1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Magnetostatic field calculations

  1. Jul 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Current, I, flows around symmetrical loop of thin wire.
    The loop comprises 2 halves (semicircle & triangle). Distance from vertex of triangle equals radius, a, of the semicircle.
    Find the mag' field at axis of loop (where axes coss)

    2. Relevant equations
    Could I use Gauss' law ∫s E.dS= 1/εov ρ(r) dv
    Amperes law ∫c B . dl
    Or is this just as easy to do wit Biot-Savart Law δB(r) = . . . . . . . . .

    I have a good feeling that I can use symmetry to simplify this without getting too bogged down in maths.

    Which of these are the ones I could use most?

    3. The attempt at a solution
  2. jcsd
  3. Jul 10, 2012 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Gauss' law involves electric field rather than magnetic field. So, punt on that one.

    I'm having some difficulty visualizing the loop. But, it appears to me that there is not enough symmetry to use Ampere's law. Ampere's law is useful if you can construct an imaginary closed path that:
    (1) passes through the point where you are trying to find the field
    (2) B[itex]\cdot[/itex]dl has a simple form for the path.
    I don't think you can find a path that meets both these criteria.

    So, you're left with Biot-Savart. However, if you've already covered the field of a complete circular loop of wire, then you can use that as a shortcut to get the field from the semicircular part of your loop. Also, if you've already covered the field of a finite, straight section of current, then you can use that result along with superposition to get the field from the straight sides of the triangle.
  4. Jul 25, 2012 #3
    Hmmm! Have tried the problem & have a solution if someone would review it.

    Thank you

    Split composite loop in to 2 parts (one semicicular one of 2 sides of a square)

    For semicircle:
    B = (μ0*I)/(4*∏*R2) ∫ dL where R = a
    = (μ0*I)/(4*∏*R2) 2*∏*R2
    = (μ0*I)/(2*R)

    For 2 side of a square:
    B = (2*μ0*I)/(4*∏*r) (sinθ1 - sinθ2) where r = 0.707*a & θ1=-θ2=/4
    = (2*μ0*I)/(4*∏*r) (sin(∏/4) - sin(∏/4))
    = (2*μ0*I)/(4*∏*r) (√2/2 + √2/2)
    = √2 ((μ0*I)/(4*∏*0.707a))

    so Btotal = (μ0*I)/(2*R) + √2 ((μ0*I)/(4*∏*0.707a))

    I feel that there's something I haven't done, or missed something or . . . . . . .
  5. Jul 25, 2012 #4


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    I believe the square on the R at the very end is just a typo. But ∫ dL should be the arc length of the semicircle, not a full circle.

    Look's good except I think you dropped a factor of 2 in going from the next-to-last to the last equation. Also, it will simplify nicely if you write the factor of .707 in the denominator as √2/2.
  6. Jul 25, 2012 #5
    Thanks for the read through &comment.
    Last edited: Jul 25, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook