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CISCEISC (Science) ISC Class 12
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अभ्यासक्रम

Self-Inductance of a Long Solenoid

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Estimated time: 8 minutes
CISCE: Class 12

Introduction

When the current in a solenoid changes, the magnetic flux linked with it also changes. By Faraday's Law, this changing flux induces an EMF in the same coil — this is self-induction. The more a coil can "link" flux per unit current, the greater its self-inductance.

 

CISCE: Class 12

Derivation: Self-Inductance of an Air-Core Solenoid

Given: Long solenoid, length l, cross-sectional area A, number of turns N, carrying current I.

Step Expression Explanation
1 B = \[μ_0\frac {N}{l}I\] Magnetic field inside a long solenoid
2 \[Φ_B\] = BA = \[μ_0\frac {NIA}{l}\] Flux through one turn
3 \[NΦ_B\] = \[\frac {μ_0N^2IA}{l}\] Total flux linkage (N turns)
4 L = \[\frac {NΦ_B}{I}\] Definition of self-inductance

Final Formula (Air-Core Solenoid): L = \[\frac {μ_0N^2A}{l}\]

CISCE: Class 12

Solenoid with a Magnetic Core

If the core has relative permeability μr, permeability becomes μ = μrμ0:

  • L = \[\frac{\mu N^2A}{l}=\frac{\mu_r\mu_0N^2A}{l}\]

Key Insight: A ferromagnetic core (high μr) drastically increases inductance — this is why inductors used in electronics have iron/ferrite cores rather than air cores.

CISCE: Class 12

Factors Affecting Self-Inductance

Factor Relationship Effect on L
Number of turns (N) L ∝ N2 Doubling turns → 4 × inductance
Cross-sectional area (A) L ∝ A Larger area → higher L
Length of solenoid (l) L ∝ \[\frac {1}{l}\] Longer solenoid → lower L
Core permeability (μr​) L ∝ μr Magnetic core → much higher L than air core

Important: Self-inductance depends only on the geometry and core material of the solenoid — it does NOT depend on the current flowing through it.

CISCE: Class 12

Example

Problem: An air-core solenoid has length l = 30 cm, area A = 25 cm2, and N = 500 turns. It carries a current of 2.5 A, which is switched off in 10−3 s. Find the average back EMF induced.

Solution:

Step Working
Given l = 0.30 m, A = 25 × \[10^{−4}m^2\], N = 500, ΔI = 2.5 A, Δt = \[10^{−3}\] s
Formula L = \[\frac {μ_0N^2A}{l}\]​, ε = −L\[\frac {ΔI}{Δt}\]
Compute L L = \[\frac {(4π×10^{−7})(500)^2(25×10^{−4})}{0.30}\] ≈ 2.62 × \[10^{−3}\] H
Compute EMF ε = L × \[\frac {ΔI}{Δt}\] = 2.62 × \[10^{−3}\] × \[\frac {2.5}{10^{−3}}\]
Answer ε ≈ 6.54 V
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