Re: Fiber Optic current measurement.

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

Chuck Britton wrote:

> Can anybody supply an order of magnitude estimate of the amount
> of rotation to be expected around a few amps of current? (i.e. should
> I even TRY detecting it with some simple polaroid filters or are we
> talking about some rather sophisticated detection system?)
>
> I've got some surplus fiber that is just BEGGING to be used.

Here is what I see in Fundamentals of Photonix (1991) by
Saleh and Teich (page 225).

"The angle of rotation is proportional to the distance, and the
rotary power, r (angle per unit length) [which] is proportional
to the magnetic flux density [component] in the direction of
wave propagation." In other words  r=V*B, where V is known
as Verdet constant.

As explained by Leigh, the length depends on the number of
loops you create around a wire. Ask for the Verdet constant of
the material from which your fiber is made (at the lambda of
your laser). This book does not have a table but it gives an
example. --> For TbAIG V=-1.16 min/(cm*Oe) at lambda of
500 nm. I suppose you can just replace Oe by Gs here. If so
then B=1000 Gs corresponds to r=1160 minutes, or 3.2 degr
for each cm. A fiber of 10 cm, made from this exotic (?),
material would produce the rotation of 32 degrees.

OK, I found a table in another book "Optoelectronics: An
Introduction", by Wilson and Hawkes (p 110). Values of
V are in radians/(meter*Tesla) at lambda of 589.3 nm.

Quartz (SiO2)  V=4.0
ZnS                  V=82.
Crown glass    V=6.4
Flint glass        V=23.
NaCl                V=9.6

THETA=V*B*L

Do not bend your fiber optics too much. This means you
must tolerate small B. I suspect that the current wire along
the axis of your "optical solenoid" will produce the largest
possible angle of rotation for a given current. But this has
to be verified.

Keep us posted on your progress
Ludwik Kowalski