Question

When both the number of turns and the core length of an inductive coil are doubled, its self-inductance will be:

• A. halved
• B. unaffected
• C. quadrupled
• D. doubled

Hint:

Inductance is directly proportional to the square of the turns ($N^2$) and inversely proportional to the length ($l$). Doubling $N$ scales the expression by 4, and doubling $l$ scales it by $\frac{1}{2}$, resulting in a net factor change of $4 \times \frac{1}{2} = 2$.

Answer 1

The Correct Option is D

Concept: The self-inductance $L$ of a long solenoid or an inductive coil wound around a core is given by the standard geometric formula: \[ L = \frac{\mu N^2 A}{l} \] Where: - $\mu$ represents the magnetic permeability of the core material. - $N$ is the total number of turns of the coil. - $A$ is the cross-sectional area of the core. - $l$ is the physical length of the core. By examining this formula, we can determine how changes to the physical dimensions scale the overall inductance value.

Step 1: Write down the initial inductance equation.
Let the initial configuration parameters be $N_1$ and $l_1$. The initial self-inductance is: \[ L_1 = \frac{\mu N_1^2 A}{l_1} \quad \cdots (1) \]

Step 2: Express the new parameters in terms of the initial ones.
According to the problem description, both the number of turns and the core length are doubled: \[ N_2 = 2N_1 \] \[ l_2 = 2l_1 \] The cross-sectional area $A$ and material permeability $\mu$ remain unchanged.

Step 3: Substitute the new parameters into the inductance formula.
\[ L_2 = \frac{\mu N_2^2 A}{l_2} = \frac{\mu (2N_1)^2 A}{2l_1} \]

Step 4: Simplify the expression algebraically.
\[ L_2 = \frac{\mu \cdot (4N_1^2) \cdot A}{2l_1} = \frac{4}{2} \cdot \left( \frac{\mu N_1^2 A}{l_1} \right) \] \[ L_2 = 2 \cdot L_1 \] Therefore, the new self-inductance is exactly doubled, which corresponds to Option (D).