A long solenoid has 400 turns. When a current of 100 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 mWb . The self-inductance of the solenoid is

Step-by-step solution

The self-inductance L of a solenoid is defined by the relation _ total = L I , where _ total is the total magnetic flux linked with all turns of the solenoid. Given the number of turns N = 400 , the current I = 100 A, and the magnetic flux linked with each turn = 4 mWb = 4 10⁻³ Wb. The total magnetic flux linked with the solenoid is _ total = N . Substituting the given values: _ total = 400 4 10⁻³ Wb = 1.6 Wb. Using the formula L = _ total I : L = 1.6 Wb 100 A = 0.016 H. Converting to millihenry (mH): L = 0.016 100