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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?

Difficulty: Easy

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
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