How Is Maximum Resistance Calculated in a Superconducting Ring Experiment?

[andrew410]

The resistance of a superconductor. In an experiment carried out by S. C. Collins between 1955 and 1958, a current was maintained in a superconducting lead ring for 2.50 yr with no observed loss. If the inductance of the ring was 4.44*10^(-8) H, and the sensitivity of the experiment was 1 part in 10^9, what was the maximum resistance of the ring? (Suggestion: Treat this as a decaying current in an RL circuit, and recall that e^(-x) = 1-x for small x.)

I know that I = ((emf)/R)*(1-e^(-Rt/L)). How do I get the maximum resistance from this? Also, what does the sensitivity of the experiment mean?
Any help would be great! thanks in advance!

 

[vincentchan]

use I(t) = I_0 e^{-Rt/L} instead, as the problem suggested, for small x, e^(-x) = 1-x
I(t) = I_0 (1-Rt/L))
I_0 - I(t) = Rt/L
the sensitivity is 1 part in 10^9 implies
\frac{I_0-I(t)}{I_0} < 10^{-9}
the rest is simple algebra

 

[andrew410]

so,
R = \frac{L} {t}(1-\frac{I} {I_0})
and \frac{I} {I_0} < 10^{-9}.
Do I substitute \frac{I} {I_0} with 1^(-9)?

 

[vincentchan]

so,
R = \frac{L} {t}(1-\frac{I} {I_0})
and \frac{I} {I_0} < 10^{-9}.
Do I substitute \frac{I} {I_0} with 1^(-9)?

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!
you substitute the whole \frac{I_0-I} {I_0} with 10^-9

 

[vincentchan]

wait a minute, where did R = \frac{L} {t}(1-\frac{I} {I_0}) came from?
The unit is not even correct... how did you get this?

 

[andrew410]

I(t) = I_0 e^{-Rt/L}
I(t) = I_0 (1-Rt/L)
\frac{I(t)}{I_0} = 1-Rt/L
\frac{Rt} {L} = 1-\frac{I} {I_0}
R = \frac{L} {t}(1-\frac{I} {I_0})

am I wrong?
and then
\frac{I_0-I} {I_0}
simplifies to -\frac{I} {I_0} so -\frac{I} {I_0}<10^{-9}

 

[vincentchan]

oh yeah, you are right, my mistake