# Current induced in concentric superconducting loops

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1. Feb 29, 2016

### Holmez2_718

1. The problem statement, all variables and given/known data
Suppose you have two superconducting loops, concentric, in a plane. Also suppose that their radii, R2 (outer) and R1 (inner) have the same order of magnitude (so you can't assume B through the inner loop is constant everywhere over the inner loop's surface.)

If a constant current I is ramped up from 0 in the outer loop, what will the induced current in the inner loop be?

2. Relevant equations
$$L = \frac{\phi}{I}$$
$$B = \frac{\mu_0 I}{2 \pi} \int_{0}^{\pi} \frac{(R_1 - rcos(\theta))R_1 d\theta}{(r^2 + R_1^2 - 2rR_1cos(\theta))^{3/2}}$$ for an arbitrary point a distance r away from the center of a loop in its plane
$$\varepsilon = -L \frac{dI}{dt}$$

3. The attempt at a solution
I begin by attempting to evaluate the mutual inductance of the loops: $$L = \frac{\phi}{I} = \mu_0 \int_{0}^{R_2} r \int_{0}^{\pi} \frac{(R_1 - rcos(\theta))R_1 d\theta}{(r^2 + R_1^2 - 2rR_1cos(\theta))^{3/2}} dr$$ where the inner integral comes from the expression for B at an arbitrary point in the plane of a current loop, and the outer integral is just the integration over the inner radius to compute the flux. Aside from the profound ugliness of that expression, I don't really have a problem with it -- it's just that with R = 0, knowing that $$\varepsilon = -L \frac{dI}{dt}$$ doesn't seem super helpful.

2. Feb 29, 2016

### Erebus_Oneiros

What is your angle theta in the expression?

3. Feb 29, 2016

### Holmez2_718

Just an angle of integration as you go from one "end" of the loop to the other, when computing B at an arbitrary point. According to Purcell, that integral can't be evaluated in closed form, so I left it like that.

4. Feb 29, 2016

I want to try to try to guess the answer here without a lot of mathematics. I think there may be some law about the total magnetic flux through a superconducting loop will always be zero. (I could be wrong). The current in the loop will adjust to keep the loop zero because of the Faraday EMF... Just a guess is that if the outer loop has current I in the counterclockwise direction, the inner loop will have current I in the clockwise direction. I'd need to check this solution for conducting loops, but if these were long coencentric cylinders instead of rings, I'm pretty sure the answer would be correct.

5. Feb 29, 2016

### Erebus_Oneiros

My solution is as follows:
As charles too said, the flux through the superconducting ring must not change, i.e. it was initially zero so at any point of time it will be zero.

$$\phi_{external} = \mu_0 I \int_{0}^{R_2} r dr \int_{0}^{\pi} \frac{(R_1 - rcos(\theta))R_1 d\theta}{(r^2 + R_1^2 - 2rR_1cos(\theta))^{3/2}}$$

Now flux due to itself will be exact same magnitude but opposite direction as the above flux. Using this with the expression to find flux from current flowing in the loop we get:

$$\phi_{induced} = \mu_0 I_{induced} \int_{0}^{R_2} r dr \int_{0}^{\pi} \frac{(R_2 - rcos(\theta))R_2 d\theta}{(r^2 + R_2^2 - 2rR_2cos(\theta))^{3/2}}$$

$$\phi_{induced} + \phi_{external} =0 \implies \phi_{induced} =- \phi_{external}$$
$$\implies I_{induced}= -I \frac{\int_{0}^{R_2} r dr \int_{0}^{\pi} \frac{(R_1 - rcos(\theta))R_1 d\theta}{(r^2 + R_1^2 - 2rR_1cos(\theta))^{3/2}}}{\int_{0}^{R_2} r dr \int_{0}^{\pi} \frac{(R_2 - rcos(\theta))R_2 d\theta}{(r^2 + R_2^2 - 2rR_2cos(\theta))^{3/2}}}$$
where negative indicates opposite direction.

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6. Mar 1, 2016

I think I was on the right track on this one above, but I don't have an answer to it... Suppose you begin with zero current in both loops and a 3rd loop of radius R2 of a room temperature conductor is placed directly adjacent to the superconducting R2 loop. Now , if you begin to ramp up a current in the room temperature loop, the resulting magnetic fields will try to cause a change in the magnetic flux everywhere including inside of R2. The result will be a small Faraday EMF in superconducting R2 which will cause a current in the opposite direction, (e.g. counterclockwise) in superconducting R2 that is equal to the amplitude of the current in the room temperature loop. These two matching (equal and opposite currents will have nearly zero magnetic field in the interior of the R2 radius. I=0 in R1 does not appear to be the correct answer though because the current I in the loop R2 was also effectively created with very small (dB/dt)*A. In the the two R2 loops, the superconducting current will be just slightly less than the room temperature one (in amplitude) to allow for a magnetic flux to generate an EMF to bring increase the current in order to negate the B field generated by the room temperature loop. I have to admit, although I gave this one some careful thought, I do not know the answer.

7. Mar 1, 2016