1. Feb 27, 2005

### 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! thx in advance!

2. Feb 28, 2005

### 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

3. Feb 28, 2005

### 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)?

4. Feb 28, 2005

### vincentchan

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

5. Feb 28, 2005

### 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?

6. Feb 28, 2005

### 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 $$-\frac{I} {I_0}<10^{-9}$$

Last edited: Feb 28, 2005
7. Feb 28, 2005

### vincentchan

oh yeah, you are right, my mistake