A solenoid core of length 10 cm provides a self-inductance of 8 mH. If the core length is doubled to 20 cm while all other quantities (turns, cross-sectional area, and permeability) remain unchanged, what will be the new self-inductance?
Electronics and Communication Engineering
Materials and Components
Difficulty: Easy
Choose an option
-
A32 mH
-
B16 mH
-
C8 mH
-
D4 mH
-
E2 mH
Answer
Correct Answer: 4 mH
Explanation
Introduction / Context:Self-inductance of a solenoid quantifies the flux linkage produced per unit current. For long solenoids, inductance depends on geometry and magnetic properties. Understanding how L varies with core length is essential in electromagnetics and electrical machine design.
Given Data / Assumptions:
- Initial length l1 = 10 cm; final length l2 = 20 cm.
- Number of turns N, cross-sectional area A, and permeability μ remain the same.
- Initial inductance L1 = 8 mH.
Concept / Approach:For a uniform solenoid, L = μ * N^2 * A / l. With N, A, and μ fixed, inductance is inversely proportional to length l. Therefore L2/L1 = l1/l2.
Step-by-Step Solution:
Write proportionality: L ∝ 1/l.Compute ratio: L2 = L1 * (l1 / l2) = 8 mH * (10 / 20).Evaluate: L2 = 8 mH * 0.5 = 4 mH.Verification / Alternative check:
Dimensional reasoning: doubling magnetic path length halves magnetizing inductance for constant N, A, μ.Why Other Options Are Wrong:
16 or 32 mH imply L increases with length, contrary to L ∝ 1/l.8 mH implies no change; not correct when l doubles.Common Pitfalls:
Confusing dependence on N (L ∝ N^2) with dependence on l (inverse).Final Answer:
4 mH