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.…

Given:

Length of the solenoid, $l = 30 \text{ cm} = 0.3 \text{ m}$l=30 cm=0.3 m

Number of turns per unit length, $n = 10 \text{ turns/cm} = 1000 \text{ turns/m}$n=10 turns/cm=1000 turns/m

Area of cross-section, $A = 5 \text{ cm}^2 = 5 \times 10^{-4} \text{ m}^2$A=5 cm2=5×10−4 m2

Change in current, $\Delta I = 4 \text{ A} - 2 \text{ A} = 2 \text{ A}$ΔI=4 A−2 A=2 A

Time interval, $\Delta t = 3.14 \text{ s}$Δt=3.14 s

The self-inductance of the solenoid is given by:

$$L = \mu_0 n^2 A l$$L=μ0​n2Al

Substituting the given values:

$$L = 4\pi \times 10^{-7} \times (1000)^2 \times 5 \times 10^{-4} \times 0.3$$L=4π×10−7×(1000)2×5×10−4×0.3

$$L = 4\pi \times 10^{-7} \times 10^6 \times 1.5 \times 10^{-4}$$L=4π×10−7×106×1.5×10−4

$$L = 6\pi \times 10^{-5} \text{ H}$$L=6π×10−5 H

The magnitude of the induced e.m.f. is:

$$e = L \dfrac{\Delta I}{\Delta t}$$e=LΔtΔI​

Substituting the values of $L$L, $\Delta I$ΔI, and $\Delta t$Δt:

$$e = 6\pi \times 10^{-5} \times \dfrac{2}{3.14}$$e=6π×10−5×3.142​

Using $\pi \approx 3.14$π≈3.14:

$$e = 6 \times 10^{-5} \times 2 = 12 \times 10^{-5} \text{ V}$$e=6×10−5×2=12×10−5 V

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

$$\alpha = 12$$α=12

Answer: $12$12