0
$\begingroup$

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.

  • 0
    write out the components of the curl and then form a dot product, then see what happens2012-10-22

1 Answers 1

1

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.