Find Distance of Separation for Magnetic Repulsion of Two Wires

Click For Summary

Homework Help Overview

The problem involves determining the distance of separation between two wires, where one wire is suspended above another and experiences magnetic repulsion due to the currents flowing through them. The context includes concepts from electromagnetism and forces acting on the wires.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply formulas related to magnetic forces and questions whether the scenario should be treated as a parallel or perpendicular wire problem. Some participants discuss the role of gravity and the need to balance forces acting on the suspended wire.

Discussion Status

Participants are exploring different interpretations of the problem, with some clarifying the relevance of gravity and the magnetic forces involved. There is an ongoing dialogue about the correct approach to take, but no explicit consensus has been reached.

Contextual Notes

There is a mention of the weight per unit length of the top wire and its implications for the calculations. Some participants express confusion about the provided information and its redundancy in the context of the problem.

Aerospace
Messages
39
Reaction score
0
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 separation 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.
 
Physics news on Phys.org
Aerospace said:
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.
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.
 
I still don't get why it's a parallel wire question. And I am not sure what you mean by using gravity.
could you please elaborate?
 
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.

[tex]\Sigma F = F_m - mg = \frac{\mu I_1I_2}{2\pi d}l - mg = 0[/tex]

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:
Chen said:
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.
The length of the top wire is given. :smile:

Aerospace, listen to Chen!
 
Doc Al said:
The length of the top wire is given. :smile:
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.
 
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 :)
 

Similar threads

Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
3K
Replies
11
Views
5K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 23 ·
Replies
23
Views
6K
Replies
3
Views
3K