## Energy Stored in Magnetic Field

\[L\]

in series with a resistance \[R\]

and an EMF \[\mathcal{E}\]

and a switch \[S\]

, with each component being ideal.Initially the switch is open, so no current flows. When the switch is closed a current

\[I\]

flows and there is voltage \[V_R=IR\]

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

\[I\]

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

\[n\]

turns per unit length, carrying a current \[I\]

is \[B= \mu_0 nI\]

. The self inductance of the coil is \[L=\mu_0 n^2Al\]

where \[A\]

and \[l\]

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.: \[u_E=\frac{\epsilon_0 E^2}{2}\]

.