An inductor of inductance L is first unwound and then rewound. The radius of the new loop is half and has same length. Find new self inductance.

Question : An inductor of inductance L is first unwound and then rewound. The radius of the new loop is half and has same length. Find new self inductance.

Doubt by Kulwinder

Solution : 
Here 
Initial Length (l) = Final Length (l) (Given)
N × 2πr = N' × 2πr'
Nr = N'r'
Nr = N' × (r/2)
2N = N'
N' = 2N

Initial Area of cross section of each loop, A = πr²
Final Area of cross section of each loop,
A' = πr'²
A' = π(r/2)²
A' = πr²/4
A' = A/4

Self Inductance of an Inductor, L = μ₀N²A/l

New Self Inductance of the coil,
L' = μ₀N'²A'/l
L' = μ₀(2N)²(A/4)/l
L' = μ₀4N²(A/4)/l
L' = μ₀N²A/l
L'=L

∴ New Self inductance will be same as the initial one i.e. equal to L.