- #1
GoldWing
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Homework Statement
Three long parallel wires are a distance L = 5.71 cm from one another. (Looking at them, they are at three corners of an equilateral triangle.) The top wire has a diameter of 1.8 mm and is made of copper; it is suspended in air due to the magnetic forces from the bottom two wires. The current in each of the bottom two wires is I0 = 2.08 A into the page. Calculate the magnitude of the required current I in the suspended wire. (The density of copper is 8.96 g/cm3.)
Homework Equations
Force per length
F/L= (μ0/2Π)*(I1*I2/d)
Magnetic Force
FB = I*L*B
Magnetic Field
B=μ0/2Π*I/r,
OR
B= 2Πx10-7(I/d)
The Attempt at a Solution
I know that the Fg will have to equal the Fb to keep the wire suspended. First, I found the mass of the wire by using the density,
D=M/V
DV=M
D*A*L=M
(8.96E3kg)/m3(Π*.001882m)*L=m
.0995kg/m*L=m
Then I can find Fg, so Fg=mg
Fg = (.0995kg/m*L)(9.80m/s)= .975kg/s*L
Then FB = I*L*B is equal to Fg, but first I need to find the magnetic field.
B = 2Πx10-7(2.08/.0571)*2
= 4.58*10-5kg/A*s
I multiplied by two because the magnetic fields add up, since both fields are going clockwise.
So now FB=Fg
.975kg*L = I*L(4.58*10-5kg/A*s)
Solving for I, I get 2.13*104A, even before I check the answer I know that it's wrong because it's a ridiculously large number. Please help me, I'm not sure where I went wrong.