Question

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

Answer 1

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

Answer 2

Thank you Elena your a life saver :)

Answer 3

my brain just died lol