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Imagine the situation below where a steady current $I$ goes through the line. The radius of the semicircle is $r$. How do I compute the magnetic flux in the point $P$, marked by a black dot?

Schematic

Furthermore, I don't know the answer, and the two attempts that I have made gave me different results. I want to have this inconsistency resolved:

Solution 1. Biot-Savart law:

$$ \mathbf{B} = \frac{\mu_0 I}{4\pi r^2}\int_C\rm{d}\mathbf{l}\times \mathbf{\hat{r}} = \frac{\mu_0 I}{4\pi r^2}\int_C \rm{dl}=\frac{\mu_0 I}{4r} $$

since $\rm{d}\mathbf{l}\times \mathbf{\hat{r}} = dl\cdot 1\cdot \rm{sin}(\frac{\pi}{2})=dl$ and $\int_C dl=\pi r$ since the curve is the semicircle.

Solution 2. Ampere's lag $$ \oint \mathbf{B}\cdot \rm{d}\mathbf{l} = B\cdot 2\pi r = I\mu_0\Rightarrow B = \frac{I\mu_0}{2\pi r} $$

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  • $\begingroup$ Could you explain the second method? I don't understand it. What curve are you integrating around and how did you do the integral? $\endgroup$ – Brian Moths Nov 13 '14 at 15:31
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs In the second integral I am integrating along a circle. The circle passes through the point $P$ and the current passes through its center. You can think of the circle has going around the middle of the semicircle. Since everything is symmetric, I expected the magnetic flux to be constant along this circle. $\endgroup$ – user714 Nov 13 '14 at 15:36
  • $\begingroup$ Which, given Alfred's answer, I now understand is wrong of course... $\endgroup$ – user714 Nov 13 '14 at 15:46
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One can make an educated guess that the magnetic field at $P$ is one-half the value of the magnetic field at the center of a current loop which is given by:

$$B = \frac{\mu_0 I}{2R} $$

So your first approach is correct.

The approach of your solution 2 doesn't make any sense to me. The integral is along a closed path and, evidently, you're assuming $\mathbf B \cdot d\mathbf l$ is a constant along a closed path of constant $r$?

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  • $\begingroup$ Please see my comment to the question which explains the second integral. Although I am sure you have the right reason for it being wrong. Why did you get divided by $2R$ instead of $4R$ which is what I got? $\endgroup$ – user714 Nov 13 '14 at 15:39
  • $\begingroup$ @Althalos, as I wrote, the result I give is for a current loop (a full circle). $\endgroup$ – Alfred Centauri Nov 13 '14 at 15:44

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