When the number of turns of a solenoid is doubled, its self inductance becomes:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4 times
The self inductance of a solenoid of N turns, length l and area of cross-section A is \( L = \mu_0N^{2}A/l \), so \( L\propto N^{2} \) when the geometry is unchanged.
The reason is that doubling N doubles the magnetic field for the same current, and it also doubles the number of turns linked with that flux.
\( \displaystyle \frac{L'}{L} = \left(\frac{N'}{N}\right)^{2} = 2^{2} = 4 \)