Re: wire tension

[Ludwik Kowalski <kowalskiL@MAIL.MONTCLAIR.EDU>]

Carl described alternative methods of solving this problem:

> A circular loop of flexible wire is placed in a uniform
> magnetic field perpendicular to its plane. The radius of
> the loop is R and its current  is I. What is the tension?

This was very useful to me. Thanks. I like this little
problem because at first it is nearly a paradox. Not a single
radial force has a tangential component but the tangential
force, I*B*R, is not zero in the loop.

"Carl E. Mungan" wrote:

> Another standard approach is to consider a short segment of length
> ds. Draw it horizontally (curving downward at both ends) with
> magnetic force dF straight upward. The two ends each make angle dQ
> with the horizontal. Then from a vertical force balance we have 2TdQ
> = dF using the small-angle approximation. But from the definition of
> angle in radians we see that dQ = (ds/2)/R from similar triangles.
> Finally dF = IdsB. This gives the solution.
>
> Alternatively, you need not integrate for your approach. There is a
> general rule that says the magnetic force on ANY thin wire in a
> uniform field is just IBxL where L is the straight-shot displacement
> from one end of the wire to the other. In your case of a semicircle,
> the magnitude of L=2R.
>
> Proof of general rule: F = integral{I*ds_vector x B_vector}. Pull I
> and B_vector out of the integral since they're constant. Then the
> integral of the little displacements ds_vector just becomes the net
> displacement L.