electromagnetisminertial-framesmagnetic fieldsspecial-relativity
My question is why exactly the moving charge or a current carrying wire creates magnetic field around themselves?
Some of them call it as a consequence of special theory of relativity some of them say it's because of maxwell's probe etc.
And how to prove the equation used to find magnetic field given by biot-savart's law.
I have a simple argument in mind. for an infinitely long wire carrying current in the +X direction, the magnetic field at a given distance r from the wire, has 2 components(suppose) radial and tangential. Consider specifically the radial component and assume it points outwards(or inwards). Now, switch the current direction to -X. The radial component should also switch directions. But notice that this current is no different from what it was before, if you just turn the wire around(i.e) watch the wire from the other side, the situation is exactly similar to what it was before, yet the radial field direction has switched. This is an impossibility, proving that the radial field must be zero, leaving behind only circles, as the possible field shape.
EDIT: CASE 1: from side A you see the current moving to the right. and the magnetic field is (suppose) radially outward. Now you reflect yourself about the wire and reach side B. You see the current moving to the left now, but the situation is essentially same(just observing from 2 sides of a wire), so the radial field is still outward.
CASE 2:you return to side A, and now reverse the current(it now travels left), so the radial component must switch(inward). But havent you already seen this current? A current flowing left was encountered in CASE 1 from side B!, where you observed the radial field to be essentially outward. But now you observe it to be inward. CONTRADICTION. thus no radial field can exist.
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}
Best Answer
Possible duplicate. Look at the answer to this post. It's a shorter version of what I describe below. I'm going to answer the part regarding the Biot-Savart law. The derivation may be found in an advanced electromagnetics book such as Balanis's. Starting with the Maxwell's equations: $$\nabla\times \vec E=-\frac{\partial \vec B}{\partial t}$$ $$\nabla\times \vec H=\vec J+\frac{\partial \vec D}{\partial t}$$ $$\nabla\cdot \vec B=0$$ $$\nabla\cdot \vec D=\rho_{\rm free}$$ to define the magnetic vector potential $\vec{A}$ using the fact that since $\nabla\cdot \vec B=0$ we can define $\vec B=\nabla\times \vec A$ such that $$\nabla\times \vec E=-\frac{\partial}{\partial t} \nabla\times \vec A$$ and therefore, $$\nabla\times (\vec E+\frac{\partial\vec A}{\partial t})= 0$$ Since $\nabla\times\nabla f=0$ for any differentiable scalar $f$, we have $$\vec E+\frac{\partial\vec A}{\partial t}=-\nabla\Phi_{\rm e}\rightarrow \vec E=-\frac{\partial\vec A}{\partial t}-\nabla\Phi_{\rm e}$$ where $\Phi_{\rm e}$ is a scalar electric potential. Note that in free space $$\frac{\partial \vec D}{\partial t}=\epsilon_{0}\frac{\partial \vec E}{\partial t}=-\epsilon_{0}(\frac{\partial^{2}\vec A}{\partial t^{2}}+\nabla\frac{\partial}{\partial t}\Phi_{\rm e})$$
Using for instance Coulumb Gauge conditions we have $$\nabla\cdot\vec E=\frac{1}{\epsilon_{0}}\nabla\cdot\vec D=\frac{\rho_{\rm free}}{\epsilon_{0}}=-\frac{\partial}{\partial t}\nabla\cdot\vec A-\nabla^{2}\Phi_{\rm e} \rightarrow \nabla^{2}\Phi_{\rm e}=-\frac{\rho_{\rm free}}{\epsilon_{0}}$$ where we have used $\nabla\cdot \vec A=0$ and $\nabla\cdot\nabla f = \nabla^{2}f$. Similarly calculating $\nabla\times \vec B$ we have $$\nabla\times \vec B=\nabla\times \nabla\times \vec A=\nabla(\nabla\cdot\vec A)-\nabla^{2}\vec A=\mu_{0}\nabla\times \vec H=\mu_{0}(\vec J-\epsilon_{0}(\frac{\partial^{2}\vec A}{\partial t^{2}}+\nabla\frac{\partial}{\partial t}\Phi_{\rm e}))$$ Thus $$-\mu_{0}\vec J=\nabla^{2}\vec A-\frac{1}{c^{2}_{0}}(\frac{\partial^{2}\vec A}{\partial t^{2}}+\frac{\partial}{\partial t}\nabla\Phi_{\rm e})$$ where $c_{0}=\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}$ is the speed of light. Now in the static case $\frac{\partial}{\partial t}=0$ (and similar to the Poisson's equation) $$\nabla^{2}\vec A=-\mu_{0}\vec J$$ has the solution $$\vec A(\vec r)=\frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec J(\vec r\,')}{|\vec r - \vec r\,'|}dV'$$ Take the curl of the above equation (with respect to the observation coordinates, i.e., unprimed coordinates) and you will have the Biot-Savart law. Note the magnetostatic condition that I mentioned in the description!