Why electrons accelarate in a solenoid? The induced voltage in a solenoid equals to the applied voltage $U = L\frac{di}{dt}$.
From that I conclude that the electric field generated by the voltage supply and the induced electric field cancel out at each moment. So why do the electrons accelarate when there is a voltage across a solonoid?
 A: When you connect a solenoid to a battery, the net emf in the circuit will be
$$\mathscr E_{net}=\mathscr E_{batt}-L\frac{dI}{dt}.$$We know that $\mathscr E_{net}=IR$ when there is resistance $R$ in the circuit. When the resistance is negligible it is, I agree, usual to assume that $\mathscr E_{net}$ is zero.  I think this is almost, but not exactly, true. Some work is needed to accelerate the electrons.
Naïvely treating the free electrons as if they were in vacuo, we'd have $$e\frac {\mathscr E_{net}}{l}=m_e \frac{dv}{dt}$$
in which $l$ is the length of wire on the solenoid and $\frac{dv}{dt}$ is the electrons' acceleration. If $\frac{dI}{dt}=1\ \text {A s}^{-1}$ then in a copper wire of cross-sectional area 1 mm$^2$, $\frac{dv}{dt}\approx 10^{-4}\ \text {m s}^{-2}$. Even if $\ l=10$ km, we find that $\mathscr E_{net}\approx 10^{-10}$ V.
This calculation is crude in the extreme, but I think the point is clear that the net emf needed to accelerate the electrons, against their mechanical inertia, is negligible, and it is perfectly permissible to put $\mathscr E_{net}=0$ in the electrical equation. But the emfs (and the electric fields) in the wire do not exactly cancel, allowing the electrons to accelerate.
