How to work out the induced electric field of a magnet moving through solenoid? The problem's solution according to Maxwell's equations is given here.
However, using Maxwell's third law gives me simply $E = B\times v$, where did the extra gamma $\left(\gamma\right)$ factor come from?
 A: The Wiki article cited by Nat in the question is very instructive, but its title and diagram are a little misleading. The article does deal with the central issue of the magnet-and-conductor thought-experiment, but only obliquely with the thought-experiment itself. 
The 'central issue' is that e-m fields are essentially one 'composite' field, and that field components transform from the S frame to the S' frame by relationships such as this one$$E'_z=\gamma \left(E_z + v B_y \right)$$
Note that this equation is not derivable from Maxwell's equations as such, but from Maxwell's equations plus the insistence that Maxwell's equations apply in all inertial frames of reference, which implies that Lorentz transforms must be used for kinematic quantities between frames.
What I'm offering below (with no pretence to originality) is an elementary treatment of the magnet and ring problem from pretty much first principles, using Maxwell's equations only informally, as well as the most trivial of the field transformations.
Consider the magnet and a small ring on the axis of the magnet and co-axial with it approaching each other. Suppose the ring to be small enough for the axial (x direction) component of the magnetic flux density to be almost uniform over the ring's cross-section.
In the S' frame (in which the ring is at rest), the emf around the ring is given by the line integral of the induced tangential electric field around the loop, which, according to the Faraday-Maxwell equation, is$$\mathscr{E}'=2\pi r\ E'=-\pi r^2 \frac{\partial{B'}_x}{\partial{t'}}$$
In the S frame (in which the magnet is at rest) the induced emf is due to the ring cutting the radial component, $B_r$, of the magnet's flux. In a time dt, the ring sweeps out a cylindrical surface of axial length $v \text{d}t$, which, together with the end faces of the cylinder, we can use as a gaussian surface. Since the net magnetic flux leaving the surface must be zero, we have
$$2\pi r (v\text{d}t) B_r+\pi r^2 \frac{\partial{B}_x}{\partial{x}}(v\text{d}t)=0,\ \ \ \ \text{that is}\ \ \ \ \ 2\pi r  B_r=-\pi r^2 \frac{\partial{B}_x}{\partial{x}}$$
The emf in the ring is the tangential magnetic Lorentz force per unit charge acting on each free charge in the ring, multiplied by the ring's circumference, so$$\mathscr{E}=2\pi r\ B_{r}v=-\pi r^2 v\frac{\partial{B}_x}{\partial{x}}$$
Comparison between $\mathscr{E}'$ and $\mathscr{E}$
Longitudinal field components are frame-invariant, that is $$B'_x=B_x\ \ \ \text{meaning}\ \ \ B'_{x}(x',\ t')=B_{x}(x,\ t)$$
Thus relating $\frac{\partial{B'}_x}{\partial{t'}}$ to $\frac{\partial{B}_x}{\partial{x}}$ is purely a matter of kinematics:
In the magnet's frame, S, the field has no time dependence, that is$\frac{\partial{B_x}}{\partial{t}}=0$, so
$$\frac{\partial{B'}_x}{\partial{t'}}=\frac{\partial{x}} {\partial{t'}} \frac{\partial{B}_x}{\partial{x}}=\gamma v\frac{\partial{B_x}}{\partial{x}}\ \ \ \text{because}\ \ \ x=\gamma(x'+vt')$$ $$\text{Thus}\ \ \ \mathscr{E'}=\gamma \mathscr{E}$$
The $\gamma$ factor relating the emfs in the two frames can, in fact, be attributed to the time interval for a given change of flux through the ring being a proper time in the ring's frame, but an improper time (greater by a factor of $\gamma$) in the magnet's frame.
In the usual laboratory setting, the relative velocity of ring and magnet, $v << c$, so $\gamma \approx 1$; in other words we could use a Galilean rather than a Lorentz transform between frames, obtaining $\mathscr{E'}=\mathscr{E}$. 
