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Consider a circular loop of radius R that carrys a uniform current I. We know(by Biot-Savart law) that the magnetic field it produces on its axis is given by [itex] \vec{B}=\frac{\mu_0 I R^2 \hat z }{2(z^2+R^2)^\frac 3 2} [/itex].
But let's calculate its vector potential:
[itex]
\vec{A}=\frac{\mu_0}{4\pi} \int \frac{I \vec{dl}}{|\vec{r}-\vec{r}'|}=\frac{\mu_0}{4\pi} \int_0^{2\pi} \frac{I R d\varphi \hat \varphi}{\sqrt{R^2+z^2}}=\frac{\mu_0 I R}{4\pi \sqrt{R^2+z^2}}\int_0^{2\pi} \hat \varphi d\varphi \\
[/itex]
But we have [itex] \hat \varphi=-\sin\varphi \hat x +\cos\varphi \hat y [/itex] and so [itex] \int_0 ^{2\pi} \hat\varphi d\varphi=0 [/itex] which gives [itex] \vec{A}=0 [/itex] and so [itex] \vec{B}=0 [/itex] !
What's wrong?
Thanks
But let's calculate its vector potential:
[itex]
\vec{A}=\frac{\mu_0}{4\pi} \int \frac{I \vec{dl}}{|\vec{r}-\vec{r}'|}=\frac{\mu_0}{4\pi} \int_0^{2\pi} \frac{I R d\varphi \hat \varphi}{\sqrt{R^2+z^2}}=\frac{\mu_0 I R}{4\pi \sqrt{R^2+z^2}}\int_0^{2\pi} \hat \varphi d\varphi \\
[/itex]
But we have [itex] \hat \varphi=-\sin\varphi \hat x +\cos\varphi \hat y [/itex] and so [itex] \int_0 ^{2\pi} \hat\varphi d\varphi=0 [/itex] which gives [itex] \vec{A}=0 [/itex] and so [itex] \vec{B}=0 [/itex] !
What's wrong?
Thanks