Dewgale
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Homework Statement
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.
Homework Equations
Biot-Savart Law:
##dB = \frac{ \mu (I ds) \times r}{4 \pi r^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 $$B = \frac{\mu I}{8 \pi r} (\pi +4)$$
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