Magnetic field in the Centre of Circular loop wire with Current?

By biot-savart:

$$\bar{H} = \frac{I}{4\pi} \oint \frac{d\bar{l} \times \bar{r}}{r^{3}}$$

so

$$\bar{H} = \frac{I}{2a} \hat{n}$$

Please, explain the last implication. I cannot find such integral to match the results. The radius of the loop is $a$. The current is $I$. $d\bar{l}$ is a vector along the perimeter.

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For the path element $d\vec{l}$ around a circle with radius $a$ you can write $d\vec{l}=rd\phi\vec{e_\phi}$ with $\vec{r}=-r\vec{e_r}$ (note the minus sign, since the vector points from the wire to the center) you get

$$\vec{H}=\frac{I}{4\pi}\oint \frac{r^2 (-\vec{e_\phi}\times \vec{e_r})}{r^3}d\phi$$ substituting $a$ for $r$ and integrating $\phi$ from 0 to $2\pi$ and realizing $-\vec{e_\phi}\times\vec{e_r}=\vec{e_z}$

$$\vec{H}=\frac{I}{4\pi}\frac{1}{a}\vec{e_z}\int_0^{2\pi}d\phi = \frac{I}{2a}\vec{e_z}$$

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Luksen's answer is 100% correct, but depending on your background and predilections you might want a less formal statement of what's going on.

When working out a calculation like this, please, please, please start by drawing a picture. There's really no substitute for this! Here's one I stole from a web site that does a good enough job (the main difference being that it calls the loop radius $R$ instead of $a$:

The integral consists of adding up contributions all the way around the circular loop. For all these contributions, $d\vec l\times\vec r$ is a vector whose magnitude is $a\,dl$ and which points in the perpendicular direction (out of the plane of the diagram). To see this, note that the magnitude of $\vec r$ is always equal to $a$, the directions of $d\vec l$ and $\vec r$ are always orthogonal, and the right-hand rule always gives the same perpendicular direction. In short, $d\vec l\times\vec r=a\,dl\,\hat z$ where $z$ is the appropriate unit vector.

Moreover, note that since the magnitude of $r$ is always $a$, the $r^3$ in the denominator of the integrand is just $a^3$. Substitute these facts into the integral, and make use of the fact that any constant quantity can be pulled outside of an integral to write $$\vec H={I\over 4\pi}\oint {a\,dl\,\hat z\over a^3}={I\over 4\pi}{\hat z\over a^2} \oint dl.$$ The last step is to observe that $\oint dl=2\pi a$. You can prove this formally by writing things out in polar coordinates or something, or better yet just think about what this integral means. $dl$ is a unit of length around the loop. The integration just means that you're adding up all the $dl$'s. If you add up all the length elements around a loop, you get the circumference.

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+1 great answer. I'd maybe differentiate the symbols for the path element and the angle element $dl$ vs $rd\phi$ as it might be confusing to use the same symbol as in the OP – luksen Aug 28 '11 at 18:19
wait, you have one extra term there $\mu_{0}$? – hhh Aug 28 '11 at 18:30
Ted accidentally wrote the $\mu_0$ for the $B$-field, I suppose. For the $H$-field there's no $\mu_0$ – luksen Aug 28 '11 at 18:45
Sorry about that! I see someone's fixed it. Thanks! – Ted Bunn Aug 28 '11 at 19:46