# Magnetic field from a moving charge

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A moving charge with uniform velocity creates magnetic field. Here we present a way to derive the magnetic field for a point charge moving at an uniform velocity. Some authors refer to this equation as Biot-Savart law for a moving charge . The name "Biot-Savart" is misleading because we base our derivation solely on the change of flux of the electric field. Biot-Savart law is derived from a completely different point, from the point of an infinitesimal current elements.

## Maxwell's equations

We start with the with the Maxwell's addition in integral form :

\begin{equation} \oint_{\partial \Sigma} \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 \left(\iint_{\Sigma} \mathbf{J} \cdot \mathrm{d}\mathbf{S} + \varepsilon_0 \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{E} \cdot \mathrm{d}\mathbf{S} \right). \end{equation}

We apply the Ampere's law to a surface shown on the picture. The surface is a part of a sphere of radius $r$. Also, $r$ is the distance between the point charge and the point $P$ at which we want to calculate the magnetic field. The electric field is uniform on the surface of a sphere since the Gauss law is satisfied at any time. The flux of the electric field through the surface of the whole sphere of radius $r$ is:

\begin{equation} \varepsilon_0 \iint_{\Sigma} \mathbf{E} \cdot \mathrm{d}\mathbf{S} = \iiint_\Omega \rho \,\mathrm{d}V = q. \end{equation}

The electric field is normal on the surface representing part of the sphere, so we can write for the intensity:

\begin{equation} E = \frac{q}{\varepsilon_0 4 \pi r^2}. \end{equation}

Area of the part of the sphere can be calculated by integrating rings of infinitesimal width $r\mathrm{d}\varphi$ (blue ring on the picture). One such ring at position defined by the angle $\varphi$, has an area of $2r\sin{\varphi}\pi\cdot r\mathrm{d}\varphi$. Integrating from zero to the angle $\Phi$ we get the area of the surface that is a part of a sphere defined by the length $h$ from the charge to the point $R$:

\begin{equation} A = \int_0^{\Phi} 2 r^2 \pi \sin{\varphi} \mathrm{d} \varphi = 2 r^2 \pi \left( \cos{0} - \cos{\Phi} \right) = 2 r^2 \pi \left( 1 - \frac{h}{r} \right). \end{equation}

Since there is no current puncturing the surface, we can write for the magnetic field from the first equation:

\begin{equation} \oint_{\partial \Sigma} \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \mu_0 \varepsilon_0 \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{E} \cdot \mathrm{d}\mathbf{S}. \end{equation}

We implement the Gauss law and the colinearity of the vector $\mathbf{E}$ to a surface element $\mathrm{d}\mathbf{S}$ at every point:

\begin{equation} \mu_0 \varepsilon_0 \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{E} \cdot \mathrm{d}\mathbf{S} = \mu_0 \varepsilon_0 \frac{\mathrm{d}}{\mathrm{d}t} \left[ \frac{q}{\varepsilon_0 4 \pi r^2} \cdot 2 r^2 \pi \left( 1 - \frac{h}{r} \right) \right] = \frac{\mu_0 q}{2} \left( \frac{h}{r^2}\frac{\partial r}{\partial h} - \frac{1}{r}\right)\frac{\mathrm{d}h}{\mathrm{d}t}. \end{equation}

The magnetic field is uniform and tangent on the border of the surface, so we can write:

\begin{equation} \oint_{\partial \Sigma} \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = B \cdot 2 b \pi. \end{equation}

Change in flux happens due to the movement of the charge. After some infinitesimal period $\mathrm{d}t$ the parameters are $r'=\sqrt{b^2 + {h'}^2}$ and $h'=h-v\mathrm{d}t$. We get:

\begin{align} & \frac{\mathrm{d}h}{\mathrm{d}t} = -v, \\ \\ & \frac{\partial r}{\partial h} = \frac{h}{r}. \end{align}

When we plug the last three equations into Ampere's law we get:

\begin{align} B \cdot 2 b \pi = -\frac{\mu_0 q}{2} \left(\frac{h^2}{r^3} - \frac{1}{r}\right)v = \frac{\mu_0 q v}{2r^3}\left( r^2-h^2\right). \\ \end{align}

From the geometry we have $b^2 = r^2 - h^2$ and $b = r\sin{\Phi}$, which finally boils down to the famous equation (similar to Biot-Savart law), for a magnetic field of a point charge moving at an uniform velocity:

\begin{align} B = \mu_0\frac{q v \sin{\Phi}}{4 \pi r^2}. \\ \end{align}

Which can be written in a vector form as:

\begin{align} \mathbf{B} = \mu_0\frac{q \mathbf{v} \times \mathbf{r}}{4 \pi \mathbf{r}^3}. \\ \end{align}