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Magnetic Field due to a curved wire

  1. Jan 31, 2016 #1
    1. The problem statement, all variables and given/known data
    A very long straight wire carries current I. In the middle of the wire a right-angle bbend is made. The bend forms an arc of a circle of radius r, as shown in the figure below. Determine the magnetic field at the center of the arc.


    2. Relevant equations
    Biot-Savart Law:
    ##dB = \frac{ \mu (I ds) \times r}{4 \pi r^3}##

    3. The attempt at a solution

    Since $$B = \frac{\mu}{4 \pi} \int_a^b{\frac{[I ds] \times r}{r^3}},$$ notice that the curve is 1/4 of a circle. Therefore, we can integrate from 0 to ##2 \pi r## and then divide by 4. Therefore, it becomes $$B = \frac{\mu}{4 \pi} \int_0^\frac{\pi r}{2}{\frac{I r sin(90)}{r^3}ds}$$ which simplifies down to $$B = \frac{\mu I}{8 r}$$

    The issue that arises now is what to do about the parts that don't form a 90 degree angle with the centre of the arc. I'm unsure about this, but my main thought is this. The magnetic field due to an infinite length of wire is $$B = \frac{\mu I}{2 \pi r}$$ Since each of these segments is only half of such an infinite wire, their magnetic fields ought to each be $$B = \frac{\mu I}{4 \pi r}$$ But since there are two, they ought to add back together by the superposition principle, and by said principle the magnitude of the magnetic field at said point ought to be $$B = \frac{\mu I}{8 r} \ \ + \ \ \frac{\mu I}{2 \pi r}$$ or $$B = \frac{\mu I}{8 \pi r} (\pi +4)$$
    Last edited: Jan 31, 2016
  2. jcsd
  3. Jan 31, 2016 #2


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    That all looks right to me, except I think you left something out of the denominator in the final equation.
  4. Jan 31, 2016 #3
    Oh yeah, I missed the ##\pi##. Thanks!
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