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Mutual Inductance of Two Long Coaxial Solenoids

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Estimated time: 7 minutes
CISCE: Class 12

Introduction

When two coils are placed close to each other, a change in current in one coil induces an EMF in the neighbouring coil. This phenomenon is called mutual induction, and it forms the basis of transformers, induction coils, and wireless charging systems.

Mutual inductance quantifies how effectively one coil can "communicate" a changing magnetic flux to another coil.

CISCE: Class 12

Derivation: Mutual Inductance of Two Long Coaxial Solenoids

Setup:

  • Primary solenoid P: N1​ turns, length l, area A, carries current I1
  • Secondary solenoid S: N2​ turns, wound coaxially over/around P

Step 1: Magnetic field inside primary solenoid:

  • B = \[\mu_0\frac{N_1}{l}I_1\]

Step 2: Flux linked with one turn of secondary:

  • \[\Phi_2=BA=\mu_0\frac{N_1}{l}I_1A\]

Step 3: Total flux linkage with secondary (N2 turns):

  • \[N_2\Phi_2=\mu_0\frac{N_1N_2A}{l}I_1\]

Step 4: Since N2Φ2 = MI1:

  • M = \[{\frac{\mu_0N_1N_2A}{l}}\]

Alternatively, writing turns per unit length n1 = N1/l:

  • M = μ0n1N2A

Important: M depends only on the geometry of the coils (turns, area, length) — not on the current flowing.

CISCE: Class 12

Factors Affecting Mutual Inductance

  • Number of turns in both coils (N1​, N2​) — more turns → higher M
  • Common cross-sectional area A — larger overlap area → higher M
  • Distance/separation between coils — greater separation → lower M
  • Relative orientation — maximum M when coils are coaxial, and one is wound directly over the other
  • Core material — inserting a ferromagnetic (iron) core greatly increases M due to higher permeability
CISCE: Class 12

Example 1

Current changes from 2 A to 6 A in 2 s in coil P, inducing an EMF of 20 mV in coil S. Find M.

ε = \[M\frac{dI}{dt}\Rightarrow M=\frac{\varepsilon}{dI/dt}=\frac{20\times10^{-3}}{2}\] = 10 mH

Answer: M = 10 mH

CISCE: Class 12

Example 2

A solenoid has 600 turns, length 1 m, radius 2 cm; a coil of 100 turns is wound over it. Find M.

Using M = μ0n1N2A, with n1 = 600, N2 = 100, A = π(0.02)2:

Answer: M ≈ 9.47 × 10-5 H

CISCE: Class 12

Real-Life Analogies

  • Transformers use mutual inductance between primary and secondary windings to step up/down voltage.
  • Wireless charging pads transfer energy between coils via mutual induction, similar to the coaxial solenoid model.
  • Analogy: Think of the primary coil as a “broadcasting antenna” and the secondary as a “receiver” — the closer and more aligned they are, the stronger the signal (flux) received.
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