Answer by J. Manuel for What does the minus sign in Maxwell's third equation...
What should really bother you, is not the minus sign in $(3)$. Is it's absence of in $(4)$!.The minus sign in $(3)$ actually prevents a run-away effect where an induced electric current would create a...
View ArticleAnswer by Nullius in Verba for What does the minus sign in Maxwell's third...
There's an approach from geometric algebra that considers the electromagnetic field as a field of bivectors, and with this geometric interpretation the electric and magnetic components can be rotated...
View ArticleAnswer by Serge Hulne for What does the minus sign in Maxwell's third...
In plain English, it is just Lenz’s law :Lenz's law, named after the physicist Emil Lenz who formulated it in 1834, states that the direction of the electric current which is induced in a conductor by...
View ArticleAnswer by J.G. for What does the minus sign in Maxwell's third equation imply?
Duality is actually not $\mathbf{E}\leftrightarrow \mathbf{B}$ (I've used $c=1$), i.e. $(\mathbf{E},\,\mathbf{B})\to(\mathbf{B},\,\mathbf{E})$. It's...
View ArticleAnswer by JEB for What does the minus sign in Maxwell's third equation imply?
It really comes from relativity, where one uses the field strength tensor:$$ F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}=\left(\begin{array}{cccc}0 & E_x & E_y & E_z\\-E_x &...
View ArticleAnswer by Michael Seifert for What does the minus sign in Maxwell's third...
The minus sign is what makes Maxwell's equations obey causality, so it's a good thing it's there! To see this, you can write out the source-free Maxwell's equations with the sign of $\nabla \times...
View ArticleAnswer by Kurt G. for What does the minus sign in Maxwell's third equation...
In vacuum (all $\mathbf{J}$ and all $\rho$ zero) the Maxwell equations imply the wave equations$$\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial...
View ArticleWhat does the minus sign in Maxwell's third equation imply?
If we write out Maxwell's equations with magnetic charges, we get$$\begin{align}\nabla \cdot \mathbf{E} &= 4 \pi \rho_e \tag{1}\\\nabla \cdot \mathbf{B} &= 4 \pi \rho_m \tag{2}\\-\nabla \times...
View Article
