Answer to Question #132493 in Electricity and Magnetism for Janarakshaa

Question

Show that the field given below, in cylindrical coordinates, Is rotational.
F = (150/p^2)ap^ + 10 a(phi)^

Answer 1

Field is rotational, if its curl does not vanish. In cylindrical coordinates, curl is:$\nabla \times \bold F = \bold a_\rho (\frac{1}{\rho}\frac{\partial A_z}{\partial \varphi} + \frac{\partial A_\varphi}{\partial z}) + \bold a_\varphi(\frac{\partial A_\rho}{\partial z} - \frac{\partial A_z}{\partial \rho}) + \bold a_z \frac{1}{\rho}(\frac{\partial (\rho A_\varphi}{\partial \rho} - \frac{\partial A_\rho}{\partial \varphi})$

For given field, $\nabla \times \bold F = \bold a_z \frac{1}{\rho}(\frac{\partial(10 \rho)}{\partial \rho}) = \frac{10}{\rho} \bold a_z$.

From the last formula, curl does not vanish, hence the field is rotational.