A long solenoid has 500 turns. When a current of 2 A is passed through it, the resulting magnetic flux linked with each the turn of the solenoid is 4×10−3 Wb.  The self-inductance of the solenoid is

Inductance of a coil is numerically equal to the emf induced in the coil when the current in the coil changes at the rate of 1 As−1.  If I is the current flowing in the circuit, then flux linked with the circuit is observed to be proportional to I, i.e., ϕ ∝ I or ϕ=LI                 …..(i)Where I is called the self-inductance or coefficient of self-inductance or simply inductance of the coil.Net flux through solenoid,ϕ=500×4×10−3=2 Wb[after putting values in Eq. (i)]or L = 1 H