AcademyInductance
Academy
Self-Inductance and Inductors
Level 1 - Physics topic page in Inductance.
Principle
An inductor opposes changes in its own current by producing a self-induced emf.
Notation
\(L\)
self-inductance
\(\mathrm{H}\)
\(I\)
current through the inductor
\(\mathrm{A}\)
\(N\Phi_B\)
magnetic flux linkage of the coil
\(\mathrm{Wb}\)
\(\mathcal E_L\)
self-induced emf
\(\mathrm{V}\)
\(V_L\)
passive-sign voltage across an inductor
\(\mathrm{V}\)
\(l\)
solenoid length
\(\mathrm{m}\)
Method
Derivation 1: Define self-inductance
Current in a coil creates magnetic flux through the same coil. In a linear magnetic system, flux linkage is proportional to current.
Flux linkage
\[N\Phi_B=LI\]
Self-inductance
\[L=\frac{N\Phi_B}{I}\]
Derivation 2: Self-induced emf
Faraday's law applied to the coil's own flux linkage gives the back emf.
Faraday's law
\[\mathcal E_L=-\frac{d}{dt}(N\Phi_B)\]
Linear inductor
\[\mathcal E_L=-L\frac{dI}{dt}\]
Passive sign convention
\[V_L=L\frac{dI}{dt}\]
Derivation 3: Long-solenoid inductance
For a long solenoid, combine the interior field with the coil's flux linkage.
Field
\[B=\mu\frac{N}{l}I\]
Flux linkage
\[N\Phi_B=NBA=\mu\frac{N^2A}{l}I\]
Inductance
\[L=\mu\frac{N^2A}{l}\]
Rules
Self-inductance
\[\displaystyle L=\frac{N\Phi_B}{I}\]
Self-induced emf
\[\displaystyle \mathcal E_L=-L\frac{dI}{dt}\]
Inductor voltage
\[\displaystyle V_L=L\frac{dI}{dt}\]
Solenoid inductance
\[\displaystyle L=\mu\frac{N^2A}{l}\]
Henry
\[1\,\mathrm{H}=1\,\mathrm{V\,s\,A^{-1}}\]
Examples
Question
An inductor has
\[L=0.40\,\mathrm{H}\]
and current increasing at \[3.0\,\mathrm{A\,s^{-1}}\]
Find \[|\mathcal E_L|\]
Answer
\[|\mathcal E_L|=L\left|\frac{dI}{dt}\right|=(0.40)(3.0)=1.2\,\mathrm{V}\]
Checks
- Inductors resist changes in current, not current itself.
- Larger turn count strongly increases inductance because \(L\\propto N^2\).
- The sign of self-induced emf is a Lenz-law direction statement.