The self-inductance of a solenoid is:

$A$ coil of $100$ turns carries a current of $5\, mA$ and creates a magnetic flux of $10^{-5} \, Wb$. The inductance is.....$mH$.

The current in an inductor of self-inductance $L=40 \text{ mH}$ is to be increased uniformly from $2 \text{ A}$ to $12 \text{ A}$ in $8 \text{ ms}$. The emf induced in the inductor during this process is (in $\text{ V}$)

Regarding self-inductance:
$A:$ The self-inductance of the coil depends on its geometry.
$B:$ Self-inductance does not depend on the permeability of the medium.
$C:$ Self-induced e.m.f. opposes any change in the current in a circuit.
$D:$ Self-inductance is the electromagnetic analogue of mass in mechanics.
$E:$ Work needs to be done against self-induced e.m.f. in establishing the current.
Choose the correct answer from the options given below:

When the number of turns in a solenoid are doubled without any change in the length of the solenoid,its self-inductance becomes

Why is self-induced $emf$ known as back $emf$?