Energy Stored in Magnetic Field

Inductors store magnetic energy, as capacitors store electrical energy. Consider an inductor  
  in series with a resistance  
  and an EMF  
  and a switch  
, with each component being ideal.
Initially the switch is open, so no current flows. When the switch is closed a current  
  flows and there is voltage  
  across the resistor and a voltage  
\[V_L=L \frac{dI}{dt}\]
  across the inductor, so  
\[\mathcal{E} =IR + l \frac{dI}{dt}\]
If we multiply this equation by  
  we get  
\[I \mathcal{E} =I^2R + IL \frac{dI}{dt}\]
The term on the left represents the power delivered by the EMF and the terms on the left represent the power dissipated by the resistor and inductor respectively.
If we concentrate on the second term, we can write
\[Power= \frac{dU_{Magnetic}}{dt}=LI \frac{I}{dt} \rightarrow dU_{Magnetic}= LIdI \rightarrow U_{Magnetic}= \int LIdI=\frac{LI^2}{2}\]
The magnetic field at the centre of a long solenoid with  
  turns per unit length, carrying a current  
\[B= \mu_0 nI\]
. The self inductance of the coil is  
\[L=\mu_0 n^2Al\]
  are the cross sectional area and length respectively. The energy stored in the solenods magnetic field is
\[U_m= \frac{LI^2}{2}=\frac{\mu_0 n^2Al (\frac{B}{\mu_0 n})^2}{2}=\frac{B^2 Al}{2 \mu_0}\]

The energy density is  
\[\frac{U_m}{Al} = \frac{B^2}{2 \mu_0}-\]
  This result holds for magnetic fields in a region of empty space. An analogous result holds for electric fields.:  
\[u_E=\frac{\epsilon_0 E^2}{2}\]

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