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Calculating turns in an inductor

  1. Nov 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Hey everyone, I got a question wrong on my homework that I'm trying to figure out. I'm given this:

    "a 500uH solenoid(inductor) with an air core is 10cm long has an inside diameter of 1cm. How many turns of wire does it have?

    2. Relevant equations

    L = N2*u*A / l

    N is the number of turns
    u is the permeability of core material
    A is the Area of the coil in square meters
    l is the average length of the coil in meters.

    3. The attempt at a solution

    So first of all I rewrote the equation to be
    N = sqrt([L*l]/[u*A]

    L = 0.5H, l = 0.1m, u is u0 * ur, or 1 * (1.26x10-6, and my area is going to ∏*(0.005)2

    So,
    N = √(0.5*0.1)/(1 * ([1.26x10-6]*[∏*(0.005)2]

    When I calculated this however, I'm getting an answer of 22477.9 turns, which I know is wrong. The answer actually is 712.

    I'm thinking I either have my permeability of wrong (which I double checked, an air core is permeability 1 and the permeability of free space is 1.26x10-6, so the permeability is 1.26x10-6.

    Do I possibly have my area wrong? The diameter is 1cm, so we know that the area of a circle is pi * r squared, so pi * (0.01/2), which I got 0.000079m2.

    Am I missing something obvious here?

    When I solve for area, I get (0.5*0.1)/(7122*(1.26x10-6)) which gives me 0.078278, which I'm guessing that's the area I SHOULD have gotten, but how is that?
     
  2. jcsd
  3. Nov 19, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    The inductance is 500 μH, which is not 0.5H.
     
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