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.