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diredragon
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Homework Statement
Three very long parallel conductors situated in the air make a direct-symmetrical 3-Phase system. The conductors pass through the A B C points of the triangle of side ##a##. The currents in the conductor form a direct-symmetrical 3-Phase system. Effective values of currents are ##I## and the frequency is ##w##. The phase of current 1 is ##Ψ_1 = 0##.
a) Calculate the vector of magnetic induction ##B## in the point ##O##.
b) What does that vector inscribe over time?
Homework Equations
3. The Attempt at a Solution [/B]
The currents in a 3-Phase direct system have a relation
##I_1 = I_m\cos {wt} = I\sqrt2\cos {wt}##
##I_2 = I_m\cos (wt - 2\pi/3) = I\sqrt2\cos (wt - 2\pi/3)##
##I_3 = I_m\cos (wt + 2\pi/3) = I\sqrt2\cos (wt + 2\pi/3)##
Magnetic Induction of a very long straight wire is ##B=\frac{μ_0I}{2\pi r}##
The distance of every wire to the point ##O## is ##r = \frac{a}{2\sqrt2}##
From the wires we only get ##B_z## and ##B_x## since ##B_y = 0##. The angles to the axis is ##\pi/4## (This I assumed).
##B_1 = \frac{\sqrt2μ_0I\sqrt2\cos {wt}}{\pi a}##
##B_2 = \frac{\sqrt2μ_0I\sqrt2\cos (wt - 2\pi/3)}{\pi a}##
##B_3 = \frac{\sqrt2μ_0I\sqrt2\cos (wt + 2\pi/3)}{\pi a}##
Total ##B_x## should then be:
##B_x = B_2*\sqrt2/2 + B_3*\sqrt2/2 - B_1*\sqrt2/2## and ##B_z##:
##B_z = B_1*\sqrt2/2 + B_3*\sqrt2/2 - B_2*\sqrt2/2##
Inserting: ##B_x = \frac{\sqrt2μ_0I\sqrt2\cos (wt - 2\pi/3)}{\pi a}*\sqrt2/2 + \frac{\sqrt2μ_0I\sqrt2\cos (wt + 2\pi/3)}{\pi a}*\sqrt2/2 - \frac{\sqrt2μ_0I\sqrt2\cos {wt}}{\pi a}*\sqrt2/2##, ##B_y = \frac{\sqrt2μ_0I\sqrt2\cos {wt}}{\pi a}*\sqrt2/2 + \frac{\sqrt2μ_0I\sqrt2\cos (wt + 2\pi/3)}{\pi a}*\sqrt2/2 - \frac{\sqrt2μ_0I\sqrt2\cos (wt - 2\pi/3)}{\pi a}*\sqrt2/2##
This seems overly complicated and doesn't seem to match the solution which is simple
Did i assume wrongly that the angle is ##\frac{\pi}{4}##. The angle of ##\frac{\pi}{6}## has ##\sqrt3## and that could result in this ##\sqrt6## shown in the answer. Or is it something else?