In cylindrical coordinate system, the potential produced by a uniform ring charge is given by $\varphi =f\left(r,z\right)$, where $f$ is a continuous function of $r$ and $z$. Let $\vec{E}$ be the resulting electric field. Then the magnitude of $\nabla \times \vec{\mathrm{E}}$

The question asks for the magnitude of the curl of the electric field, denoted as $\nabla \times \vec{E}$, where $\vec{E}$ is the electric field produced by a uniform ring charge described by a potential $\varphi =f(r,z)$ in cylindrical coordinates.

Understanding the Curl of an Electric Field

In electrostatics, the electric field $\vec{E}$ is always conservative. A conservative field is one for which the work done in moving a charge between two points is independent of the path taken. A fundamental property of conservative vector fields is that their curl is always the zero vector.

The relationship between the electric field $\vec{E}$ and the scalar electric potential $\varphi$ is given by:

$$\vec{E}=-\nabla \varphi$$

The curl of the electric field $\vec{E}$ is calculated as $\nabla \times \vec{E}$. Substituting the relationship between $\vec{E}$ and $\varphi$:

$$\nabla \times \vec{E}=\nabla \times (-\nabla \varphi )$$

A key identity in vector calculus states that the curl of a gradient of any scalar function (like potential $\varphi$) is always zero:

$$\nabla \times (\nabla \varphi )=\vec{0}$$

Therefore, for any electrostatic field derived from a scalar potential:

$$\nabla \times \vec{E}=\nabla \times (-\nabla \varphi )=-(\nabla \times (\nabla \varphi ))=-\vec{0}=\vec{0}$$

**Magnitude of $\nabla \times \vec{E}$ **

The result $\nabla \times \vec{E}=\vec{0}$ means that the curl of the electric field is the zero vector.

The magnitude of the zero vector $\vec{0}$ is simply 0.

The specific details of the charge distribution (uniform ring charge) and the coordinate system (cylindrical) define the potential function $f(r,z)$ and the resulting electric field $\vec{E}$. However, the property that the curl of an electrostatic field is zero is universal and independent of these specifics. It stems from the fact that the electrostatic field is derivable from a scalar potential.

Conclusion

Since the electric field $\vec{E}$ originates from a scalar potential $\varphi$, it is an electrostatic field. All electrostatic fields are conservative, meaning their curl is zero. Consequently, the magnitude of $\nabla \times \vec{E}$ is 0.

Based on this, the correct option is that the magnitude of $\nabla \times \vec{E}$ is 0.