# What current will allow the wire to float ?

1. Oct 22, 2007

### Xaspire88

What current will allow the wire to "float"?

The linked image is a cross section through three long wires with linear mass density 50 g/m. They each carry equal currents in the directions shown. The lower two wires are 4.0 cm apart and are attached to a table. What current I will allow the upper wire to "float" so as to form an equilateral triangle with the lower wires?

The force that the upper wire feels is a combined force from the two other wires and the normal force of gravity. I am just not sure how to calculate the normal force of gravity with the linear mass density. Also the current that is needed will have to produce a magnetic field that combined equals the force downward on the wire from gravity, correct?

The equation for the force between two wires carrying identical currents with an unknown length is
F/L= ($$\mu$$I)/($$\mu$$r)

For gravity

F= m*9.8m/s

i have a linear mass density of 50g/m.. I can take and divide both sides by L to get

F/L = (m*9.8m/s)/L but I'm afraid that doesn't help me.. Hmmm. Ill keep trying. Please any hints are helpful

Last edited: Oct 22, 2007
2. Oct 22, 2007

### Xaspire88

3. Oct 22, 2007

### mgb_phys

If you have a linear density of M kg/m then you have a force of 9.8M N/m.
So F=9.8*0.05 N/m= 0.49N/m

4. Oct 23, 2007

### Xaspire88

so then i can set that force equal to the force between the wires and solve for I.
I get a value of 49kA when i do this.. this seems like a lot but it would also seem that an enormous amount of current would be needed to overcome the force of gravity. let me know what you think.