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: \(12\)