91Ó°ÊÓ

Electromagnetism is described as the interaction amongst the magnetic fields and the electric fields or currents. Moreover, it is also helpful for studying the interaction between the particles having electrically charged.

The equation of the vector potential in the z direction is expressed as:

$-\frac{∂A_z}{∂x}=F_y$

Here,$F_y$ is the force in the y direction and$A_z$ is the differential equation with respect to the function .

The above equation can also be expressed as:

${\mathrm{A}}_{\mathrm{z}}(\mathrm{x},\mathrm{y},\mathrm{z})=-{∫}_0^{\mathrm{x}}{\mathrm{F}}_{\mathrm{y}}({\mathrm{x}}^{\text{'}},\mathrm{y},\mathrm{z}){\mathrm{dx}}^{\text{'}}+{\mathrm{C}}_1(\mathrm{y},\mathrm{z})$ ..(i)

The equation of the vector potential in the y direction is expressed as:

$-\frac{∂A_y}{∂x}=F_z$

Here,$F_z$ is the force in the z direction and $A_y$is the differential equation with respect to the function x .

The above equation can also be expressed as:

${\mathrm{A}}_{\mathrm{y}}(\mathrm{x},\mathrm{y},\mathrm{z})=+{∫}_0^{\mathrm{x}}{\mathrm{F}}_{\mathrm{z}}({\mathrm{x}}^{\text{'}},\mathrm{y},\mathrm{z}){\mathrm{dx}}^{\text{'}}+{\mathrm{C}}_2(\mathrm{y},\mathrm{z})$ (ii)

The equations (i) and (ii) satisfies the condition (i) and (ii) for the constants $C_1$and $C_2$.

Differentiating the equation (i) with respect to the y function and differentiating the equation (ii) with respect to the function z and subtracting the equation (i) and (ii).

$-{∫}_0^x\frac{∂F_y(x^{\text{'}},y,z)}{∂yd{x}^{\text{'}}}d{x}^{\text{'}}+\frac{∂C_1}{∂y}-{∫}_0^x\frac{∂F_z(x^{\text{'}},y,z)}{∂z}d{x}^{\text{'}}+\frac{∂C_2}{∂z}=F_x(x,y,z)$

According to the condition given in the question, $\frac{∂F_x}{∂x}+\frac{∂F_y}{∂y}+\frac{∂F_z}{∂z}=0$.

$$\begin{gathered} {∫}_0^x\frac{∂F_x(x^{\text{'}},y,z)}{∂x^{\text{'}}}d{x}^{\text{'}}+\frac{∂C_1}{∂y}-\frac{∂C_2}{∂z}=F_x(x,y,z) \\ {∫}_0^x\frac{∂F_x(x^{\text{'}},y,z)}{∂x^{\text{'}}}d{x}^{\text{'}}=F_x(x,y,z)-F_x(0,y,z) \end{gathered}$$

The equation of the constants can be written as:

$\frac{∂C_1}{∂y}-\frac{∂C_2}{∂z}=F_x(0,y,z)$

From the above equation, $C_2=0$and$C_1(y,z)={∫}_0^y{F}_x(0,y^{\text{'}},z)d{y}^{\text{'}}$

Then the values of the functions$A_x$ , $A_y$and$A_z$ is expressed as:

$$\begin{gathered} A_x=0 \\ {A}_y={∫}_0^x{F}_z(x^{\text{'}},y,z)d{x}^{\text{'}} \\ {A}_z={∫}_0^y{F}_x(0,y^{\text{'}},z)d{y}^{\text{'}}-{∫}_0^x{F}_y(x^{\text{'}},y,z)d{x}^{\text{'}} \end{gathered}$$

Thus, the values of$A_x$ ,$A_y$ and $A_z$are$A_x=0,A_y={∫}_0^x{F}_z(x^{\text{'}},y,z)d{x}^{\text{'}}and{A}_z={∫}_0^y{F}_x(0,y^{\text{'}},z)d{y}^{\text{'}}-{∫}_0^x{F}_y(x^{\text{'}},y,z)d{x}^{\text{'}}$respectively.

