Calculate Induced EMF in Solenoid with Changing Current

To find the induced electromotive force (e.m.f.) in the solenoid, we use Faraday's Law of Induction, $e = L \frac{\Delta I}{\Delta t}$, where $L$ is the self-inductance of the solenoid given by $L = \mu_0 n^2 A l$.

Given values:

• Number of turns per unit length, $n = 10\text{ turns/cm} = 1000\text{ turns/m}$
• Cross-sectional area, $A = 5\text{ cm}^2 = 5 \times 10^{-4}\text{ m}^2$
• Length, $l = 30\text{ cm} = 0.3\text{ m}$
• Change in current, $\Delta I = 4\text{ A} - 2\text{ A} = 2\text{ A}$
• Time interval, $\Delta t = 3.14\text{ s} \approx \pi\text{ s}$

Calculating the self-inductance $L$: $L = (4\pi \times 10^{-7}\text{ H/m}) \times (1000\text{ m}^{-1})^2 \times (5 \times 10^{-4}\text{ m}^2) \times (0.3\text{ m}) = 6\pi \times 10^{-5}\text{ H}$

Calculating the magnitude of the induced e.m.f.: $e = L \frac{\Delta I}{\Delta t} = (6\pi \times 10^{-5}\text{ H}) \times \frac{2\text{ A}}{\pi\text{ s}} = 12 \times 10^{-5}\text{ V}$

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

Correct Option: B (12)