic field and electric field produced by the wire at a certain distance, say 5m from the wire? I am confused because when we say 230V, we do not mention, with respect to what this is stated. It becomes easier if current instead of voltage is stated for the wires which however is never the case. Can anybody please help?

2. Nov 11, 2011

### yungman

I think this is a hard problem since it is not static. I can only make a guess from material of antennas. And I don't think you can calculate from the voltage.

We start by finding the vector magnetic potential:

$$\vec A=\frac {\mu_0\;I}{4\pi}\oint_c \frac {e^{-j\beta R}}{R}d\vec l' \;\hbox { where } R \;\hbox { is distance from dl' to the observation point and }\; \beta =\frac {2\pi}{\lambda}$$

The line integral integrate along the transmission line. First pass assume it to be a straight line and make it easy. But it is really like a hanging chain problem and the equation of the line is more complicate.

Then find magnetic field B using $\vec B = \nabla \times \vec A$

Since it is a time varying field, there is always an electric field accompany with the magnetic field. We find the electric field by:

$$\vec E =\frac 1 {j\omega \epsilon_0}\nabla \times \vec H$$

Since this is power line, the wave length is very long, so $\beta \approx 0$. This will simplify the calculation. But still when you work the whole problem out, it is going to be in three space. Further approximation on how far the observation point from the line and simplify the $\nabla \times\;$ result. This is basically the near and far field approximation. In your case of 5m distance only, a near field approx should be good enough.

I don't claim this is the answer, I just want to joint in and put in my piece.

Last edited: Nov 11, 2011
3. Nov 11, 2011

### jim hardy

hmmmm

here in US the wires entering a house are run as a pair, two at nominal 115V , 180 degrees out of phase. They are insulated.
Both of them are twisted around a bare metal supporting cable,
forming a twisted pair, if unshielded.

Yungman's formulae should be applied to all three conductors and the results added.

i think, since at any instant one wire has opposite polarity from the other,
that beyond a few wire diameters the two fields will so nearly cancel out that the net electric field is for all practical purposes,,, zero.
The magnetic field would be also very nearly zero because any unbalance in current between the two insulated wires returns through the bare one.
So, since the currents add to zero, [h dot dl ] around the wire bundle is zero. No magnetic field ??

.........................................

"I am confused because when we say 230V, we do not mention, with respect to what this is stated."

walk out in your back yard and study the transformer.
here in US, the low voltage winding is the one with shorter insulators and your two insulated wires going to the house come from the ends of that winding. It is 230 volts between those two insulators.
The center-tap of that winding is connected to the bare support wire and to earth.
Each of the insulated wires is 115V respect to centertap (and therefore to earth), but they're of exactly opposite phase
so the difference between them ( + 115 to - 115) is 230.
i think of the centertap as my reference. Many folks use earth instead, note the centertap and earth are are connected so we'll get same answer.

go back about a month in this forum and look for threads on "ground" - some of the folks who are more capable than me put up excellent drawings. Good reading!

old jim

4. Nov 11, 2011

### yungman

I only think of it is a single wire!!! If it is two wires twisted together, theoretically it has no emission as the return is 180 deg out of phase.