Further Background
As seen in the previous problems, the solenoid (or inductor) acts to resist any changes occuring in the magnetic field flux produced by the current that is passing through its loops. Remember that an emf is ONLY induced when there is a change in the flux; moreover, it is as if the inductor acts as a "temporary variable battery" whose orientation opposes the flux change.
The degree to which it provides resistance is called its inductance, L. The emf produced by the inductor is calculated with the formula .

So, if the current is decreasing, , the induced emf in the loops will be positive syncing the inductor with the direction of the battery so as to temporarily attempt to replace the decreasing flux through the inductor.

Conversely, if the current is increasing, , the induced emf in the loops will be negative which reverses the inductor's emf with that o the battery so as to temporarily attempt to decreases the increasing flux through the inductor.
NOTE: if you did not view the solutions to the previous problems, do so now so you can visualize how the conditions pair together. 
Refer to the following information for the next two questions.
We are now going to calculate some numerical values for the inductance of a solenoid and induced emf. To determine the numerical value of the geometric inductance of a solenoid we will use the equation where
 N is the total number of loops (coils)

is the length of the solenoid
 A is the crosssectional area of the solenoid's core

µ_{o} is the permeability of free space (or in our case, air)
Also recall that inductance, L, is measured in Henries where 1 H = 1 volt per (amp/sec). Another definition of this unit is 1 H = 1 weber of flux per amp of current. For the solenoid used in the previous sections we now are given that it has a total number of 2000 loops (or turns of wire), its length is 16 cm, and its crosssectional area has a diameter of 6 cm.
