Current induced in concentric superconducting loops

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SUMMARY

The discussion centers on the induced current in an inner superconducting loop when a constant current I is ramped up in an outer concentric superconducting loop. The mutual inductance is evaluated using the formula L = φ/I, where φ is the magnetic flux. The participants conclude that the induced current in the inner loop will be opposite in direction to the current in the outer loop due to the principle of conservation of magnetic flux, leading to the relationship I_induced = -I. The discussion also explores the effects of an adjacent room temperature conductor on the superconducting loops, suggesting that the induced current will adjust to maintain zero magnetic flux within the inner loop.

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  • Understanding of mutual inductance and its calculation
  • Familiarity with the principles of superconductivity
  • Knowledge of Faraday's law of electromagnetic induction
  • Ability to perform integrals involving magnetic fields from current loops
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  • Study the derivation of mutual inductance for concentric superconducting loops
  • Learn about the implications of superconductivity on magnetic flux and induced currents
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Holmez2_718
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Homework Statement


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?

Homework 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}

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.
 
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Holmez2_718 said:

2. Homework 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}
What is your angle theta in the expression?
 
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.
 
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.
 
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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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.
 
I'd be very curious to know if anyone might have a good solution for this one. If it is an introductory problem, I don't expect the answer to be so mathematically complex. The B field from a loop of current is much more complicated than that from a solenoidal/cylindrical shape in which the B field is completely uniform and points along the z-axis. I don't think that the current I=0 in the inner loop is necessarily the correct answer- it remains a puzzle what the correct answer might be. Perhaps this problem is really much too difficult. A better question might be to take a long (hollow) superconducting cylindrical shape (R1) inside a long room temperature cylinder/solenoid (of radius R2) (current I, n turns per unit length, K=n*I) , make the current per unit length in the solenoid change from K=0 to K=Ko (in the clockwise direction), and ask how does the current per unit length in the superconducting (R1) cylinder change? That question has a simple answer- the current per unit length of the R1 superconducting cylinder would be Ko in the counterclockwise direction. Another simpler problem would be to have an ordinary loop and a superconducting loop, both at the same radius. That would make a very good introductory homework question.
 
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