Question

The self-inductance of an air core solenoid is \( L \). If the number of turns in the solenoid is doubled, keeping all other factors constant, then its self-inductance will be

• A. \(L \)
• B. \( \frac{L}{2} \)
• C. \(2L \)
• D. \(4L \)
• E. \(8L \)

Hint:

Inductance $\propto$ (number of turns)$^2$.

Answer 1

The Correct Option is D

Step 1: Understanding Self-Inductance of a Solenoid:
Self-inductance (\(L\)) is a property of an electrical circuit (like a solenoid) that describes its opposition to a change in current flowing through it. For a long air-core solenoid, its self-inductance depends on its physical characteristics: the number of turns, its length, its cross-sectional area, and the permeability of the core material.
Step 2: Key Formula or Approach:
The formula for the self-inductance of a long solenoid is given by:
\[ L = \frac{\mu_0 N^2 A}{l} \] where:
\(\mu_0\) = permeability of free space (a constant for an air core)
\(N\) = total number of turns in the solenoid
\(A\) = cross-sectional area of the solenoid
\(l\) = length of the solenoid
Step 3: Detailed Explanation:
From the formula, we can establish the relationship between the self-inductance \(L\) and the number of turns \(N\). All other factors (\(\mu_0\), \(A\), \(l\)) are kept constant.
\[ L \propto N^2 \] This means that the self-inductance is directly proportional to the square of the number of turns.
Let the initial number of turns be \(N_1 = N\) and the initial inductance be \(L_1 = L\).
The number of turns is then doubled, so the new number of turns is \(N_2 = 2N\).
The new inductance, \(L_2\), can be found by setting up a ratio:
\[ \frac{L_2}{L_1} = \frac{(N_2)^2}{(N_1)^2} \] Substituting the values:
\[ \frac{L_2}{L} = \frac{(2N)^2}{(N)^2} = \frac{4N^2}{N^2} = 4 \] \[ L_2 = 4L \] Step 4: Final Answer:
If the number of turns is doubled, the self-inductance becomes four times its original value.