Question #132493

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

Expert's answer

Field is rotational, if its curl does not vanish. In cylindrical coordinates, curl is:∇×F=aρ(1ρ∂Az∂φ+∂Aφ∂z)+aφ(∂Aρ∂z−∂Az∂ρ)+az1ρ(∂(ρAφ∂ρ−∂Aρ∂φ)\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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