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

Solution :
Net flux through solenoid is, $${\phi _{{\text{net}}}} = N\phi $$
$$\eqalign{ & \therefore {\phi _{{\text{net}}}} = 500 \times 4 \times {10^{ - 3}} \cr & = 2\,Wb \cr} $$
where, $$\phi =$$  flux through each turn, and
$$N =$$  total number of turns
Also, $${\phi _{{\text{net}}}} = Li = 2\,Wb$$
$$\eqalign{ & {\text{Now,}}\,\,L \times 2 = 2 \cr & \Rightarrow {\text{Self - inductance,}}\,L = 1\,H \cr} $$