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Using Stokes theorem I seem to find the integral value as $2\pi r^2$. Can anyone help me if I am right? How do i do it without using Stokes theorem?

$$C=\oint\limits_K \mathrm d \mathbf r\cdot \mathbf A$$

Where $\displaystyle \mathbf A=\frac{1}{2}\mathbf n\times\mathbf r,\quad \mathbf n\cdot\mathbf n=1$, and $K$ is a circle with radius $R $ and $\mathbf n$ is the normal to the plane where the circle lives.

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migrated from Oct 23 '12 at 4:16

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Is this HW or are you just curious for some reason? – DJBunk Oct 22 '12 at 23:15
@DJBunk : This question is something that i got from my friend. It looks interesting and he told me that it appears frequently in Electrodynamics in its general form . Can you help me? – Theorem Oct 22 '12 at 23:18
@DJBunk : I did some calculation assuming that the circle is in $xy$ plane , now again i found it to be zero . using line integral and stokes theorem. – Theorem Oct 22 '12 at 23:19
write out the components of the curl and then form a dot product, then see what happens – Timtam Oct 22 '12 at 23:52

In cylindrical polar-coordinates ($\rho$, $\phi$, $z$) with $\mathbf{n}$ along the $z-$axis, we would get $$\mathbf{A} = \frac{\rho}{2} \hat{\phi}$$ and along the circle $K$ of radius $\rho=R$, $$d\mathbf{r} = R d\phi$$ Then $$\oint_K d\mathbf{r} \cdot \mathbf{A} = \frac{R^2}{2} \int_{0}^{2\pi} d\phi = \pi R^2$$

Now using a vector identity $$\nabla \times \mathbf{A} = \frac{1}{2} \nabla \times ( \mathbf{n} \times \mathbf{r} ) = \frac{1}{2} \Big[ \hat{z} (\nabla \cdot \mathbf{r}) - (\hat{z} \cdot \nabla) \mathbf{r} \Big] = \hat{z}$$ So that $$\int_{S} d\mathbf{S} \cdot \nabla \times \mathbf{A} = \pi R^2$$ where the surface $S$ is a cylinder oriented along the $z-$axis with one (open) end on the circle $K$ and the other end closed at an arbitrary distance $z$ from the circle.

In the form presented, $A$ is the vector potential for a uniform magnetic field along the $z-$direction (and in this case of unit strength). I think this is called the radial gauge.

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