Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 314 / 3 / a

Table of contents
- Solution a

Solution

Since

\nabla ϕ = e_{ϕ} / r

, the vector field has cylindrical components

A = - \frac{α _{z}}{r} e_{r} + \frac{β}{r} e_{ϕ} + \frac{α _{r}}{r} e_{z} .

The divergence in cylindrical coordinates is therefore

\nabla \cdot A = - \frac{1}{r} \partial_{r} α_{z} + \frac{1}{r} \partial_{z} α_{r} = 0.

Direct use of the curl in cylindrical coordinates gives

(\nabla \times A)_{r} = - \frac{β _{z}}{r}, (\nabla \times A)_{z} = \frac{β _{r}}{r},

and

(\nabla \times A)_{ϕ} = \frac{1}{r} [- r \partial_{r} (r^{- 1} α_{r}) - α_{zz}] .

Thus, defining

Lα = - r \partial_{r} (r^{- 1} \partial_{r} α) - \partial_{z}^{2} α,

we obtain

\nabla \times A = \nabla β \times \nabla ϕ + (Lα) \nabla ϕ .

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version