Magnetic Fields

1. Mar 27, 2004

Aerospace

A 1.59 m long wire weighing 0.0668 N/m is suspended directly above an infinitely straight wire. The top wire carries a current of 32.4 A and the bottom wire carries a current of 24.5 A. Find the distance of seperation between the wire so that the top wire will be held in place by the magnetic repulsion.

Ok. So the formula that I wanted to use was
F/l = (u x I1 x I2) / (2 x pi x a)
where F/l is 0.0668 N/m
l = 1.59 m
u = 4 x pi e - 7 T.m/A
I1 = 32.4 A
I2 = 24.5 A
a = ?

But this is not a parallel wire question, it's a perpendicular wire question. If I use the formula B = (u x I) / (2 x pi x r) where r is the perpendicular distance to the wire, then will I have to ignore the current of the bottom wire? and I don't have the B.

I'm confused. Can anyone help? Thanks guys.

2. Mar 27, 2004

Staff: Mentor

Sure it's a parallel wire question. The current in the bottom wire creates the magnetic field that repels the current in the top wire. You have to figure out how far apart the wires must be so that the net force on the top wire segment is zero. Don't forget about gravity.

3. Mar 28, 2004

Aerospace

I still don't get why it's a parallel wire question. And I am not sure what you mean by using gravity.

4. Mar 28, 2004

Chen

The two wires are placed one above another, with the infinitely long wire fixed in its place. The top wire, which weighs 0.0668 N/m, is supposed to stay suspended above the bottom wire. This would only happen when the repulsion force between the two wires cancels the weight of the top wire, making the net force on it equal to zero.

$$\Sigma F = F_m - mg = \frac{\mu I_1I_2}{2\pi d}l - mg = 0$$

Of course you don't have the length of the top wire, so you also don't have its weight, so just divide both terms by the length.

Last edited: Mar 28, 2004
5. Mar 28, 2004

Staff: Mentor

The length of the top wire is given.

Aerospace, listen to Chen!

6. Mar 28, 2004

Chen

Oh right... I didn't think it would given because it's redundant, since you have the weight per length unit and not just the weight.

7. Mar 28, 2004

Aerospace

Thank You!

Thank you so much. It still took me a while to get it, but I finally got it!! Thanks to both of you :)