# Magnetic field due to a single moving charge

The Biot-Savart law can only be used in the case of magnetostatics (constant current) so how do we calculate the magnetic field of a single charge moving at constant velocity at a distance r. I tried by calculating the displacement current using but i was not sure wether the biot savart law can be applied to displacement currents. Please don't use relativity if possible because i have no experience with relativity yet.

• use $idl = qdv$ Dec 30, 2019 at 14:34 A point charge $$\:q\:$$ is moving uniformly on a straight line with velocity $$\:\boldsymbol{\upsilon}\:$$ as is the Figure. The electromagnetic field at a point $$\:\mathrm{P}\:$$ with position vector $$\:\mathbf{x}\:$$ at time $$\:t\:$$ is

\begin{align} \mathbf{E}_{_{\mathbf{LW}}}\left(\mathbf{x},t\right) & \boldsymbol{=}\dfrac{q}{4\pi \epsilon_{\bf 0}}\dfrac{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\right)}{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\sin^{\bf 2}\!\phi\right)^{\boldsymbol{3/2}}}\dfrac{\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{\bf 3}},\quad \beta\boldsymbol{=}\dfrac{\upsilon}{c} \tag{01a}\\ \mathbf{B}_{_{\mathbf{LW}}}\left(\mathbf{x},t\right) & \boldsymbol{=}\dfrac{1}{c^{ \bf 2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\right)\vphantom{\dfrac{a}{\dfrac{}{}b}}\boldsymbol{=}\dfrac{\mu_{0}q}{4\pi }\dfrac{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\right)}{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\sin^{\bf 2}\!\phi\right)^{\boldsymbol{3/2}}}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{\bf 3}} \tag{01b} \end{align} Equations (01) are relativistic. They come from the Lienard-Wiechert potentials.

Biot-Savart Law

After a quick calculation with Biot-Savart Law (using the Dirac $$\:\delta\:$$ function) I found the solution $$\begin{equation} \mathbf{B}_{_{\mathbf{BS}}}\left(\mathbf{x},t\right) \boldsymbol{=}\dfrac{\mu_{0}q}{4\pi }\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{\bf 3}} \tag{02} \end{equation}$$ which compared with that from the Lienard-Wiechert potentials, see above equation (01b) $$\begin{equation} \mathbf{B}_{_{\mathbf{LW}}}\left(\mathbf{x},t\right)\boldsymbol{=}\dfrac{\mu_{0}q}{4\pi }\dfrac{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\right)}{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\sin^{\bf 2}\!\phi\right)^{\boldsymbol{3/2}}}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{\bf 3}} \tag{03} \end{equation}$$ it looks as an approximation for charges whose velocities are small compared to that of light $$\:c$$ $$\begin{equation} \mathbf{B}_{_{\mathbf{BS}}}\left(\mathbf{x},t\right)\boldsymbol{=} \lim_{\beta \boldsymbol{\rightarrow} 0}\mathbf{B}_{_{\mathbf{LW}}}\left(\mathbf{x},t\right)\boldsymbol{=} \lim_{\beta\boldsymbol{\rightarrow} 0}\left[\dfrac{\mu_{0}q}{4\pi }\dfrac{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\right)}{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\sin^{\bf 2}\!\phi\right)^{\boldsymbol{3/2}}}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{\bf 3}}\right]\boldsymbol{=}\dfrac{\mu_{0}q}{4\pi}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{3}} \tag{04} \end{equation}$$

(1) EDIT Answer to OP's comment :

