Number of turns on the solenoid 15 turns /cm=1500 turns /m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area,
A=2.0 cm2=2×10−4m2
The current carried by the solenoid changes from 2 A to 4 A.
Therefore, change in current in the solenoid,
di=4−2=2 A
Change in time, dt=0.1 s
Therefore, rate of change in current, didt=20A/s
According to Faraday's law, induced emf in the solenoid is given by:
e=dϕdt....(i) Where,
ϕ = induced flux through the small loop
ϕ=BA....(ii) B = Magnetic field due to a solenoid
B=μ0niμ0 = Permeability of free space
μ0=4π×10−7Hm
Hence, equation (i) can be rduced to:
e=ddt(BA)
Put B=μ0ni in the given equation,
e=Aμ0n×(didt)
e=2×10−4×4π×10−7×1500×20.1
e=7.54×10−6v Hence, the induced voltage in the loop is 7.54×10−6V
Final answer: 7.5×10−6V