Magnetic Levitation magnitude of current

[GoldWing]

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
F_B = 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 F_B = 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 F_B=F_g
.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.

 

[mfb]

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.

It is a ridiculously large number, but here it tells you that levitation does not work with currents of 2 A in the bottom wires. I didn't check if every number is right but the order of magnitude is correct.

(the system would also be unstable, without suspension the wire would escape towards one side quickly.)

 

[GoldWing]

It is a ridiculously large number, but here it tells you that levitation does not work with currents of 2 A in the bottom wires. I didn't check if every number is right but the order of magnitude is correct.

(the system would also be unstable, without suspension the wire would escape towards one side quickly.)

I'm not sure what you mean. Wasn't I supposed to use 2.08A as my currents?

 

[mfb]

Yes, and your answer is correct.
"does not work" was meant in terms of a practical realization.

Hmm.. where did you consider the direction of the force? I don't see that part in your calculation.
The magnetic fields don't add up linearly, as they point in different directions.

 

[GoldWing]

Yes
, and your answer is correct.
"does not work" was meant in terms of a practical realization.

Hmm.. where did you consider the direction of the force? I don't see that part in your calculation.
The magnetic fields don't add up linearly, as they point in different directions.

I considered the direction of to be going up (according to the right hand rule, current goes out of page, filed goes counterclockwise so the force is up) , and please correct me if I'm wrong.

 

[mfb]

The direction is not "up". It is angled, corresponding to the lines "L" between the wires for each lower wire.

 

[GoldWing]

That makes sense and I wasn't sure when that would come into play, so would I take the sin of 60 since it's an equilateral triangle and multiply it by the force? And add the forces from both wires?

 

[mfb]

Right.

 

[GoldWing]

Thank you so much!