Tension at which a loop of wire will break

[Signifier]

Homework Statement

Note that in the following, 4[tex]\pi[/tex] means 4 * pi not 4 ^ pi...

A circular loop of wire of self-inductance L and radius r carries a current I. If T is the tension in the wire for which it will break, show that T must be greater than (I²/4[tex]pi[/tex])(dL/dr)

Homework Equations

Well, the magnetic energy is U = (1/2)I²L. F equals grad U. The circumference of a circle of radius r is 2[tex]\pi[/tex]r. We are assuming a constant current I and a deformable wire.

The Attempt at a Solution

The force that tends to increase the radius of the loop is F = (1/2)I²(dL/dr). My problem is that I am not sure how to relate this isotropic outward force to tension in the wire. If I have a wire loop, and I exert a "magical force" F that is a function of the radius of the loop r, F(r), what is the tension that develops in the loop?

T must be greater than (I²/4[tex]\pi[/tex])(dL/dr)
This is given. I can rewrite this with my force result to show that T must be greater than F / 2[tex]\pi[/tex].

I think I just need someone to talk to me about tensions in curves. I know I have a book that talks about this somewhere, but I can't find it. BTW this homework problem is from Wangness, Electromagnetic Fields 2nd edition, chapter 18 page 296.

Thank you!
(Please forgive me for the latex formatting issues)

 

[NateD]

The tension T in a loop of wire is related to the force F that is applied to the loop. The tension is equal to the force divided by the circumference of the loop. In this case, the force is given as (1/2)I^2(dL/dr), so the tension must be greater than (I^2/4\pi)(dL/dr)/2\pi.