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.

  • $\begingroup$ use $idl = qdv$ $\endgroup$ Commented Dec 30, 2019 at 14:34

4 Answers 4


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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

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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}

  • 1
    $\begingroup$ how did you get equation 02 when v << c. $\endgroup$
    – DHYEY
    Commented Jun 29, 2018 at 11:49
  • $\begingroup$ 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 $\endgroup$
    – Kyle Kanos
    Commented Dec 30, 2019 at 16:24
  • $\begingroup$ 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? $\endgroup$
    – Ashpect
    Commented Nov 29, 2020 at 4:50
  • $\begingroup$ @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. $\endgroup$
    – Frobenius
    Commented 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


two vectors u and t and θ angle between them, |uXt| = |u|*|t|*sin(θ)

when we have a function the equation dB = Constant*dq when you use integration you can get B=C *integral dq = C *q

  • $\begingroup$ Welcome to PSE. Your answer is not readable. I suggest you to learn editing your equations with MathJax. $\endgroup$
    – Frobenius
    Commented Aug 19, 2021 at 10:53

You can use the Biot-Savart formula to find magnetic field with very good approximation, if electric field everywhere is a potential field (it can be expressed as gradient of a potential function). This is because the Biot-Savart field obeys the Ampere-Maxwell law with displacement current due to that potential electric field. In other words, for systems where electric field is a potential field, correction to Biot-Savart magnetic field due to displacement current is zero.

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.

  • $\begingroup$ This answer is incorrect, simply compare lienard wichert potentials and the biotsavart equivelent to a point charge. At any instant, there is an associated de/dt assigned to a point in space. $\endgroup$ Commented Mar 19, 2022 at 0:51
  • $\begingroup$ @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. $\endgroup$ Commented Mar 19, 2022 at 1:53

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