Posted by John on Monday, February 25, 2013 at 10:28pm.
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
PLEASE COULD SOMEONE HELP ME OUT  Ms. Sue, Monday, February 25, 2013 at 5:54pm
If you really want help, you'd type in your subject name so that an expert would see your post.
PLEASE COULD SOMEONE HELP ME OUT  Elena, Monday, February 25, 2013 at 6:10pm
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)=..
PLEASE COULD SOMEONE HELP ME OUT Physics  John, Monday, February 25, 2013 at 7:33pm
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

PLEASE COULD SOMEONE HELP ME OUT Physics  Elena, Tuesday, February 26, 2013 at 4:57am
Separation between two wires is
b=2Lsinα=2•0.5•sin3⁰=5.23•10⁻² m
The mass of 1meter of the wire is
m₀=ρπd²/4=2700•3.14•(0.3•10⁻³)²/4=1.9•10⁻⁴ kg
where ρ =2700 kg/m³ is the density of Al
The force between two wires (per 1 meter) is
F=μ₀I²/2πd
F=Tsinα
mg=Tcosα
F/mg=tanα
μ₀I²/2π•b•m₀•g= tanα
I=sqrt{2π•b•m₀•g •tanα/ μ₀}=
=sqrt{2π •5.23•10⁻²•1.9•10⁻⁴•9.8•5.2•10⁻²/4π•10⁻⁷}=
=5.03 A

PLEASE COULD SOMEONE HELP ME OUT Physics  John, Tuesday, February 26, 2013 at 8:41am
Thank you Elena your a life saver :)
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