Since $\nabla\phi=\mathbf e_\phi/r$, the vector field has cylindrical components
$$
\mathbf A=-\frac{\alpha_z}{r}\mathbf e_r
+\frac\beta r\mathbf e_\phi
+\frac{\alpha_r}{r}\mathbf e_z.
$$
The <divergence> in cylindrical coordinates is therefore
$$
\nabla\cdot\mathbf A
=-\frac1r\partial_r\alpha_z
+\frac1r\partial_z\alpha_r=0.
$$
Direct use of the <curl> in cylindrical coordinates gives
$$
(\nabla\times\mathbf A)_r=-\frac{\beta_z}{r},
\qquad
(\nabla\times\mathbf A)_z=\frac{\beta_r}{r},
$$
and
$$
(\nabla\times\mathbf A)_\phi
=\frac1r\left[-r\partial_r(r^{-1}\alpha_r)-\alpha_{zz}\right].
$$
Thus, defining
$$
\boxed{L\alpha=-r\partial_r(r^{-1}\partial_r\alpha)-\partial_z^2\alpha},
$$
we obtain
$$
\boxed{\nabla\times\mathbf A
=\nabla\beta\times\nabla\phi+(L\alpha)\nabla\phi}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-314/3/a/solution