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Let's say we have a loop with a clockwise current, and my angle increases in the counter-clockwise direction (that is, $\hat{\phi}$ is counter-clockwise).

I have $$B=\frac{\mu_{0} i\vec{dl} \times \vec{r}}{4 \pi |\vec{r}|^{3}}$$

If I were to evaluate the line integral counter-clockwise, then when I change this integral in terms of $\phi$, my integral goes from $0$ to $2 \pi$.

If I evaluatethis integral clockwise, when I switch this integral in terms of $\phi$, my integral goes from $2 \pi$ to $0$ $-$ but now my $\vec{dl}$ is $-R d\phi \hat{\phi}$, so the two negative signs just compensate.

However, in one case, the current is in the same direction as $\vec{dl}$ and in the other, it is antiparallel, so it looks like I will get two different answers (although i expect the same answer)!

Which, of course, just leaves me confused.

Thanks for any help!

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  • $\begingroup$ Why do you think the direction of the magnetic field should depend on the mechanics of the integration? The current itself does not change direction. $\endgroup$ – Triatticus May 5 '19 at 16:14
  • $\begingroup$ My question wasn't very clear, sorry! I've edited it a bit. $\endgroup$ – Rahul Arvind May 5 '19 at 16:27
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The actual Biot-Savart's law for a wire carrying a steady current reads

$$B(\vec{r})=\frac{\mu_0}{4\pi}\int\frac{\vec{I}\times\vec{r'}}{r'^3}d\ell'$$

where $\vec{r'}$ is a vector pointing from the charge to the point in space $\vec{r}$. What you have written is a simplification that we can always make for the steady current carrying wire because $I$ is just some constant. So we define a vector $d\ell'$ that satisfies the condition

$$\vec{d\ell'}=d\ell \hat{I}$$

where

$$\hat{I} = \frac{\vec{I}}{|{I}|}$$

and thus it is equivalent to write this as

$$B(\vec{r})=\frac{\mu_0 I}{4\pi}\int\frac{\vec{d\ell'}\times\vec{r'}}{r'^3}$$

which is convenient. The answer for your question is then that it absolutely matters which way you choose to integrate: you must integrate in the direction that the current flows. You're correct to think that the direction is not arbitrary.

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  • $\begingroup$ Thanks for the detailed response! $\endgroup$ – Rahul Arvind May 8 '19 at 13:01
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The convention is to parametrize $\vec{d\ell}$ in the direction that the current is flowing so that there is no ambiguity. When you stick to this convention the sign ambiguity goes away.

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