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Magnetic field at the centre of current carrying spiral

  1. Dec 8, 2015 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    A thin insulated wire forms a plane spiral of N turns carrying a current ##i##. The inner radius is ##b## an outer radius is ##a##. Find magnetic field at centre of spiral

    2. Relevant equations
    $$B=\frac{n\mu_0i}{2R}$$

    3. The attempt at a solution
    For an area of ##\pi(b^2-a^2)##, number of turns is N.
    So for an elemental area of ##2\pi r dr##,number of turns ##n## is ##\frac{2rNdr}{(b^2-a^2)}##

    After substing ##n##, $$dB=\frac{\mu_0 I 2Ndr}{2(b^2-a^2)}$$
    But on integrating, I am getting wrong answer

    Instead of taking number of turns per unit area, if I took number turns per unit width, I got the correct answer. That is if $$n=\frac{N}{b-a}$$.

    Why does the answer change?
     
  2. jcsd
  3. Dec 8, 2015 #2

    TSny

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    Did you mean to have ##a## as the inner radius and ##b## as the outer radius?
    OK. What did you get for your final answer for ##B##?
     
  4. Dec 9, 2015 #3

    Titan97

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    typo. Inner radius is ##a## and outer radius is ##b##.

    Integrating from a to b, $$B=\frac{\mu_0 i N}{b+a}$$ This is not the answer given.
     
  5. Dec 9, 2015 #4

    TSny

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    That looks right to me. It has the correct limit for ##a## →##b##. Hopefully someone else will confirm this or else point out our mistake.

    What answer was given?
     
  6. Dec 9, 2015 #5

    Titan97

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  7. Dec 9, 2015 #6

    TSny

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    Ah, good. Thank you. Now I see the mistake. There are, in fact, a constant number of turns per unit radial distance. Just picture the individual wraps of wire. The wire has a fixed thickness. So, there will be a fixed number of turns per unit radial distance. So, it is not the area ##2 \pi r dr## that determines the number of wraps of wire in a distance ##dr##, but rather its just the length ##dr## itself that determines the number of wraps. Sorry for not catching that.
     
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