If a uniform magnetic field ${\bf{B}}=B_{z}{\bf{\hat{z}}}$ exists in a hollow cylinder (with the top and bottom open) with a radius $R$ and axis pointing in the $z$-direction, then the vector potential
$${\bf{A}}=\frac{BR^{2}}{2r}{\bf{\hat{\phi}}}?$$
using Stokes's theorem.
How can you prove this?
Best Answer
You are given a uniform magnetic field $\vec{B}=B_z\hat{z}$
We have the relation connecting the magnetic field vector $\vec{B}$ and the vector potential $\vec{A}$
$$\vec{B}=\nabla\times\vec{A}\tag 1$$
Now, according to Stoke's theorem, we have
$$\int_S\left(\nabla\times\vec{A}\right)\cdot d\vec{S}=\oint_C\vec{A}\cdot d\vec{r}\tag 2$$
The theorem can be stated as follows: The surface integral of the curl of a vector field is equal to the line integral of the vector field around the boundary of the surface. So, in the above equation, the surface integral is across the surface $S$ bounded by a boundary curve $C$, along which the line integral is taken.
Now, from $(1)$, we can write $(2)$ as
$$\oint_C\vec{A}\cdot d\vec{r}=\int_S\vec{B}\cdot d\vec{S}\tag 3$$
Since the magnetic field is uniform and points in a direction along the axis of the cylinder, the magnetic vector potential lies along the radial direction, which is clear from equation $(1)$. At the end, the magnetic field points in the same direction as the cross sectional area of the hollow cylinder. Hence the dot product simplifies as
$$\oint_C\vec{A}\cdot d\vec{r}=B_z\int_S d{S}\tag 4$$
If we choose our origin on an axis of symmetry, so that we can take $A$ as circumferential ($\hat{\phi}$ component only) and a function only of $r$, then
$$\oint_C\vec{A}\cdot d\vec{r}=A_{\phi}\left(2\pi r\right)\tag 5$$
That's all. Now, check $(4)$ and $(5)$; and using appropriate substitutions, you can reach your final answer.