# Why Moving charge and a current carrying wire creates magnetic fields?

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.

• "Why" in physics is always non sencicle. We can only describe nature how we see it. We go by experimentation. Search up amperes experiments. Here we can experimentally derive amperes law. And then mathematically with divergence of B we can then "derive " biot savart" Commented Jan 10, 2022 at 23:25
• Suggest you to read the answers here: physics.stackexchange.com/questions/65335/… Commented Jan 11, 2022 at 7:50
• Is there a magnetic field around an electron beam inside a vacuum tube? Commented Jan 13, 2022 at 5:07
• @jensenpaull Not quiet. There are many why questions that can be answered like why does light follow the law of reflection. It can be proved from Huygens and Fermat's principle. What would you say about this? Commented Jan 16, 2022 at 10:53
• It can also be proved more fundamentally from maxwells equations and the EM boundary conditions along with the fact that the principle of superposition applies / freznel equations. All of which are experimentally derived. Commented Jan 16, 2022 at 11:22

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!