Given:
Length of the solenoid, l=30 cm=0.3 m
Number of turns per unit length, n=10 turns/cm=1000 turns/m
Area of cross-section, A=5 cm2=5×10−4 m2
Change in current, ΔI=4 A−2 A=2 A
Time interval, Δt=3.14 s
The self-inductance of the solenoid is given by:
L=μ0n2Al
Substituting the given values:
L=4π×10−7×(1000)2×5×10−4×0.3
L=4π×10−7×106×1.5×10−4
L=6π×10−5 H
The magnitude of the induced e.m.f. is:
e=LΔtΔI
Substituting the values of L, ΔI, and Δt:
e=6π×10−5×3.142
Using π≈3.14:
e=6×10−5×2=12×10−5 V
Comparing this with e=α×10−5 V, we get:
α=12
Answer: 12