Question #132493
Show that the field given below, in cylindrical coordinates, Is rotational.
F = (150/p^2)ap^ + 10 a(phi)^
1
Expert's answer
2020-09-10T15:49:07-0400

Field is rotational, if its curl does not vanish. In cylindrical coordinates, curl is:×F=aρ(1ρAzφ+Aφz)+aφ(AρzAzρ)+az1ρ((ρAφρAρφ)\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, ×F=az1ρ((10ρ)ρ)=10ρaz\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.


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