Question

The self-inductance of a solenoid of length $l$, area of cross-section $A$ and number of turns $N$ is proportional to

• A. $N^2 A / l$
• B. $N A / l$
• C. $N^2 A l$
• D. $N A l$

Hint:

Remember that the self-inductance \( L \) depends on the square of the number of turns and is inversely proportional to the length of the solenoid.

Answer 1

The Correct Option is A

Step 1: Recall the solenoid inductance.
The self-inductance of a solenoid is \[ L = \mu_0 N^2 \frac{A}{l} \] where $N$ is the number of turns, $A$ is the cross-section area, and $l$ is the length.

Step 2: Look at the turns.
The number of turns appears as $N^2$. So the inductance grows with the square of the turns. \[ L \propto N^2 \] More turns means much more inductance.

Step 3: Look at the area.
The area $A$ is on top, so a wider solenoid has more inductance. \[ L \propto A \]

Step 4: Look at the length.
The length $l$ is on the bottom, so a longer solenoid has less inductance. \[ L \propto \frac{1}{l} \]

Step 5: Combine them.
Putting all parts together, \[ L \propto \frac{N^2 A}{l} \]

Step 6: State the answer.
The inductance is proportional to $N^2 A / l$. \[ \boxed{L \propto \frac{N^2 A}{l}} \]