arroy_0205 said:
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?
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:
[tex]\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}[/tex]
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 [itex]\vec B = \nabla \times \vec A[/itex]
Since it is a time varying field, there is always an electric field accompany with the magnetic field. We find the electric field by:
[tex]\vec E =\frac 1 {j\omega \epsilon_0}\nabla \times \vec H[/tex]
Since this is power line, the wave length is very long, so [itex]\beta \approx 0[/itex]. 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 [itex]\nabla \times\;[/itex] 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.