How does one show that the curl of magnetic field is equal to $\mu_0\mathbf J+\mu_0\epsilon_0\frac{\partial \mathbf E}{\partial t}$? Normally, the curl is supposed to be equal to $\mu_0\mathbf J$. However, when checking for invariance for Maxwell's equations under duality transformations, the term $\mu_0\epsilon_0\frac{\partial \mathbf E}{\partial t}$ is introduced. How does it come about?
 A: I do not see what you mean by "normally, the curl is supposed to be equal to $\mu_0\vec J$". I will assume that you mean: experiments show that $${\rm curl}\ \!\vec B=\mu_0\vec J,$$ i.e. an electric current generates a magnetic field turning around the wire. However, this equation cannot be true: taking the divergence, the l.h.s vanishes $${\rm div}\ \!{\rm curl}\ \!\vec B=0$$ while the r.h.s. is (up to a factor $\mu_0$)
     $${\rm div}\ \!\vec J+{\partial\rho\over\partial t}=0$$
(charge continuity equation, i.e. conservation of the total charge). Since the Maxwell-Gauss equation reads
    $${\rm div}\ \!\vec E={\rho\over\varepsilon_0}$$
it is easy to see how to modify our original equation:
    $${\rm curl}\ \!\vec B=\mu_0\vec J+\mu_0\varepsilon_0
{\partial\vec E\over\partial t}$$
The divergence of the two sides of the equation now vanishes. Physically, you should imagine that when electric wires are finite, there is an accumulation of charge at the two ending points of these wires. These varying charges also contribute to the magnetic charge.
