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Energy stored in two loops of current

  1. Feb 7, 2015 #1
    1. The problem statement, all variables and given/known data

    Derive equation 8.43 (given below). The equation shows the internal energy stored in a system of two current loops, one of radius b with current Ib and one of radius a with current Ia. The currents are spinning in the same direction, the loops are parallel, and they share the same axis. The distance between them is h.

    Hint: Build the two currents up from zero to their final values.

    2. Relevant equations

    ##U = \frac{1}{2} L_{a}I_{a}^2 + \frac{1}{2} L_{b}I_{b}^2 + MI_{a}I_{b} ##

    M is also given:

    ## M = \frac{\mu_{0} \pi b^2 a^2}{2(h^2+b^2)^{3/2}} ##

    An earlier part of the question required that I calculate B on the larger wire due to the smaller (which I assumed to be a dipole. I know I'm right here:

    ##B = \frac{\mu_{0} I_{a} a^2}{8(h^2+b^2)^{5/2}} (2h^2-b^2)\hat{r} - 3hb\hat{z}##

    3. The attempt at a solution

    The first two terms seem simple enough to me - the energy stored in a loop of current of radius a is ##U = \frac{1}{2} L_{a}I_{a}^2 ##. But where the third term comes from, I haven't the slightest of clues. I did find the B-field due to the smaller loop (which I assumed to be the dipole) at the larger loop. This was for an earlier part of the problem which I know I got right.
     
  2. jcsd
  3. Feb 7, 2015 #2
    Hi. The third term is called the mutual inductance: you can look it up in your text book or online, it's usually introduced with two concentric solenoids as an example.
     
  4. Feb 7, 2015 #3
    I know what mutual inductance is, I'm not sure how to derive the third term of the energy stored in the system and how it relates to mutual inductance.
     
  5. Feb 7, 2015 #4
    M is the mutual inductance and the Ls are the self-inductances... Physically, you can visualize the contribution of M by field lines from one loop going through the other loop.
     
  6. Feb 7, 2015 #5
    I know what M and L are. I know how to calculate them, how to visualize them, and what they mean "physically". What I don't know is why M multiplied by the two currents is some sort of energy stored in the system.
     
  7. Feb 7, 2015 #6
    Ok, then remember that the energy stored is also an integral of B2 over all space: here, B has two components so its square gives you a cross-term which you can relate to the mutual inductance by going back to its definition...
     
  8. Feb 7, 2015 #7
    Oh... duh, I didn't even think of that. However I'll need to integrate over a volume, and which one do I integrate over?

    Edit: actually, scratch that, B is constant
     
  9. Feb 7, 2015 #8
    In principle, all space. But you won't necessarily have to carry out the integration if you can formally relate it to the mutual induction... (just a guess, i didn't go through the problem but it's pretty clear where M is coming from)
    Edit: no, B is not constant on all space!
     
  10. Feb 7, 2015 #9
    Ah.... so I'll need to re-derive my B-field for a generic point in space, and then integrate it?
     
  11. Feb 7, 2015 #10
    Well, there's a lot of symmetry in the problem so you can probably look at the fields along the central axis and express them as integrals; then go back to the definitions of L and M and what they would be here, so you can relate everything... Again i didn't go through the whole thing and it's not trivial so it might take you a little time and efforts to get through it but that's part of solving a good problem!
    At least i think you have an outline for the method and a rationale for the physics now...
     
  12. Feb 8, 2015 #11
    So I really feel quite stupid, but do I take into account both currents? I'm just not sure what you mean by "look at the fields along the central axis and express them as integrals".
     
  13. Feb 8, 2015 #12
    Each current produces a magnetic field and, fortunately, they're additive so you can determine them independently; then yes, looking at the total field at any point in space there's a contribution from both currents so they have to be taken into account.
    Now i realize that although all the above would lead you to the correct answer it seems quite tedious and formidable due to the integral expressions involved, so there's a more direct and general route to proving your equation:
    The work done on a charge around a given loop is –ε (this is the emf, i assume you know what it is);
    the current on the loop (say, loop 1) is I1, so the total work done per unit time there is: dW/dt = –εI1.
    Now the emf has two components because if you vary the current in loop 2 you'll vary the flux through loop 1 and vice versa, so you have the usual emf of a circuit with self-inductance L plus a contribution from Faraday's law.
    If you express dW/dt in terms of L1, L2, I1, I2 and M, how do you get the total energy stored in the system?
     
  14. Feb 8, 2015 #13
    Thank you for your response. I feel silly, but I'm still not getting it. My issue is understanding why there is an induced EMF at all when the currents (and therefore B, and therefore flux) aren't changing. In a similar problem done in class, we assumed that the currents aren't changing (in reality, they would, but for the purposes of this calculation I don't think that is the case).
     
  15. Feb 8, 2015 #14
    The answer lies in the hint you are given: "Build the two currents up from zero to their final values."
    It relates to my last line: "If you express dW/dt in terms of L1, L2, I1, I2 and M, how do you get the total energy stored in the system?"
    The total energy stored in the system corresponds to the amount of work needed to put it together; doing the above, you'll see that you can integrate the right side (of dW/dt) easily from t = 0 to t = t.
    Currents are not changing in this problem and if they were the total energy would be changing too, but that doesn't prevent you from looking at how much work would be done if they were and infer your total energy... The expression you used for the energy due to a single loop (LI2/2) comes from this type of calculation.
     
  16. Feb 8, 2015 #15
    AHA! I looked over your posts so many times I practically have them memorized, but I think I have something.

    I calculated dW/dt for the loop of radius a. It equals -εIa. Then I said ε = -La dIa/dt - dΦ/dt. Like you said, a term for the self-industance, and a term for Faraday's law. I then integrated over time and used the fact that Φa = M Ib, and I got

    W = 1/2 La Ia2 + M Ia Ib. However, now I don't have a term involving the self-inductance of b. Do I just say that the W I calculated is the "work" required to assemble the loop of radius a, and then add a term for the work required to assemble the loop of radius b? But then why wouldn't I include the Faraday's law term into this? (since I know the answer is stuff + stuff + M Ia Ib and not 2 M Ia Ib? Or did I even do this right at all?
     
  17. Feb 8, 2015 #16
    On second thought, if I first assemble loop a, and then I assemble loop b (and take into account the flux caused by loop a), I get the right answer, and that makes sense.
     
  18. Feb 8, 2015 #17
    Exactly, and it completes the mutual inductance term because you integrate MI1dI2/dt + MI2dI1/dt, which correctly gives MI1I2.
     
  19. Feb 8, 2015 #18
    Yay! Thanks so much for your help!
     
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