PLEASE COULD SOMEONE HELP ME OUT
posted by John on .
Two long straight aluminum wires, each of diameter 0.30mm, carry the same current but in opposite directions. They are suspended by 0.50m long strings...If the suspension strings make an angle of 3.0 degree with the vertical, what is the current in the wires?
Thank you...please help i don't get this

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Separation between two wires is
b=2Lsinα=2•0.5•sin3⁰=5.23•10⁻² m
The mass of 1 meter of the wire is
m₀=ρπd²/4,
where ρ is the density of Al
The force between two wires (per 1 meter) is
F=μ₀I²/2πd.
F=Tsinα
mg=Tcosα
F/m₀g=tanα
μ₀•I²/2π•b•m₀•g= tanα
I=sqrt{2π•b•m₀•g •tanα/ μ₀}=
=sqrt{2π•b•ρ•π•d²•g/4•4π•10⁻⁷)=.. 
Hi Elena thank you, but this is what I did and i got it wrong
sqrt(2π*5.23x10^2*2700*π*2.25x10^6*9.8/4*4π*10^7)= 1.38x10^4 but it is wrong
what did i do wrong aluminum density is 2700 kg/m^3 and 2.25x10^6 is .30 mm divided by 100 to get meters an then divided by 2 to get r and square that so gives me 2.25x10^6