While keeping the area of cross-section of a solenoid the same, the number of turns and length of the solenoid are both doubled. The self-inductance of the coil will be?

The self-inductance of a coil is given by:
L = NΦ/I
where N = number of turns on the coil, A = cross-sectional area, and I = current passing through the coil.
Now, Φ = BA = (μ₀NI/l)A
where B = magnetic field inside the solenoid and l = length of the solenoid.
Thus, L = μ₀N²A/l
So, L' = μ₀(2N)²A/(2l) = 2μ₀N²A/l = 2L
Therefore, the self-inductance is doubled.