The equation of the cross product of the curl of the vector potential can be expressed as:

$∇×\mathrm{A}=\left(\frac{∂{\mathrm{A}}_{\mathrm{z}}}{∂\mathrm{y}}-\frac{∂{\mathrm{A}}_{\mathrm{y}}}{∂\mathrm{z}}\right)\hat{\mathrm{x}}+\left(\frac{∂{\mathrm{A}}_{\mathrm{x}}}{∂\mathrm{z}}-\frac{∂{\mathrm{A}}_{\mathrm{z}}}{∂\mathrm{x}}\right)\hat{\mathrm{y}}+\frac{∂{\mathrm{A}}_{\mathrm{y}}}{∂\mathrm{x}}-\frac{∂{\mathrm{A}}_{\mathrm{x}}}{∂\mathrm{y}}\hat{\mathrm{z}}$

The above equation can be solved as:

$∇×A=\left[F_x(0,y,z)-{∫}_0^x\frac{∂F_y(x^{\text{'}},y,z)}{∂y}d{x}^{\text{'}}-{∫}_0^x\frac{∂F_z(x^{\text{'}},y,z}{∂z}d{x}^{\text{'}}\right]\hat{x}+[0+F_y(x,y,z\left)\right]\hat{y}+\left[F_z\right(x,y,z)-0]\hat{z}$

Here, the function ,$∇×F=0$ hence the equation of the coordinate is expressed as:

$\left[{\mathrm{F}}_{\mathrm{x}}(0,\mathrm{y},\mathrm{z})+{∫}_0^{\mathrm{x}}\frac{∂{\mathrm{F}}_{\mathrm{x}}({\mathrm{x}}^{\text{'}},\mathrm{y},\mathrm{z}}{∂\mathrm{z}}\right]={\mathrm{F}}_{\mathrm{x}}(0,\mathrm{y},\mathrm{z})+{\mathrm{F}}_{\mathrm{x}}(\mathrm{x},\mathrm{y},\mathrm{z})-{\mathrm{F}}_{\mathrm{x}}(0,\mathrm{y},\mathrm{z})$

Hence, $∇×\mathrm{A}=\mathrm{F}$is verified.

The equation of the dot product curl of the vector potential can be expressed as:

role="math" localid="1657614217314" $$\begin{gathered} ∇×\mathrm{A}=\frac{∂{\mathrm{A}}_{\mathrm{x}}}{∂\mathrm{x}}+\frac{∂{\mathrm{A}}_{\mathrm{y}}}{∂\mathrm{y}}+\frac{∂{\mathrm{A}}_{\mathrm{z}}}{∂\mathrm{z}} \\ =0+{∫}_0^{\mathrm{x}}\frac{∂{\mathrm{F}}_{\mathrm{z}}({\mathrm{x}}^{\text{'}},\mathrm{y},\mathrm{z})}{∂\mathrm{y}}{\mathrm{dx}}^{\text{'}}+{∫}_0^{\mathrm{x}}\frac{∂{\mathrm{F}}_{\mathrm{x}}({\mathrm{x}}^{\text{'}},\mathrm{y},\mathrm{z})}{∂\mathrm{z}}{\mathrm{dy}}^{\text{'}}-{∫}_0^{\mathrm{x}}\frac{∂{\mathrm{F}}_{\mathrm{y}}({\mathrm{x}}^{\text{'}},\mathrm{y},\mathrm{z})}{∂\mathrm{z}}\mathrm{dx}\text{'} \\ ≠0 \end{gathered}$$

Thus, $∇×A=F$is verified and A is divergenceless.

The equation of the vector potential in the y direction is expressed as:

${\mathrm{A}}_{\mathrm{y}}={∫}_0^{\mathrm{x}}{\mathrm{x}}^{\text{'}}{\mathrm{dx}}^{\text{'}}$

Hence, the value of the vector potential is:

$A_y=\frac{x^2}{2}$

The equation of the vector potential in the z direction is expressed as:

$A_z={∫}_0^x{y}^{\text{'}}d{y}^{\text{'}}-{∫}_0^{x}zd{x}^{\text{'}}$

Hence, the value of the vector potential is:

$A_z=\frac{y^2}{2}-zx$

From the above equations, the value of the vector potential can be calculated as:

$A=\frac{x^2}{2}\hat{y}+\left(\frac{y^2}{2}-zx\right)\hat{z}$

The value of the cross product of the curl and the vector potential is expressed as:

$$\begin{gathered} ∇×\mathrm{A}=\left|\begin{array}{ccc}\hat{\mathrm{x}}& \hat{\mathrm{y}}& \hat{\mathrm{z}}\\ ∂\mathrm{l}∂\mathrm{x}& ∂\mathrm{l}∂\mathrm{y}& ∂\mathrm{l}∂\mathrm{z}\\ 0& {\mathrm{x}}^2/2& \left({\mathrm{y}}^2/2-\mathrm{zx}\right)\end{array}\right| \\ =\mathrm{y}\hat{\mathrm{x}}+\mathrm{z}\hat{\mathrm{y}}+\mathrm{x}\hat{\mathrm{z}} \\ =\mathrm{F} \end{gathered}$$

Thus, the value of A isrole="math" localid="1657542045516" $\frac{x^2}{2}\hat{y}+\left(\frac{y^2}{2}-zx\right)\hat{z}$and $∇×A=F$is proved.