Example 3. An electric current of strength $I$ flows from bottom to top along an infinite wire coinciding with the $O z$ axis. Find the vector $\mathbf{H}$ of the magnetic field strength created by this current at an arbitrary point $M(x, y, z)$ in space (Fig. 6).
Solution. Consider a sufficiently small element $P P_{1}=d \zeta$ of the $O z$ axis. According to the Biot-Savart law, the magnetic field intensity $d \mathbf{H}$ created at point $M$ by the current flowing through the wire element $d \zeta$ is in the direction of the vector product $\left[d \zeta, \mathbf{r}_{1}\right]$, where $d \zeta=\overrightarrow{P P_{1}}$, $|d \zeta|=d \zeta$, $\mathbf{r}_{1}=\overrightarrow{P M}$. According to the same law, the magnitude of the vector $d \mathbf{H}$ is

$$
|d \mathbf{H}|=\frac{I}{r_{1}^{2}} \sin \left(\overline{d \zeta, \mathbf{r}_{1}}\right) d \zeta
$$

where $\left(\widehat{d \zeta, \mathbf{r}_{1}}\right)$ is the angle formed by the vectors $d \zeta$ and $\mathbf{r}_{1}$. Since

$$
\left|\left[d \zeta, \mathbf{r}_{1}\right]\right|=r_{1} d \zeta \sin \left(\widehat{d \zeta, \mathbf{r}_{1}}\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-06.jpg?height=605&width=543&top_left_y=979&top_left_x=2787)

Figure 6

$$
d \mathbf{H}=\frac{I}{r_{1}^{3}}\left[d \zeta, \mathbf{r}_{1}\right]
$$

To obtain the desired vector $\mathbf{H}$ at point $M$, we need to sum all the vectors $d \mathbf{H}$ corresponding to different elements $P P_{1}$ of the wire, i.e., integrate expression (4) over the entire $O z$ axis:

$$
\mathbf{H}=\int_{-x}^{4 x} \frac{I}{r_{1}^{3}}\left[d \zeta, \mathbf{r}_{1}\right]
$$

We have

$$
\mathbf{r}_{1}=\overrightarrow{O M}-\overrightarrow{O P}
$$

However,

so

$$
\begin{gathered}
\overrightarrow{O M}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}, \quad \overrightarrow{O P}=\zeta \mathbf{k} \\
\mathbf{r}_{1}=x \mathbf{i}+y \mathbf{j}+(z-\zeta) \mathbf{k}
\end{gathered}
$$

thus

$$
r_{1}=\left|\mathbf{r}_{1}\right|=\sqrt{x^{2}+y^{2}+(z-\zeta)^{2}}=\sqrt{\rho^{2}+(z-\zeta)^{2}}
$$

where $\rho=\sqrt{x^{2}+y^{2}}$ is the distance from point $M$ to the axis of the wire.

For the vector product $\left[d \zeta, \mathbf{r}_{1}\right]$, we have

$$
\left[d \zeta, \mathbf{r}_{1}\right]=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 0 & d \zeta \\
x & y & z-\zeta
\end{array}\right|=-\mathbf{i} y d \zeta+\mathbf{j} x d \zeta
$$

and formula (5) becomes (point $M(x, y, z)$ is fixed, $I=$ const)

$$
\mathbf{H}=I(-y \mathbf{i}+x \mathbf{j}) \int_{-\infty}^{+\infty} \frac{d \zeta}{\left[\rho^{2}+(z-\zeta)^{2}\right]^{3 / 2}}
$$

To compute the integral on the right-hand side of (6), we make the substitution

$$
\zeta-z=\rho \operatorname{tg} t, \quad d \zeta=\frac{\rho d t}{\cos ^{2} t}
$$

We will have

$$
\int_{-\infty}^{+\infty} \frac{d \zeta}{\left[\rho^{2}+(z-\zeta)^{2}\right]^{3 / 2}}=\int_{-\pi / 2}^{\pi / 2} \frac{\rho d t}{\cos ^{2} t\left[\rho^{2}+(z-\zeta)^{2}\right]^{3 / 2}}=\frac{1}{\rho^{2}} \int_{-\pi / 2}^{\pi / 2} \cos t d t=\frac{2}{\rho^{2}}
$$

Thus, the vector of magnetic field intensity $\mathbf{H}$ in our case is given by the formula

$$
\mathbf{H}=\frac{2 I}{\rho^{2}}(-y \mathbf{i}+x \mathbf{j}), \quad \text { or } \quad \mathbf{H}=\frac{2}{\rho^{2}}[\mathbf{I}, \mathbf{r}]
$$

where $\mathbf{I}=\mathbf{I} \cdot \mathbf{k}$ is the current vector, $\mathbf{r}$ is the radius vector of point $M(x, y, z)$ in the field, and $\rho$ is the distance from point $M$ to the axis of the wire.