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
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
Thank you Elena your a life saver :)
my brain just died lol
Well, John, it looks like you got yourself tangled up in the calculations. But don't worry, I'm here to help you untangle the mess.
Let's start from the beginning. The equation you used to calculate the current is correct:
I = sqrt((2π•b•m₀•g •tanα) / μ₀)
Now let's plug in the values correctly:
b = 2Lsin(α) = 2 * 0.50 * sin(3°) = 5.23 * 10^-2 m
m₀ = ρπd²/4 = (2700 kg/m³) * π * (0.30 mm / 1000)² / 4 = 2.8365 * 10^-7 kg/m
g = 9.8 m/s²
μ₀ = 4π * 10^-7 T*m/A
Now let's put it all together:
I = sqrt((2π * 5.23 * 10^-2 * 2.8365 * 10^-7 * 9.8 * tan(3°)) / (4π * 10^-7))
= sqrt(7.36 * 10^-17 / 5.02654824 * 10^-7)
= sqrt(1.46484184 * 10^-24)
= 1.21 * 10^-12 A
So, the current in the wires is approximately 1.21 * 10^-12 Amperes.
To solve this problem, we can use the concept of the magnetic force between the two wires to determine the current in the wires. The equation for the magnetic force between two wires per meter is given by:
F = μ₀ * I² / (2πd)
Where F is the force, I is the current, μ₀ is the permeability of free space (4π × 10^-7 Tm/A), and d is the distance between the wires.
Given that the suspension strings make an angle of 3.0 degrees with the vertical, we can determine the separation between the wires, b, using the equation:
b = 2L * sin(α)
Where L is the length of the strings (0.50 m) and α is the angle (3.0 degrees). Plugging in the values, we have:
b = 2 * 0.50 * sin(3°) = 5.23 x 10^-2 m
Next, we need to calculate the mass per meter of the wire, m₀. The equation for the mass per meter of the wire is:
m₀ = ρ * π * (d/2)^2
Where ρ is the density of aluminum (2700 kg/m³) and d is the diameter of the wire (0.30 mm). Converting the diameter to meters and plugging in the values, we have:
d = 0.30 mm = 0.30 × 10^-3 m
m₀ = 2700 * π * (0.30 × 10^-3 / 2)^2 = 2.25 x 10^-6 kg/m
Now we can calculate the current, I, using the equation derived earlier:
I = sqrt((2π * b * m₀ * g * tanα) / μ₀)
Plugging in the values, we have:
I = sqrt((2π * 5.23 x 10^-2 * 2.25 x 10^-6 * 9.8 * tan(3°)) / (4 * 4π * 10^-7))
Evaluating this expression, we find that the current in the wires is approximately 1.38 x 10^-4 A.
It seems like you made a calculation error in your equation. Make sure to use the correct units and remember to square the diameter when calculating the mass per meter of the wire.