Ampere's law to calculate the B field from a balanced three-phase system

[jim hardy]

Everything is clear until page 4, up til they write "We now need to resolve each current into an in-phase and an out-of-phase component. We do this using a phasor diagram:" Why is this step necessary?

[ I haven't forgot you, just this one was going to take some study and i had to use daylight to finish putting side back on the barn. ]

I think Charles did a better job with the algebra and Latex than i could.

If i understand your question the answer is simple -

It's difficult to add in polar notation . I don't know how to do it.

One of the three flux's X components is of the form Asin(ωt) , one is Bsin(ωt-120°) and one is Csin(ωt+120°).
A, B and C differ because of the different radii from observer to each phase of the transmission line..
likewise their Y components.

To get the magnitude we can do:
⃗Bresultant(t)=⃗Ba(t)+⃗Bb(t)+⃗Bc(t)

once for X components and once for Y components ? Sure,
but because they're 120 degrees apart
I'd have to convert them EDIT to polar from polar to rectangular notation to add them.

(please pardon misspelling of rectangular , )

same exercise for Y components.

Maybe you know some math shortcut that i don't.

old jim

 

[rude man]

Thanks for the replies. I looked at a table which showed typical values of B at a distance x from a 420 kV transmission line, and they were not zero. So if I would want to calculate the value of B below the transmission line at ground level, I would have to use Biot-Savart equation on each conductor and sum them up?

Yes.
Ampere's law gives the total integral of the B field (actually ∫B⋅ds) surrounding one or more current-carrying wires. But if the wires are separated you lose symmetry and the value of B will vary as you go around any amperian loop.