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Induction in two coils that are connected :S

  1. Aug 29, 2014 #1
    I'm a third year physics undergrad and I still massively struggle with electromagnetism - especially induction. Here's the opening part of the question:

    A loop is placed in an external alternating magnetic field that is perpendicular to the plane of the loop, with B=B0sin(ωt). Consider first the case where the loop makes a circle of radius r and calculate the EMF induced in the loop in terms of the magnetic field B0.

    This seemed simple, and I just use Faradays law and the fact that the magnetic field is the only variable changing to give

    EMF = -πr2ωcos(ωt).

    The next part is trickier.

    The loop is now folded to make two rings of radii r1 and r2, as shown in the
    figure with corresponding resistances R1 and R2. Derive the formula for the
    current flowing in the wire in terms of the electromotive forces E1 and E2
    induced in the two parts of the loop, and their resistances R1 and R2.

    The figure just shows a smaller circular loop within the larger circular loop.

    Now I assumed because the current must be the same at all places in the loop (Kirchoff's laws) then E1/R1=E2/R2

    Of course as the current is changing and presumably inductance is involved, I'm not sure this is correct. It's worth 3 marks and it seems too easy.

    The final part asks to Derive the potential difference U between the two parts of the loop at their
    crossing point in terms of the same variables used above (i.e., E1,E2,R1,R2).

    I don't even know where to start for this! I assumed it would just be E1 - E2? Again! Way too easy. I'm missing something fundamental! It's worth 4 marks. What am I doing wrong?

    Any help would be greatly appreciated
     
  2. jcsd
  3. Aug 29, 2014 #2

    Orodruin

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    This is not what Kirchoff's current law says - it says that the current must be the same in both loops. This does not mean that the EMF generated in both loops are related by this relation. You can use the current law to relate the voltage drops across the resistances, but you will also need to use the voltage law to relate the potential drop across the loop resistances with the EMF generated in the two loops.

    The problem as you have defined it is not completely unambiguous as you did not attach the picture. Do the loops go in the same or in opposite directions? This will determine whether or not the EMFs of the loops work against each other or together in order to drive the current through the loop resistances.
     
  4. Aug 29, 2014 #3
    Thanks for the reply, I've attached a screenshot of the Q. Hope this helps. Yeah, the current goes in the same direction.
     

    Attached Files:

  5. Aug 29, 2014 #4

    Orodruin

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    So what is the total EMF induced in the loop? What is the loop resistance?
     
  6. Aug 29, 2014 #5
    Presumably you could represent as a series circuit with two separate resistances so their resistances add? R1 + R2?
     
  7. Aug 29, 2014 #6
    So is EMF = i(R1 + R2)? Or am I still missing the point?
     
  8. Aug 29, 2014 #7

    Orodruin

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    You are still missing it. How do you compute EMF induced by a magnetic flux through a loop?

    Edit: Well, what you wrote must also hold true ... But you need to find out what the induced EMF is ...
     
  9. Aug 29, 2014 #8
    using Faraday's law, but this is where I got stuck. Do I treat each loop separately? Do I work out the EMF in loop 1 and then loop 2? I'm guessing there is self interaction going on here (well that's what induction is), but I don't know how to set up the equation. If I take the larger loop I can calculate the EMF as before, but when I add another loop, the current in that loop will also contribute a magnetic field and thus subtract or add to the other loop's EMF?

    Sorry, I'm aware I'm missing something obvious
     
  10. Aug 29, 2014 #9

    Orodruin

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    Self inductance would typically be considered negligible. If you put two batteries in series, what would be the voltage across them? The approach here is essentially the same.
     
  11. Aug 29, 2014 #10
    V = V1 + V2? So E1 + E2 = Etotal ?
     
  12. Aug 29, 2014 #11

    Orodruin

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  13. Aug 29, 2014 #12
    Thanks. Does that mean I equate E(total) with the E I got for the single loop?
     
  14. Aug 29, 2014 #13

    Orodruin

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    No, the EMF induced will be different in the two loops. The big loop will have an induced EMF proportional to its area and the small one proportional to its area. The total will be the sum of the two.
     
  15. Aug 29, 2014 #14
    Hmmm ok, but then if I can't use EMF = i×R how can I write the equation in terms of R?
     
  16. Aug 29, 2014 #15

    Orodruin

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    Yes you can, it is just that the EMF will be different from that of the single loop. You already computed the EMF for a single loop to be EMF = -πr2ωcos(ωt), which is true apart from the missing amplitude B0. So what would be the total EMF of both of the loops?
     
  17. Aug 30, 2014 #16
    EMF1 = -π(r1)2B1ω1cos(ω1t)

    EMF2 = -π(r2)2B2ω2cos(ω2t)

    if ω1=ω2


    EMFTotal = -πωcos(ωt){B1(r1)2 + B2(r2)2)

    Is this the right approach?
     
  18. Aug 30, 2014 #17

    Orodruin

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    The magnetic field is the same in both loops so both frequencies and both Bs are the same.
     
  19. Aug 30, 2014 #18
    Ah of course! So the total emf is equivalent to that of a single loop with radius r1 + r2? Still struggling to bring in resistance.
     
  20. Aug 30, 2014 #19

    vanhees71

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    Well, if you have conceptual problems, it's good to go back to the basics (at least this always helps me, when I get stuck). The basics in this case are Faraday's Law (in SI units),
    [tex]\vec{\nabla} \times \vec{E}=-\partial_t \vec{B},[/tex]
    [CORRECTED!]
    and Ohm's Law,
    [tex]\vec{j}=\sigma \vec{E}.[/tex]
    Now you have a situation, where the integral form of the laws are simpler than the local ones. Then we also can simplify the wires to be very thin. Let them be in the [itex]xy[/itex] plane and the magnetic field along the [itex]z[/itex] axis. Now do the following:

    (a) integrate [itex]\vec{E}[/itex] along the two closed loops and express this integral with help of Ohm's Law. Split the result in the contribution from the two loops separately. This gives you the law for the resitance of two resistors in series of course. For this you must relate the current [itex]I[/itex] through the loop with the current-density vector [itex]\vec{j}[/itex] to express everything in terms of current.

    (b) Use Stokes's Law to relate the loop integral(s) to the magnetic flux through the loops. Then you are done.

    Concerning (b) you are on the right track already!
     
    Last edited: Aug 30, 2014
  21. Aug 30, 2014 #20
    I think Faraday's law has B in it? I'm sure that was just a typo. I'm not sure how to integrate E though as it's not a static E field?
     
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