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Inductance of a coil

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data

    How will a coil's self inductance change if you unwind it and then re-wind half of the length of wire into a coil with the same diameter but with half of the number of turns?

    The answer is that it will decrease by 1/4, but I don't understand why it isn't 1/2 instead.

    2. Relevant equations


    phi (magnetic flux) = LI (where L is inductance, I is current)
    n = N/l where "l" is length
    B = mew naught * n * I

    3. The attempt at a solution

    original: L = phi/I = (NBAcostheta) / I = (N* mew naught *n*I*costheta) / I
    Now the L with the proposed change: L' = (N/2)* mew naught * ((N/2) / (l/2))* costheta = 1/2 L
     
  2. jcsd
  3. Nov 20, 2008 #2
    I think it's because you have confused the length of the solenoid, l, with the length of the wire, 2*pi*R*l, where R is the radius of the coil. The length of the solenoid hasn't changed, only the number of turns per unit length (and the wire's length) have dropped.

    Edit:Woopsey, no solenoid mentioned, sorry I've been doing a lot of these and went tunnel vision. Your right the length is halved. Where does your N come from in the NBAcostheta/I?
     
    Last edited: Nov 20, 2008
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