Concepts of Physics

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Self Induction

Class XII • Electromagnetic Induction

Self Induction

The property of a coil by which it opposes any change in current flowing through it, by inducing an emf in itself.

Definition

When the current in a coil changes, the magnetic flux linked with the coil also changes. According to Faraday’s law, this change induces an emf in the same coil. This phenomenon is called self-induction, and the emf is called self-induced emf.

Mathematical Expression

  • Magnetic flux linkage through a coil of \(N\) turns:
    \[ \Phi_B = L I \]
    where \(L\) is called the self-inductance of the coil.
  • According to Faraday’s law:
    \[ \varepsilon = – \frac{d\Phi_B}{dt} \]
  • Substituting \(\Phi_B = L I\):
    \[ \varepsilon = – L \frac{dI}{dt} \]

Thus, self-induced emf is proportional to the rate of change of current.

Physical Meaning of L

  • \(L\) depends on geometry of the coil (number of turns, area, length) and the medium inside (permeability).
  • SI unit of \(L\) is henry (H).
  • If current of 1 A in a coil produces flux linkage of 1 weber-turn, its inductance is 1 H.

Self-Inductance of a Solenoid

For a long solenoid of cross-sectional area \(A\), length \(l\), and \(n\) turns per unit length:

  • Magnetic field inside: \( B = \mu_0 n I \)
  • Total flux linkage: \( N \Phi_B = \mu_0 n^2 A l I \)
  • Hence, self-inductance:
    \[ L = \mu_0 n^2 A l \]
    If the core has relative permeability \(\mu_r\), then
    \[ L = \mu_0 \mu_r n^2 A l \]

Energy Stored in an Inductor

Work is done against the back emf to establish current in a coil. This energy is stored in the magnetic field.

  • Energy stored:
    \[ U = \frac{1}{2} L I^2 \]
  • Energy density in magnetic field:
    \[ u = \frac{B^2}{2\mu_0} \]

Analogy with Mechanics

Self-inductance \(L\) is analogous to mass in mechanics:

  • Just as mass opposes change in velocity, inductance opposes change in current.
  • Self-induced emf acts like “electrical inertia.”

Applications

  • Chokes in fluorescent tubes.
  • Energy storage in inductors.
  • Filters in AC circuits.
  • Transformers (along with mutual induction).

Worked Example

A solenoid of length \(0.5\,\text{m}\), cross-sectional area \(2\times10^{-3}\,\text{m}^2\), with 1000 turns is wound on a core of \(\mu_r = 1000\). Calculate its inductance.

Using \( L = \mu_0 \mu_r \dfrac{N^2 A}{l} \):

\( L = 4\pi \times 10^{-7} \times 1000 \times \dfrac{(1000)^2 \times 2 \times 10^{-3}}{0.5} \approx 5.0\,\text{H} \)

Answer: Inductance = 5 henry.

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