4.2 Ampere's law

Ampere's law (with displacement current)

For a steady current flowing through a straight wire, the magnetic field at a point at a perpendicular distance $r$ from the wire, has a value $$B=\frac{{\mu }_{0}I}{2\pi r}$$ If we integrate around the wire in a circle, then clearly we get $${\displaystyle \begin{array}{c}\oint \vec{B}\cdot d\vec{l}=\frac{{\mu }_{0}I}{2\pi r}\oint dl\\ =\frac{{\mu }_{0}I}{2\pi r}2\pi r\\ ={\mu }_{0}I\end{array}}$$ This is true for irregular paths around the wire

$$\oint \vec{B}\cdot d\vec{l}=\oint B\cdot dl\cos \theta$$ but for small $dl$ $dl\cos \theta =rd\phi$ $${\displaystyle \begin{array}{c}\oint \vec{B}\cdot d\vec{l}=\oint Brd\phi \\ =\frac{{\mu }_{0}I}{2\pi }\oint d\phi \\ ={\mu }_{0}I\end{array}}$$ In fact instead of current we use the surface integral of the current density $J$ , which is the current per unit area $$\oint \vec{B}\cdot d\vec{l}={\mu }_0{\int }_S\vec{J}\cdot d\vec{A}$$ Maxwell's great insight was to realize that this was incomplete. He reasoned that $\frac{d{\phi }_B}{dt}$ gives a $\vec{E}$ field so we should expect that $\frac{d{\phi }_E}{dt}$ gives a $\vec{B}$ field.

Again it is left as an exercise to show that $$\vec{\nabla }\times \vec{B}={\mu }_0\left(\vec{J}+{\epsilon }_0\frac{\partial \vec{E}}{\partial t}\right)$$

Maxwell's equations

Lets recall Maxwell's equations (in free space) in differential form $${\displaystyle \begin{array}{c}\vec{\nabla }\times \vec{E}=-\frac{\partial \vec{B}}{\partial t}\\ \vec{\nabla }\times \vec{B}={\mu }_0\left(\vec{J}+{\epsilon }_0\frac{\partial \vec{E}}{\partial t}\right)\\ \vec{\nabla }\cdot \vec{E}=\frac{\rho }{{\epsilon }_0}\\ \vec{\nabla }\cdot \vec{B}=0\end{array}}$$