Answer:
Induced emf is equal to 0.05 volt
Explanation:
We have given length of the solenoid l = 25 cm = 0.25 m
Radius of the solenoid r = 2.3 cm = 0.023 m
Area of the solenoid [tex]A=\pi r^2=3.14\times 0.023^2=0.00166m^2[/tex]
Number of turns N = 580
Inductance of the solenoid [tex]L=\frac{\mu _0N^2A}{l}=\frac{4\pi \times 10^{-7\times 580^2\times 0.00166}}{0.25}=0.0028H[/tex]
Rate of change of current [tex]\frac{di}{dt}=17.8A/sec[/tex]
Induced emf in inductor is equal to [tex]e=L\frac{di}{dt}=0.0028\times 17.8=0.05volt[/tex]
