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Self-inductance of a rectangular loop

  1. Dec 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Given a rectangular loop of length C by length D (C >> D) through which a current I flows, calculate the self-inductance L of the loop (consider only the long sides of the rectangle).

    2. Relevant equations

    Self-inductance: [itex]L = \frac{N \cdot \Phi_B}{I}[/itex]

    magnetic flux: [itex]\Phi_B = \int{B \cdot dA}[/itex]

    magnetic field for a long conductor: [itex]B = \frac{\mu_0 \cdot I}{2\pi r}[/itex]

    3. The attempt at a solution

    sketch

    A single loop: N = 1

    C >> D, so we approximate the magnetic field in the loop by one made by 2 long conductors, meaning B is only dependent on the distance from the long sides.

    Each of the 2 long sides contributes to the magnetic field constructively:

    [itex]\Phi_B = 2 \int{\frac{\mu_0*I}{2\pi y} dx dy} = \frac{2*C*\mu_0*I}{2\pi} \int_0^D{\frac{1}{y}dy}[/itex]

    This is where I'm stuck; the integral isn't convergent.
     
  2. jcsd
  3. Dec 21, 2011 #2

    Redbelly98

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    You're correct; if the wire is approximated as being infinitesimally thin, then the integral to calculate the flux diverges. My best guess is that either your professor wanted you to figure that out for yourself, or you have to do the calculation for a nonzero wire diameter -- which I don't think is an easy calculation.

    p.s. Welcome to Physics Forums.
     
  4. Dec 22, 2011 #3
    Thank you so much! Yes, in the original question there are 2 long copper cylinders, but since it was a 2-part question, I thought that detail was irrelevant for the fist part (second part is on how much magnetic energy is left 20µs after turning off the power source). My course is in Dutch; that's why I didn't just copy the question.

    Anyway, the only thing that would change are the boundaries of the integral: y = r..(D-r). This gives:

    [itex]L = \frac {C \cdot \mu_0 \cdot \ln{\frac{D-r}{r}}}{\pi}[/itex]

    Or should I account for the magnetic field within the cylinders as well? That integral would be divergent.
     
  5. Dec 22, 2011 #4

    Redbelly98

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    Well, if the current is spread out over the cross-section of the wire, then B does not go to infinity inside the wire -- it actually becomes zero at the wire's central axis -- and the integral would not diverge. I have thought about this before, but I confess I have never come to a satisfactory answer for myself: if the current is distributed over the wire's cross-section, then where exactly is the "boundary" for the area over which flux is to be calculated?

    I guess go ahead with your current approach, and use the wire's outer surface to do the integral. I'd be interested in hearing what your professor has to say about it! Meanwhile, I'm going to move this to the Advanced Physics subforum, in hopes that somebody else may chime in.
     
  6. Dec 22, 2011 #5
    Right, I meant converge...

    magnetic field inside a long cylindrical conductor with radius R (r: distance to axis): [itex]B = \frac{\mu_0 I \cdot r}{2 \pi R^2}[/itex]
     
  7. Dec 23, 2011 #6

    Redbelly98

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    :smile:
    Yes. The way I see it, there are two ways one might include the current inside the wire in calculating the flux:

    1. Do the flux integral from 0 to r (the radius of the wire), add the result to what you have, and decide if it's a significant or a negligible contribution to the total flux;

    or

    2. Do the flux integral from -r to +r, so that the entire wire is included inside the area over which flux is to be calculated. Again, decide if it's a significant or a negligible contribution :wink:.

    Hope that helps. Sorry I can't be more definitive in my suggestions.
     
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