Q188 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Physics
The self inductance L of a solenoid of length $l$ and area of cross-section A, with a fixed number of turns N increases as
- Aboth $l$ and A increase.
- Correct answerB$l$ decreases and A increases.
- C$l$ increases and A decreases.
- Dboth $l$ and A decrease.
Solution
The self-inductance of a long solenoid is given by the formula:
$L = \dfrac{\mu_0 N^2 A}{l}$
where $N$ is the total number of turns, $A$ is the cross-sectional area, and $l$ is the length of the solenoid.
Since the number of turns $N$ is fixed, the self-inductance $L$ is directly proportional to the area of cross-section $A$ and inversely proportional to the length $l$.
$L \propto \dfrac{A}{l}$
Thus, to increase the self-inductance $L$, the area $A$ must increase and the length $l$ must decrease.
Answer: $l$ decreases and A increases.
$L = \dfrac{\mu_0 N^2 A}{l}$
where $N$ is the total number of turns, $A$ is the cross-sectional area, and $l$ is the length of the solenoid.
Since the number of turns $N$ is fixed, the self-inductance $L$ is directly proportional to the area of cross-section $A$ and inversely proportional to the length $l$.
$L \propto \dfrac{A}{l}$
Thus, to increase the self-inductance $L$, the area $A$ must increase and the length $l$ must decrease.
Answer: $l$ decreases and A increases.
Hence, the correct answer is option: 'B'.