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Calculating a magnetic field at a given point

  1. Oct 31, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data

    For a picture of the problem, look at the attached picture.
    There's a current i through the wire.
    1)Calculate the magnetic field (at point C) produced by each segment of length L.
    2)The semi-circular segment of radius R.
    3)The entire wire.
    2. Relevant equations
    None given. Biot-Savart and Ampère's law.


    3. The attempt at a solution
    1)Using Biot-Savart law, I have that [tex]\vec B = \frac{\mu _0 i}{4 \pi} \int d \vec l \times \frac{\vec r}{r^3}[/tex] but [tex]\vec l[/tex] and [tex]\vec r[/tex] are parallel, so the cross product is worth [tex]0[/tex] and so the contribution of the 2 segments of length [tex]L[/tex] to the magnetic field at point [tex]C[/tex] is null.
    2) and 3) : Using Ampère's law I get that [tex]B=\frac{\mu _0 i}{\pi R}[/tex]

    Is that right? I find my answer to 1) strange.
     

    Attached Files:

  2. jcsd
  3. Oct 31, 2009 #2

    ideasrule

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    Your answer to 1) is right: charges moving directly towards or away from a point don't induce any magnetic field at that point. Your answer to 2) is a bit strange, though. How did you use Ampere's law? Which closed loop did you integrate around? I find it easier to just use the Biot-Savart law; the integral isn't necessary because the vector dl is always perpendicular to the vector r.
     
  4. Oct 31, 2009 #3

    fluidistic

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    Thank you very much for helping!
    Ok, nice to know I got part 1) right.
    2) [tex]\oint \vec B d \vec l = \mu _0i[/tex] but as you pointed out [tex]\vec B \perp \vec l[/tex] for the entire semi circle loop so that [tex]B \cdot \pi R = \mu _0 i[/tex], hence my result.
    I can say I've took a surrounding to the semi circle loop, closed loop. Does it matter? The current enclosed will always be worth [tex]i[/tex].
     
  5. Oct 31, 2009 #4

    ideasrule

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    There are a few problems with this. First, there's no obvious symmetry that suggests B is the same along the entire loop. Second, even if B doesn't vary, you're trying to find the magnetic field at point C; finding the field along the loop doesn't help. Third, i represents the current enclosed within the loop; in your scenario, the current is flowing parallel to the loop you chose. Finally, you chose a semi-circle, not a closed loop. Ampere's law only works for closed loops (notice the circle on the integral sign in your equation). There are absolutely no exceptions.
     
  6. Nov 1, 2009 #5

    fluidistic

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    First, thanks for helping.

    hmm I don't understand why not. The semi-circular loop is the only part of the wire that create the magnetic field at point C, so enclosing it by a loop sounded good to me.

    I haven't been clear enough. :shy: sorry. I meant I can say I have chosen a closed loop which has the same shape as the semi-circular part of the wire. For example a loop that has a width of [tex]\varepsilon[/tex] arbitrary small or big, instead of no width as is considered the wire. But a closed loop that enclose absolutely all the semi-circular wire.

    Lastly, I will see a friend today, I will ask him if he knows how to solve this problem. I will post back here anyway, whether or not I have to procedure to solve the problem. Answer to my misconception if you like though, I'd be glad to learn.
     
  7. Nov 1, 2009 #6

    Doc Al

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    As ideasrule states, Ampere's law is not useful in solving this problem. Instead, use the Biot-Savart law for all parts of the problem. The integration is trivial.
     
  8. Nov 1, 2009 #7

    fluidistic

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    You're right, I've underestimated this part.
    [tex]\vec B = \frac{\mu _0 i}{4 \pi} \oint d\vec l \times \frac{d \vec r}{r^3}[/tex] but as [tex]\vec l \perp \vec r[/tex], the path integral is worth [tex]\frac{\pi R}{R^2}=\frac{\pi}{R}[/tex], thus [tex]B= \frac{\mu _0 i}{4 R}[/tex].
    The direction of the magnetic field must point upward C, I believe. (for symmetrical reasons).

    I hope I got it right.
     
  9. Nov 1, 2009 #8

    Doc Al

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    Looks good. (The direction of the field depends on the direction of the current. It's either into or out of the paper.)
     
  10. Nov 1, 2009 #9

    fluidistic

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    Ah I see.

    Thanks a lot to both!
     
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