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Homework Help: Length of a Solenoid

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data

    (a) What would have to be the self-inductance of a solenoid for it to store 9.3 J of energy when a 1.56-A current runs throught it?

    (b) If this solenoid's cross-sectional diameter is 3.85 cm, and if you could wrap its coils to a density of 10 coils/mm, how long would the solenoid be?

    2. Relevant equations


    3. The attempt at a solution

    I solved part a, I am only having trouble with part b.
    The problem is, I only have the diameter, but I don't have N, the number of coils.

    Here's how I tried to get N:

    coils/m = 10/1*10^-3 = 10000 coils/m

    N = (10000 coils/m) * (0.0385 m) = 385 coils

    l = pi*(0.0385 m) * 385 coils = 46.57m
  2. jcsd
  3. Nov 30, 2013 #2

    Simon Bridge

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    0.0385m is the diameter - are there 10000coils per meter across the diameter?! Usually the coils are across the length aren't they?

    You are close:
    If n=#turns per unit length, and N is the number of turns in the solenoid, and L is the length of the solenoid, then n=N/L

    You have a formula for the inductance in terms of the L and N don't you?
    So use some algebra to express it in terms of n and L.
  4. Nov 30, 2013 #3
    I got 7.643H for the inductance from part a

    So if n=N/L, then N=n*L:
    N = (10/(1*10^-3)) * 7.643 = 76430 turns

    l = pi*0.0385 * 76430 = 9244 m

    It this correct?
  5. Dec 1, 2013 #4

    Simon Bridge

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    ... um... no: not what I wrote.
    Hmmm... L was not a good choice for a variable-name for length was it?
    You pick one.
  6. Dec 1, 2013 #5
    Actually I don't. I couldn't find one in my book, but searching online I believe I found it:
    L = μ0*A*l*n2

    rearranging it gives:
    l = L/(μ0*A*n2)

    l = (7.643)/([4*π*10^-7]*[π*(0.0385/2)2]*100002) = 52.24m

    Is this right?
  7. Dec 1, 2013 #6

    Simon Bridge

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    ... hmmm, it's usually the first thing they do right after the Biot-Savart law. As in these utex lecture notes eq.909.
    "Self Inductance" doesn't mean much without it.

    Oh well...
    You got the formula in the end.

    Code (Text):

    > 7.643/((4*pi*10^-7)*(pi*(0.0385/2)^2) *(10000^2))
    ans =  52.245
    ... which is rather long so probably impractical.
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