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Magnetic Field of wire carrying current

  1. Apr 3, 2008 #1
    A wire carrying a current I is bent into the shape of an equilateral triangle of side L. Find the magnitude of the magnetic field at the center of the triangle.

    Since the vector of the magnetic field in respect to each side of the triangle points in the same direction at the center, I should be able to find the magnetic field generated by one side of the triangle and multiply it by 3. After trying several times, I keep getting
    B = 1.5uI/(pi*L), in which u = permeability of of free space...but the actual answer is
    B = 4.5uI/(pi*L), which is 3 times greater than my answer. After looking over my work, nothing seems to be wrong...but my answer is down by a factor of 3 from the one I'm supposed to get. What am I doing wrong??
  2. jcsd
  3. Apr 3, 2008 #2
    Maybe you are not multiplying it by 3 as you said you would...
  4. Apr 3, 2008 #3
    Nah...I have...still doesn't come out right. I'm pretty sure I've done things right...maybe the question meant length L rather than side L...with that, I actually get the right answer.
  5. Apr 3, 2008 #4
    Well.. i tried doing it... For a triangle, you get the perpendicular distance from a side to the centroid to be:

    \frac{L}{2} \tan{\left(\frac{\pi}{6}\right)}

    on using [itex]B = \frac{\mu_o I}{2 \pi r}[/itex]

    i'm getting

    B = \frac{1.724 \mu_o I}{\pi L}

    and for three sides i'm getting it as:

    B = \frac{5.16 \mu_o I}{\pi L}

    which is again.. not the right answer?
    Last edited: Apr 3, 2008
  6. Apr 3, 2008 #5
    Hmmm...I suppose it really just might be length L instead of side L....oh well...
  7. Apr 3, 2008 #6
    are you sure we can approximate the magnetic field using the formula we've used? If not try using the Biot-Savart law and see what comes up.
  8. Apr 4, 2008 #7
    Thats actually the formula I used...I'm sure I also checked earlier with Ampere's Law...
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