Re: wire tension

["Carl E. Mungan" <mungan@USNA.EDU>]

> 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?
> The end of the book answer is: T=I*B*R.
>
> How can this be derived? I know that each dL element
> is subjected to the radial outward force B*I*dL. But
> tension is a tangential force. Intuitively I do expect tension
> to be proportional to the radial force per unit length (I*B)
> but I do not know how to demonstrate that T=I*B*R.
> Ludwik Kowalski

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.

HTH, Carl
--
Carl E. Mungan, Asst. Prof. of Physics  410-293-6680 (O) -3729 (F)
      U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5026
mungan@usna.edu    http://physics.usna.edu/physics/faculty/mungan/