A long solenoid has 500 turns. When a current of 2 ampere

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

Total number of turns in the solenoid, N = 500
Current, I = 2A.
Magnetic flux linked with each turn
= 4 × 10^–3 Wb
As, φ = LI or N φ = LI

⇒ L =	Nφ
	1

=	500 × 4 × 10^-3	henry = 1 H.
	2