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Homework Help: What is the flux density at point p?

  1. Apr 21, 2008 #1
    1. The problem statement, all variables and given/known data

    What is the flux density at point p in Fig "see link below". due only to the 3-A current in the portion of wire 2 shown?

    1) 1.64 x 10^-5 Wb/m^2
    2) 2.45 x 10^-5 Wb/m^2
    3) 3.11 x 10^-5 Wb/m^2
    4) 3.65 x 10^-5 Wb/m^2

    In which direction does the flux at point p of Question 1 point
    1) Out of the page.
    2) To the left of the page.
    3) To the right of the page.
    4) Into the page.

    Fig Link
    ignore the red marks.

    Answers 2 and 2.. Is this correct?


    2. Relevant equations

    B=I x 10^-7 (Integral sign) sinΘ/r^2 dl

    3. The attempt at a solution

    B= I x 10-7 (Integral sign) sin Θ /r^2 rdΘ/sin0

    B= I x 10^-7/0.0200 (Integral sign) sinΘ dΘ

    B=.5 x 10^-7 [-(-2)-(-5)] = ?????????????????????

    Does anyone have a real good document on how to Find flux density at a point? My current book does not fully explain it. Is there a online calculator that would help?

  2. jcsd
  3. Apr 22, 2008 #2


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    Homework Helper

    Hi Adventure123,

    I think you have a couple of problems with your integral. It's difficult for me to tell because I don't know how you set up the geometrey, but here are some notes:

    In the first line of section 3, it looks like you are writing dl in terms of d(theta). That's good, but since your expression for dx has r in it, it looks like you forgot to take into account dr (since r is also variable). If so, it would probably be better to write dl in terms of theta and the perpendicular distance from the wire to the point (2 cm in your diagram). If we call that distance x, then dx would be zero.

    In your second line of section 3, you brought r out of the integral, but it depends on theta and so cannot be removed like that. (If you write dl in terms of x, that constant x value would come out of the integral.)

    But like I said, I could just be misinterpreting your method.
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