Answer:
14.17H
Explanation:
Energy stored in an inductor is expressed as;
E = 1/2LI²
L is the inductance of the inductor
I is the current flowing through the inductor
Given
E = 10.2J
I = 1.20
Required
Inductance L
Substitute the given parameters into the formula;
10.2 = 1/2L(1.2)²
10.2 = 1/2*1.44L
10.2 = 0.72L
L = 10.2/0.72
L = 14.17H
Hence the self inductance of the solenoid is 14.17H
