A $30$ cm long solenoid has $10$ turns per cm and area of $5$ cm$^2$. The current through the solenoid coil varies from $2$ A to $4$ A in $3.14$ s. The e.m.f. induced in the coil is $\alpha\times 10^{-5}$ V. The value $\alpha$ is ________.

Given:
Length of the solenoid, $l = 30 \text{ cm} = 0.3 \text{ m}$
Number of turns per unit length, $n = 10 \text{ turns/cm} = 1000 \text{ turns/m}$
Area of cross-section, $A = 5 \text{ cm}^2 = 5 \times 10^{-4} \text{ m}^2$
Change in current, $\Delta I = 4 \text{ A} - 2 \text{ A} = 2 \text{ A}$
Time interval, $\Delta t = 3.14 \text{ s}$

The self-inductance of the solenoid is given by:
$L = \mu_0 n^2 A l$

Substituting the given values:
$L = 4\pi \times 10^{-7} \times (1000)^2 \times 5 \times 10^{-4} \times 0.3$
$L = 4\pi \times 10^{-7} \times 10^6 \times 1.5 \times 10^{-4}$
$L = 6\pi \times 10^{-5} \text{ H}$

The magnitude of the induced e.m.f. is:
$e = L \dfrac{\Delta I}{\Delta t}$

Substituting the values of $L$, $\Delta I$, and $\Delta t$:
$e = 6\pi \times 10^{-5} \times \dfrac{2}{3.14}$

Using $\pi \approx 3.14$:
$e = 6 \times 10^{-5} \times 2 = 12 \times 10^{-5} \text{ V}$

Comparing this with $e = \alpha \times 10^{-5} \text{ V}$, we get:
$\alpha = 12$

Answer: $12$