how did you get equation 02 when v << c. – DHYEY Jun 29 '18 at 11:49 From Jackson's : Biot and Savart Law $$\begin{equation} \mathrm d\mathbf{B}=\dfrac{\mu_{0}}{4\pi}I\dfrac{\left(\mathrm d\boldsymbol{\ell}\boldsymbol{\times}\mathbf{r'}\right)}{\:\:\Vert\mathbf{r'}\Vert^{3}} \tag{BS-01} \end{equation}$$ $$\begin{equation} I=q\upsilon\delta\left(x'-r\cos\phi\right), \qquad \mathrm d\boldsymbol{\ell}=\mathbf{i}\mathrm dx', \qquad \mathbf{r'}=x'\mathbf{i}\boldsymbol{+}\alpha\mathbf{j}\boldsymbol{+}0\mathbf{k} \tag{BS-02} \end{equation}$$ $$\begin{equation} \mathrm d\mathbf{B}=\dfrac{\mu_{0}q}{4\pi}q\upsilon\delta\left(x'\!\boldsymbol{-}\!r\cos\phi\right)\dfrac{\left(\mathbf{i}\boldsymbol{\times}\mathbf{r'}\right)}{\:\:\Vert\mathbf{r'}\Vert^{3}}\mathrm dx'=\dfrac{\mu_{0}q}{4\pi}q\upsilon\delta\left(x'\!\boldsymbol{-}\!r\cos\phi\right)\dfrac{\left(\alpha\mathbf{k}\right)}{\:\:\left(x'^2\!\boldsymbol{+}\!\alpha^2 \right)^{3/2}}\mathrm dx' \tag{BS-03} \end{equation}$$ $$\begin{equation} \mathbf{B}=\dfrac{\mu_{0}}{4\pi}q\upsilon\alpha\mathbf{k}\int\limits_{\boldsymbol{-}\boldsymbol{\infty}}^{\boldsymbol{+}\boldsymbol{\infty}}\dfrac{\delta\left(x'\!\boldsymbol{-}\!r\cos\phi\right)}{\:\:\left(x'^2\!\boldsymbol{+}\!\alpha^2 \right)^{3/2}}\mathrm dx'=\dfrac{\mu_{0}q}{4\pi}\dfrac{\upsilon\alpha\mathbf{k}}{\:\:\left(r^2\cos^2\phi\!\boldsymbol{+}\!\alpha^2 \right)^{3/2}}= \dfrac{\mu_{0}q}{4\pi}\dfrac{\left(\upsilon\mathbf{i}\right)\boldsymbol{\times}\left(\alpha\mathbf{j}\right)}{\:\:\left(r^2\cos^2\phi\!\boldsymbol{+}\!\alpha^2 \right)^{3/2}} \tag{BS-04} \end{equation}$$ $$\begin{equation} \mathbf{B} =\dfrac{\mu_{0}q}{4\pi }\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{3}} \tag{BS-05} \end{equation}$$

• how did you get equation 02 when v << c. Jun 29, 2018 at 11:49
• Why on Earth would you use bold symbol for things like the equals sign and cross product? And mix in the deprecated \bf with \mathbf? That's just weird Dec 30, 2019 at 16:24
• but finding magnetic field due to single moving charge using biot savart law(which is applicable for curent) should be wrong since its the avg current ehich we can determine for a single charge ? where is the argument wrong? Nov 29, 2020 at 4:50
• @Ashish Kumar : In "Jackson's : Biot and Savart Law" read the paragraph beginning with "A word of caution about (5.4)...". Equation (5.4) therein is equation (BS-01) in my answer. Nov 29, 2020 at 5:25

Actually there is only a slight modification in the form of the potential otherwise its the same.

The magnetic vector potential gets modified to $$\mathbf{A}(\mathbf{r}, t) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}', t_r')}{|\mathbf{r} - \mathbf{r}'|} d^3\mathbf{r}$$

where $$t_r' = t - \frac{1}{c}|\mathbf{r} - \mathbf{r}'|$$ is the retarded time.

There is no relativity in here but causality is only taken into account.

Calculating the magnetic fields is a bit more difficult you can reffer to Lienard Wiechert Potential

dB = (µ/4π)(i/r²)(dlXr'), dl and r' are vectors, |r'| = 1 (r with ^)

i = dq/dt

dl/dt = v

dB = (µ/4π)(dq/r²)(vXr'), v is vector and |v| = v

B = (µ/4π)(q/r²)(vXr') = (µ/4π)*(q/r²)vsin(θ)

B = µqvsin(θ)/4πr²

A charged particle moving with constant velocity has electric field that moves in space but if the speed is much lower than speed of light, at any instant electric field can be expressed as gradient of a potential function (giving a $$\gamma$$ - contracted Coulomb field). So you can use the Biot-Savart formula if the charge speed is low enough.
• @jensenpaull Yes the Biot-Savart formula isn't exactly solution of Maxwell's equation. But for constant (no acceleration thus no radiation) and low (compared to $c$) velocity it is most often good enough. This is because for such motion electric field is almost a gradient of a scalar function and its contribution to magnetic field via "displacement current magnetic field" is zero. Mar 19, 2022 at 1:53