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B How is current induced in superconductors?

  1. Nov 7, 2018 #1
    Assume resistor R/heated segment is 5 Ohms, while the path in the right hand side remains 0, thus, total resistance is 0 Ohms, my question is how could current be induced if potential difference is 0(due to zero total resistance) Or is it probably because of the batteries internal resistance? bS73Z.jpeg
     
  2. jcsd
  3. Nov 7, 2018 #2
  4. Nov 7, 2018 #3

    Dale

    Staff: Mentor

    You are correct that the superconducting side does have 0 resistance, however resistance is not the only means of creating a voltage. The superconducting side also has a non zero inductance which means: ##v=L \frac{d}{dt}i ##

    Typically the inductance L is quite large and v is small so the current increases slowly, often taking hours to ramp up to the desired current. Once the desired current is obtained then the voltage can be reduced to 0 and the current will not change. The resistive section can be cooled and the current source removed.
     
  5. Nov 7, 2018 #4
    Really? They don't use resistance for creating a voltage, then why is the heater even placed there?
     
  6. Nov 7, 2018 #5

    Dale

    Staff: Mentor

    There needs to be a return path for the current. At the beginning all of the current goes through the resistor. As the inductor current ramps up less and less of the current goes through the resistor.
     
  7. Nov 7, 2018 #6
    If I somehow manage creating an ideal current source, will there still be an induced current?
     
  8. Nov 7, 2018 #7

    Dale

    Staff: Mentor

    Yes. In this circuit the difference between ideal and nearly ideal is small.

    For a current source the internal resistance is a large resistance in parallel with the source. So the large internal resistance is in parallel with the small external resistance (heated superconductor material). The parallel equivalent of the two is almost the same as the small resistance alone.
     
    Last edited: Nov 7, 2018
  9. Nov 8, 2018 #8
    Why should there be a heater there, and what's the point of parallel circuit, why don't use a series circuit and use the source's internal resistance to produce potential difference? (I'm still new to this area) bS73Z.jpeg
     
  10. Nov 8, 2018 #9

    Dale

    Staff: Mentor

    Physically that just wouldn’t be possible. This is a superconducting magnet, so like all electromagnets it is wound in a big loop, usually with multiple turns. Further, when you disconnect the power you want the entire loop to be superconducting so the magnetic field can be permanent, so the loop is completely closed. Anywhere you connect an external circuit would therefore have two paths through the loop, which makes it a parallel circuit. There simply is no way around that.

    Now, given that we must have a parallel circuit then the goal is to get a large current into the high inductance side. To do that requires a voltage, which we cannot get across the low inductance side without making it resistive.
     
  11. Nov 8, 2018 #10
    But doesn't it matter how much resistance is in the segment heated by the heater, like I asked on my short circuit problem, it doesn't matter how much resistance you have in this path(5 Ohms), because the actual current would be the voltage divided by the internal resistance (10V/2Ohms), and what's the difference between the superconducting induction and the circuit I draw, doesn't it work the same way?(or am I confusing voltage source and current source?)
    20181107_222219.jpeg
     
    Last edited: Nov 8, 2018
  12. Nov 8, 2018 #11

    Dale

    Staff: Mentor

    The short circuit problem is a different circuit. It does not behave the same. In the short circuit there is no resistance and no inductance. In the superconductor there is no resistance but a very large inductance. That makes a huge difference in the circuit behavior. Again, resistance is not the only way to get a voltage.
     
  13. Nov 8, 2018 #12
    So it behaves totally different? And If you don't mind, can you explain briefly how the superconducting loop works? Just briefly. Thankyou
     
    Last edited: Nov 8, 2018
  14. Nov 8, 2018 #13

    Dale

    Staff: Mentor

    The loop has a very high inductance, L, which means ##v=L\frac{d}{dt}i##. So a voltage does not cause an infinite current, it causes a slowly increasing current.
     
    Last edited: Nov 8, 2018
  15. Nov 8, 2018 #14
    What do you mean by the term 'very high inductance'? (I'm kind of new) and I can you explain me in details how a permanent current is formed in the loop?
     
    Last edited: Nov 8, 2018
  16. Nov 9, 2018 at 5:44 AM #15

    Dale

    Staff: Mentor

    You should start with the hyperphysics page and maybe the Wikipedia page on inductance

    First we heat up the small segment and then we turn on the current source. At the beginning all of the current goes through the resistive side, but over time the current slowly increases through the loop until all of the current is going through the loop. Then the heater is turned off, the resistive segment becomes superconducting, the current becomes permanent, and the current source is disconnected.
     
  17. Nov 9, 2018 at 7:23 AM #16
    How could the 'current slowly goes to the loop, until no current passes the resistance'? Is there a method doing that?
     
  18. Nov 9, 2018 at 7:29 AM #17

    Dale

    Staff: Mentor

    The inductance. Please read the hyperphysics and wiki pages. You may also want to search for a page on “RL circuits”
     
  19. Nov 9, 2018 at 7:30 AM #18
    *and why does the current moves through the resistive side 'first'?
     
  20. Nov 9, 2018 at 7:31 AM #19

    Dale

    Staff: Mentor

    Again, search for an introductory page on “RL circuits”
     
  21. Nov 9, 2018 at 7:34 AM #20
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