a solenoid question

Flux linkage throught the coil
= 150 x (pi x 0.045^2) x B
where B is the magnetic flux density through the coil

Magnetic flux density B produced by the solenoid is
B = (uo) x (230 x 100) x I, where I is the current through the solenoid, and uo is the permeability of free space (=4 x pi x 10^-7 H/m)
hence, dB/dt = 2300.uo.dI/dt = 2300.uo.(2/0.1) T/s

since induce emf at the coil E, by Faraday's Law, is
E = d[150 x (pi x 0.045^2) x B]/dt = 150 x (pi x 0.045^2) x dB/dt
hence E = 150 x (pi x 0.05^2) x 2300.uo.(2/0.1) volts

Induced current, by Ohm's Law, is = E/12 A
The direction of the induced current is from left to right through the resistor.