2.27. Check whether the field $\dot{A}$ is solenoidal $(\operatorname{div} \dot{A}=0)$. If it is, find its vector potential:

a) $\dot{A}=(y+z) \dot{i}+(x+z) \vec{j}+(x+y) \vec{k}$;

b) $\dot{A}=y e^{x^{2}} \dot{i}+2 y z \dot{j}-\left(2 x z y e^{x^{2}}+z^{2}\right) \ddot{k}$.
Solution.

a) $\operatorname{div} \ddot{A}=\frac{\partial(y+z)}{\partial x}+\frac{\partial(x+z)}{\partial y}+\frac{\partial(x+y)}{\partial z}=0$.

Therefore, the field $\vec{A}$ is solenoidal. Let's find its vector potential $\vec{B}$. As is known ( $\$ 2.4$ ), the vector potential is defined up to an additive term $\operatorname{grad} \varphi$, where $\varphi(x, y, z)$ is an arbitrary differentiable function. Therefore, we will seek the simplest expression for the vector potential $\vec{B}\left(B_{1}, B_{2}, B_{3}\right)$. In particular, let $B_{3}=0$. The components $B_{1}$ and $B_{2}$ will be determined from the equation

$$
\begin{aligned}
& \operatorname{rot} \vec{B}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
B_{1} & B_{2} & B_{3}
\end{array}\right|=-\frac{\partial B_{2}}{\partial z} \vec{i}+\frac{\partial B_{1}}{\partial z} \dot{j}+\left(\frac{\partial B_{2}}{\partial x}-\frac{\partial B_{1}}{\partial y}\right) \vec{k}= \\
& =(y+z) \vec{i}+(x+z) \vec{j}+(x+y) \vec{k}
\end{aligned}
$$

From this, we obtain:

$$
\frac{\partial B_{2}}{\partial z}=-y-z \Rightarrow B_{2}=\int(-y-z) d z+f(x, y)
$$

where $f(x, y)$ is an arbitrary differentiable function;

$$
\frac{\partial B_{1}}{\partial z}=x+z \Rightarrow B_{1}=\int(x+z) d z+g(x, y)
$$

where $g(x, y)$ is an arbitrary differentiable function.

After integration, we determine $B_{1}$ and $B_{2}$ :

$$
\begin{aligned}
& B_{1}=x z+\frac{z^{2}}{2}+g(x, y) \\
& B_{2}=-y z+\frac{z^{2}}{2}+f(x, y)
\end{aligned}
$$

To determine the functions $g$ and $f$, recall that there is still one unsatisfied condition:

$$
\frac{\hat{\partial} B_{2}}{\partial x}-\frac{\hat{c} B_{1}}{\partial y}=x+y \Rightarrow \frac{\partial}{\partial x}\left(-y z+\frac{z^{2}}{2}\right)-\frac{\hat{c}}{\partial y}\left(x z+\frac{z^{2}}{2}\right)+\frac{\hat{\partial} g}{\partial x}-\frac{\hat{c} f}{\partial y}=x+y
$$

We are interested in the simplest partial expression for the functions $g$ and $f$. For example, we can set

$$
g=\frac{x^{2}}{2} \text { and } f=-\frac{y^{2}}{2},
$$

then $\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}=x+y$. Thus, the sought vector potential $\vec{B}$ has the form:

$$
\vec{B}=\left(x z+\frac{z^{2}}{2}+\frac{x^{2}}{2},-y z-\frac{z^{2}}{2}-\frac{y^{2}}{2}, 0\right)+\operatorname{grad} \varphi
$$

b) Let's calculate $\operatorname{div} \vec{A}$ :

$$
\begin{aligned}
\operatorname{div} \vec{A} & =\frac{\partial\left(y e^{x^{2}}\right)}{\partial x}+\frac{\partial(2 y z)}{\partial y}-\frac{\partial\left(2 x z y e^{x^{2}}+z^{2}\right)}{\partial z}= \\
& =2 x y e^{x^{2}}+2 z-2 z-2 x y e^{x^{2}}=0
\end{aligned}
$$

The field $\vec{A}$ is solenoidal. The vector potential $\vec{B}$ will be sought in the same way as in case a):

$$
\begin{aligned}
& \operatorname{rot} \dot{B}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
B_{1} & B_{2} & 0
\end{array}\right|=-\frac{\partial B_{2}}{\partial z} \vec{i}+\frac{\partial B_{1}}{\partial z} \vec{j}+\left(\frac{\partial B_{2}}{\partial x}-\frac{\partial B_{1}}{\partial y}\right) \vec{k}= \\
& =y e^{x^{2}} \vec{i}+2 y z \vec{j}-+\left(2 x y z e^{x^{2}}+z^{2}\right) \dot{k} .
\end{aligned}
$$

$$
\begin{aligned}
& B_{2}=-\int y e^{x^{2}} d z+f(x, y)=-y z e^{x^{2}}+f(x, y) \\
& B_{1}=\int 2 y z d z+g(x, y)=y z^{2}+g(x, y) \\
& \frac{\partial B_{2}}{\partial x}-\frac{\partial B_{1}}{\partial y}=\frac{\hat{c}}{\partial x}\left(-y z e^{x^{2}}+f\right)-\frac{\hat{c}}{\partial y}\left(y z^{2}+g\right)=-2 x y z e^{x^{2}}-z^{2}+f_{x}^{\prime}-g_{y}^{\prime} .
\end{aligned}
$$

The simplest choice for the functions $f$ and $g$ is zero, i.e., $f=g=0$. Finally, we obtain:

$$
\vec{B}=\left(y z^{2},-y z e^{x^{2}}, 0\right)+\operatorname{grad} \varphi
$$

where $\varphi=\varphi(x, y, z)$ is an arbitrary differentiable scalar field.