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Solenoid Magnetic Field

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data
    You want to construct a solenoid 20cm long that has an interior magnetic field Strength B (Bc) of about 1.5MN/C. The coil has to be wound as a single layer of wire around a form whose diameter is 3.0 cm. You have two spools of wire handy. #18 wire has a diameter of about 1.02mm and can carry a current of 6.0A before overheating. #26 wire has a diameter of 0.41mm and can carry up to 1.0A. Which kind of wire should you use and why? What potential difference do you want to put across the coil's end?



    2. Relevant equations

    B=NI/(cLE0)...or simplified as about 377NI/L.

    3. The attempt at a solution

    377NI/L must equal about 1.5X10^6N/C.......so NI/L must equal 3979. I Have an I....and don't really know where to go from here...and not even sure if this is the right path to take. Not really understanding the question? So I unwind the wires and see how many of them I'd need to fill the 3cm diameter? Any help would be appreciated.
     
  2. jcsd
  3. Feb 9, 2009 #2

    LowlyPion

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    Your L is constant isn't it, so I think you are primarily concerned with the N*I product.

    So how many turns can you get along 20 cm using each wire?
    Then which of those has the rating to satisfy the current requirement for your target B?
     
  4. Feb 9, 2009 #3
    ah thanks. Ok so I got N of #18 to be 133, and N of #26 to be 797.
    So that's the number of turns. I'm now stuck with how to advance from here.
    I assume it has something to do with diameters and am probably missing the obvious here but not seeing how I can relate # of turns to satisfying the requirements of what I'm assuming are diameters.
    Thanks again.
     
  5. Feb 9, 2009 #4

    LowlyPion

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    So where did that come from?
     
  6. Feb 9, 2009 #5

    LowlyPion

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    Why isn't the number of turns of #18 = 20 cm/1.02mm ?
    Similarly for the #26.
     
  7. Feb 9, 2009 #6
    Those numbers came from B=NI/(cLE0).
    Anyway, using those number of turns I'd get 196 and 488 for 18 and 26, respectively.
    I guess I just don't see WHAT i'm supposed to be solving for here?
    Using those values for N, I get a B of 1.1MN/C for 18 and 0.18MN/C for 26. Neither of which are, or even realistically close to, the 1.5MN/C.
     
  8. Feb 9, 2009 #7

    LowlyPion

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    I guess I'm not familiar with what that equation is.
     
  9. Feb 9, 2009 #8
    It's an equation given in my book for the magnetic field inside a solenoid, where B is some special magnetic Field the author uses to make units N/C...and is basically the regular magnetic field times the speed of light.
    E0 is epsilon zero too if that wasn't clear.
     
  10. Feb 10, 2009 #9

    LowlyPion

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    OK. I get what you're doing now.

    You've calculated the B field correctly from the units given.

    So set 3981 = N/L*I = n*I where n is your turn density.

    The turn densities of the 2 wires are 1/D

    For the 1.02 mm that comes to 980.4 turns/m and the .41 mm gives 2439 turns/m

    One equation then is I = 3981/980.4 < 6a

    The other is I = 3981/2439 > 1a
     
  11. Feb 10, 2009 #10
    Lovely. Wow was I stupid. Guess I got so flustered I didn't even understand I was looking for a current below the heating level instead of just solving for B with the info I had like I was doing.
    Thanks a lot. Truly appreciated.
     
  12. Mar 5, 2011 #11
    I was following this problem until the concept of turn density was introduced. How does the turn density N/L equal 1/D, assuming that D stands for density of the wire?
     
  13. Feb 8, 2012 #12
    Hey.

    I noticed that the complete question was not answered. Anyone know how to answer what the potential difference across the wire needs to be?

    Thanks.
     
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