AcademyInductance
Academy
Magnetic Field Energy
Level 1 - Physics topic page in Inductance.
Principle
Inductors store energy in the magnetic field produced by their current.
Notation
\(U_B\)
magnetic energy stored
\(\mathrm{J}\)
\(L\)
inductance
\(\mathrm{H}\)
\(I\)
current
\(\mathrm{A}\)
\(u_B\)
magnetic energy density
\(\mathrm{J\,m^{-3}}\)
\(B\)
magnetic field magnitude
\(\mathrm{T}\)
\(\mu\)
magnetic permeability of the medium
\(\mathrm{N\,A^{-2}}\)
Method
Derivation 1: Power into an inductor
Using the passive sign convention, the voltage across an inductor is \(V_L=L\\,dI/dt\).
Electrical power
\[P=IV_L\]
Inductor voltage
\[P=IL\frac{dI}{dt}\]
Energy rate
\[\frac{dU_B}{dt}=LI\frac{dI}{dt}\]
Derivation 2: Stored energy
Integrate the power input from zero current to final current.
Differential energy
\[dU_B=LI\,dI\]
Stored energy
\[U_B=\int_0^I LI'\,dI'\]
Inductor energy
\[U_B=\frac{1}{2}LI^2\]
Derivation 3: Field energy density
For a long solenoid, combine \(U_B=\\frac12LI^2\), \(L=\mu N^2A/l\), and \(B=\mu NI/l\).
Energy density
\[u_B=\frac{U_B}{Al}\]
Field form
\[u_B=\frac{B^2}{2\mu}\]
Rules
Inductor energy
\[U_B=\frac{1}{2}LI^2\]
Energy density
\[\displaystyle u_B=\frac{B^2}{2\mu}\]
Vacuum density
\[\displaystyle u_B=\frac{B^2}{2\mu_0}\]
Power into inductor
\[\displaystyle P=LI\frac{dI}{dt}\]
Examples
Question
A
\[0.20\,\mathrm{H}\]
inductor carries \[3.0\,\mathrm{A}\]
Find stored energy.Answer
\[U_B=\frac12LI^2=\frac12(0.20)(3.0)^2=0.90\,\mathrm{J}\]
Checks
- Stored magnetic energy is always nonnegative.
- Doubling current quadruples the stored energy.
- Energy can return to the circuit when the current decreases.