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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.

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  • $\begingroup$ use $idl = qdv$ $\endgroup$ Dec 30 '19 at 14:34
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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}

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  • $\begingroup$ how did you get equation 02 when v << c. $\endgroup$
    – DHYEY
    Jun 29 '18 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
    Dec 30 '19 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? $\endgroup$ Nov 29 '20 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
    Nov 29 '20 at 5:25
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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

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You can use the Biot-Savart formula to find magnetic field, 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, including 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 at any instant can be expressed as gradient of a potential function (it is a $\gamma$ - contracted Coulomb field). So you can use the Biot-Savart formula.

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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

comments:

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

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

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