A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 10-3 Wb. The self-inductance of the solenoid is
- 1.0 henry
- 4.0 henry
- 2.5 henry
- 2.0 henry
Answer
(1) 1.0 henry
Solution
Total flux linkage N = LI
L = N I = 500 4 10-32
= 1.0 H
Answer: (1)