Question: An infinite wire carrying a constant current in the direction is moving in the direction at a constant speed . Find the electric field, in the quasistatic approximation, at the instant the wire coincides with the axis (Fig. 7.54).

The constant current in the wire is I.

The constant speed is v

The expression for the magnetic field is given as follows.

$\int B·dl={\mu }_{0}I$

Here, is the differential length and is the free space permeability.

The expression for differential form of the maxwell’s equation is given as follows.

$\nabla \times E=\frac{-\partial B}{\partial t}$ …… (1)

The expression for the quasistatic approximation in the wire is given as follows.

$\mathit{B}\mathbf{=}\left(\frac{{\mu }_{0}I}{2\pi s}\right)\hat{\mathbf{\varphi }}$

Here, $\varphi$ is the angle between the direction of the magneti field and axis.

Form the figure,7.54,

$$\begin{gathered} \cos \varphi =\frac{x}{s} \\ \sin \varphi =\frac{y}{s} \\ s=\sqrt{x^2+y^2} \end{gathered}$$

The magnetic field in the cartesian system is given by,

$B=\left(\frac{{\mu }_{0}I}{2\pi s}\right)\left(-\sin \varphi \hat{x}+\cos \varphi \hat{y}\right)$

Substitute $\frac{x}{s}for\cos \varphi and\frac{y}{s}for\sin \varphi$into above equation.

$$\begin{gathered} B=\left(\frac{{\mu }_{0}I}{2\pi s}\right)\left(-\frac{y}{s}\hat{x}+\frac{x}{s}\hat{y}\right) \\ B=\left(\frac{{\mu }_{0}I}{2\pi }\right)\left(\frac{-y\hat{x}+x\hat{y}}{s^2}\right) \end{gathered}$$

Substitute$x^2+y^2$ for into above equation.

$B=\left(\frac{{\mu }_{0}I}{2\pi }\right)\left(\frac{-y\hat{x}+x\hat{y}}{x^2+y^2}\right)$

The current carrying wire is moving in the direction, therefore the displacement along direction is$y\to y-vt$

$B=\left(\frac{{\mu }_{0}I}{2\pi }\right)\left(\frac{-\left(y-vt\right)\hat{x}+x\hat{y}}{x^2+{\left(y-vt\right)}^2}\right)$

Calculate the curl of the electric field.

Substitute$\left(\frac{{\mu }_{0}I}{2\pi }\right)\left(\frac{-\left(y-vt\right)\hat{x}+x\hat{y}}{x^2+{\left(y-vt\right)}^2}\right)$ for B into equation (1).

$$\begin{gathered} \nabla \times E=-\frac{\partial }{\partial t}\left[\left(\frac{{\mu }_{0}I}{2\pi }\right)\left(\frac{-\left(y-vt\right)\hat{x}+x\hat{y}}{x^2+{\left(y-vt\right)}^2}\right)\right] \\ \nabla \times E=\left(\frac{{\mu }_{0}I}{2\pi }\right)\left\{\frac{v\hat{x}\left(x^2+{\left(y-vt\right)}^2\right)-\left[-\left(y-vt\right)\hat{x}+x\hat{y}\right]\left(-2v\right)\left(y-vt\right)}{{\left[x^2+{\left(y-vt\right)}^2\right]}^2}\right\} \\ \nabla \times E=\left(\frac{{\mu }_{0}I}{2\pi }\right)\left\{\frac{v\hat{x}}{x^2+{\left(y-vt\right)}^2}+\frac{2v\left[-\left(y-vt\right)\hat{x}+x\hat{y}\right]\left(y-vt\right)}{{\left[x^2+{\left(y-vt\right)}^2\right]}^2}\right\} \end{gathered}$$

At t = 0 the curl of the electric field would be zero that means the above expression would be zero.

$$\begin{gathered} \nabla \times E=\left(\frac{{\mu }_{0}I}{2\pi }\right)\left\{\frac{v\hat{x}}{x^2+{\left(y-v\left(0\right)\right)}^2}+\frac{2v\left[-\left(y-v\left(0\right)\right)\hat{x}+x\hat{y}\right]\left(y-v\left(0\right)\right)}{{\left[x^2+{\left(y-v\left(0\right)\right)}^2\right]}^2}\right\} \\ \nabla \times E=\left(\frac{{\mu }_{0}I}{2\pi }\right)\left\{\frac{v\hat{x}}{x^2+y^2}+\frac{2v\left[-\left(y\right)\hat{x}+x\hat{y}\right]\left(y\right)}{{\left[x^2+y^2\right]}^2}\right\} \\ \nabla \times E=\left(\frac{{\mu }_{0}Iv}{2\pi }\right)\left\{\frac{\hat{x}}{x^2+y^2}+\frac{2\left[-y^2\hat{x}+xy\hat{y}\right]}{{\left(x^2+y^2\right)}^2}\right\} \end{gathered}$$

Change the above expression from the cartesian coordinates to cylindrical coordinates.

$\nabla \times E=\left(\frac{{\mu }_{0}Iv}{2\pi s^2}\right)\left(\cos \varphi \hat{s}+\sin \varphi \hat{\varphi }\right)$

The expression for the electric field in the cylindrical coordinate is given by,

$E\left(s,\varphi \right)=E_s\left(s,\varphi \right)\hat{s}+E_{\varphi }\left(s,\varphi \right)\hat{\varphi }+E_z\left(s,\varphi \right)\hat{z}$

The divergence of the electric field is zero and curl of electric field is goes to zero at the large value of the s. Therefore, the expression of divergence of electric field is given by,

$$\begin{gathered} \nabla ·E=\frac{1}{s}\frac{\partial \left(s{E}_s\right)}{\partial s}+\frac{1}{s}\frac{\partial E_{\varphi }}{\partial s} \\ \nabla ·E=0 \end{gathered}$$

The expression for the curl of electrical field along the direction.

$$\begin{gathered} {\left(\nabla \times E\right)}_s=\frac{1}{s}\frac{\partial E_z}{\partial \varphi } \\ {\left(\nabla \times E\right)}_s=-\frac{{\mu }_{0}Iv}{2\pi s^2}\cos \varphi \end{gathered}$$

The expression for the curl of electrical field along the direction.

$$\begin{gathered} {\left(\nabla \times E\right)}_{\varphi }=\frac{\partial E_z}{\partial s} \\ {\left(\nabla \times E\right)}_{\varphi }=-\frac{{\mu }_{0}Iv}{2\pi s^2}\sin \varphi \end{gathered}$$

The expression for the curl of electrical field along the direction.

$$\begin{gathered} {\left(\nabla \times E\right)}_z=\frac{1}{s}\left(\frac{\partial E_z}{\partial s}-\frac{\partial E_s}{\partial \varphi }\right) \\ {\left(\nabla \times E\right)}_z=0 \end{gathered}$$

The first and last term of the curl of electric field is satisfied if and

The middle two expression satisfied,

$E_z=\frac{-{\mu }_{0}Iv\sin \varphi }{2\pi s}$

Hence the expression for the electric field at the instant the wire coincide with axis is .

$\frac{-{\mu }_{0}Iv\sin \varphi }{2\pi s}$