Biot-Savart law

The Biot-Savart law is an equation describing the magnetic field generated by a current segment.

Consider a point $\small P$ in the vicinity of a wire carrying a current $\small I$ (see diagram below). The magnitude of $\small I$ represents the rate at which electric charge flows through a cross-section of the wire. For a steady current, a larger current produces a proportionally larger magnetic field. Suppose an infinitesimal segment of the wire is represented by the vector $\small d\vec{l}$ . The contribution of this current element to the magnetic field at $\small P$ is proportional to $\small Id\vec{l}$ . Furthermore, this contribution depends on the orientation of the current element relative to the position vector $\vec{R}$ , which points from the current element towards $\small P$ . Specifically, only the component of $\small d\vec{l}$ perpendicular to $\vec{R}$ contributes to the magnitude of the magnetic field. This perpendicular component is $dlsin\theta$ , where $\theta$ is the angle between $\small d\vec{l}$ and $\vec{R}$ . Hence,

$dB\propto Idlsin\theta$

Additionally, empirical observations show that the magnitude of the magnetic field contribution decreases in proportion to the square of the distance between the current element and $\small P$ . Therefore,

$dB\propto\frac{Idlsin\theta}{R^2}$

The magnetic field also depends on the permeability of the surrounding medium. Assuming that the system is in a vacuum, the experimentally established proportionality constant is $\frac{\mu_0}{4\pi}$ , where $\mu_0$ is the vacuum permeability. Thus,

$dB=\frac{\mu_0Idlsin\theta}{4\pi R^2}\;\;\;\;\;\;\;\;150$

To express this relationship in vector form, we use the cross product, which accounts for both the angular dependence and the direction of the magnetic field:

$d\vec{B}=\frac{\mu_0Id\vec{l}\times\vec{R}}{4\pi R^3}\;\;\;\;\;\;\;\;151$

Therefore, the total field at $\small P$ is

$\vec{B}=\frac{\mu_0I}{4\pi}\int_{wire}\frac{d\vec{l}\times\vec{R}}{R^{3}}\; \; \; \; \; \; \; \; 152$

Eq150, eq151 and eq152 are different forms of the Biot-Savart law.

If the wire is a circle, the direction of $\small d\vec{B}$ at $\small P$ is determined by the right hand rule and is perpendicular to the plane formed by $\small d\vec{l}$ and $\small \vec{R}$ , where the angle $\small \theta$ between $\small d\vec{l}$ and $\small \vec{R}$ is $\small 90^{\circ}$ (see diagram below).

From eq150,

$dB=\frac{\mu_0Idlsin90^{\circ}}{4\pi R^{2}}=\frac{\mu_0Idl}{4\pi R^{2}}\; \; \; \; \; \; \; \; 153$

Let’s analyse the components of $\small d\vec{B}$ . If we sum all the vectors of $\small d\vec{B}_\perp$ at $\small P$ from all current elements around the wire, they cancel out. The vectors of $\small d\vec{B}_\parallel$ at $\small P$ , however adds. Therefore, in scalar form, $\small B=\int dB_\parallel$ , where $\small dB_\parallel=dBcos\alpha$ , which when combined with eq153 gives

$dB_\parallel=\frac{\mu_0Icos\alpha}{4\pi R^{2}}dl\; \; \; \; \; \; \; \; 154$

Since $\small R=\sqrt{r^{2}+z^{2}}$ and $\small cos\alpha=\frac{r}{\sqrt{r^{2}+z^{2}}}$ , eq154 becomes $\small dB_\parallel=\frac{\mu_0Ir}{4\pi(r^{2}+z^{2})^{3/2}}dl$ and

$B=\int dB_\parallel=\frac{\mu_0Ir}{4\pi(r^{2}+z^{2})^{3/2}}\int dl$

Substituting $\small \int dl=2\pi r$ into the above equation gives

$B=\frac{\mu_0Ir^{2}}{2(r^{2}+z^{2})^{3/2}}\; \; \; \; \; \; \; \; 155$

For the case of $\small B$ at point $\small O$ , we have $\small z$ , which is the length $\small OP$ , equals to zero. So,

$B=\frac{\mu_0I}{2r}\; \; \; \; \; \; \; \; 156$

For the case of $\small z\gg r$ , eq155 becomes $\small B=\frac{\mu_0IA}{2\pi z^{3}}$ , where $\small A=\pi r^{2}$ . Substituting eq60 and eq61 into $\small B$ yields,