Electric Field in conductor near Solenoid My question is essentially a follow up to this question. It's answer says, the mobile charges in the conductor distribute themselves to make E = 0 inside the conductor. But, I don't understand how a conservative E field (from the charge distribution) cancels out the non conservative E field (from the time varying magnetic field) to produce no net E field in the conductor?
 A: First of all , when current is inside the conductor , it is driven by voltage and not by an electric field . When current starts flowing , the coil  wants to build a magnetic field . While doing , it creates a back emf which resists the current . Once the magnetic field is created it allows the current to pass normally. The coil just resists the flow of current . 
You should study more Faraday's laws.
A: 
I don't understand how a conservative E field (from the charge distribution) cancels out the non conservative E field

Conservative field can have lines of force of any shape provided they do not form closed loops. Think of conservative field of electric dipole: its lines of force look like closed loops far from the dipole, but they are not, because they start and end at the charges.
Similar thing is happenning with the conservative field of that half ring conductor; due to induced electric field, electric charges accumulate on extremities of the conductor to form conservative field that cancels the induced field inside the half ring. Of course, the conservative (Coulomb) field can't cancel the induced field both inside and outside the conductor. The cancellation happens only inside the perfect conductor.
