Answer:
Current induced in the loop = 0.032 mA
Explanation:
emf induced in the solenoid using Faraday law e = [tex]\frac{-d\phi}{dt}=-\pi R^{2} \frac{dB}{dt}[/tex]
here, R is the radius of solenoid which is constant & [tex]\frac{dB}{dt}[/tex] is the change in magnetic field.
Magnetic field inside the solenoid from B=μ0 n i -----(i)
Where, i is the current in the coil.
n= numbers of turn in per meter length.
n=19 x 100 turns/m
Differentiating equation (i)
[tex]\frac{dB}{dt} =[/tex] μ0 n [tex]\frac{dI}{dt}[/tex]
= 4 π 10-7 x 1900 x 8 =0.019
[tex]E = 3.14 X 0.42^{2} X 0.019[/tex] =0.095 m V
Hence electrical current induced in the loop = [tex]\frac{E}{R} =\frac{0.095}{3}[/tex] = 0.032 mA