Additional: Energy Stored in Inductors

Energy Stored in an Inductor ($ U_B = \frac{1}{2}LI^2 $)

We have seen that an inductor opposes the change in current flowing through it by inducing a back EMF. To establish a current through an inductor, work must be done by the source against this back EMF. This work done is not dissipated as heat (assuming a pure inductor with zero resistance); instead, it is stored in the inductor's magnetic field. This stored energy can be recovered when the current in the inductor decreases.

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Derivation of Energy Stored

Consider a pure inductor of inductance $L$ connected to a voltage source. Let the current in the inductor be changing at a rate $di/dt$. The self-induced back EMF in the inductor is $\mathcal{E} = -L \frac{di}{dt}$.

If the voltage source provides an instantaneous voltage $v$ to drive the current, then this voltage must be equal and opposite to the back EMF (in the ideal case, ignoring any resistance in the external circuit or source internal resistance):

$ v = -\mathcal{E} = -(-L \frac{di}{dt}) = L \frac{di}{dt} $

The instantaneous power delivered by the source to the inductor is $p = vi$.

$ p = \left(L \frac{di}{dt}\right) i = Li \frac{di}{dt} $

This power is the rate at which energy is being delivered to the inductor and stored in its magnetic field. Let $dU_B$ be the infinitesimal amount of energy stored in a small time interval $dt$.

$ dU_B = p \, dt = \left(Li \frac{di}{dt}\right) dt = Li \, di $

To find the total energy $U_B$ stored in the magnetic field when the current is built up from 0 to a final value $I$, we integrate this expression:

$ U_B = \int dU_B = \int_0^I Li \, di $

Since $L$ is constant:

$ U_B = L \int_0^I i \, di $

The integral of $i$ with respect to $i$ is $\frac{i^2}{2}$.

$ U_B = L \left[ \frac{i^2}{2} \right]_0^I = L \left( \frac{I^2}{2} - \frac{0^2}{2} \right) $

$ U_B = \frac{1}{2} L I^2 $

This formula gives the energy stored in the magnetic field of an inductor when a current $I$ flows through it. The unit of energy is the joule (J).

This energy is stored in the magnetic field itself. When the current decreases, the magnetic field collapses, and the stored energy is released back into the circuit (e.g., driving current through a resistance, charging a capacitor).

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Answer:

Energy Density of Magnetic Field ($ u_B = \frac{B^2}{2\mu_0} $)

The energy stored in an inductor is actually stored in the magnetic field created by the current. It is often useful to describe this energy in terms of the energy stored per unit volume in the region where the magnetic field exists. This is called the energy density of the magnetic field.

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Derivation of Energy Density

Let's derive the energy density formula using the example of a long solenoid, where the magnetic field is uniform inside and approximately zero outside.

Consider a long air-cored solenoid of length $L$, cross-sectional area $A$, and $N$ turns. Its self-inductance is $L_{solenoid} = \frac{\mu_0 N^2 A}{L}$.

When a current $I$ flows through the solenoid, the magnetic field inside is $B = \mu_0 n I = \mu_0 \frac{N}{L} I$.

From this equation, we can express the current $I$ in terms of the magnetic field $B$: $I = \frac{B L}{\mu_0 N}$.

The total energy stored in the solenoid is $U_B = \frac{1}{2} L_{solenoid} I^2$.

Substitute the expressions for $L_{solenoid}$ and $I$:

$ U_B = \frac{1}{2} \left(\frac{\mu_0 N^2 A}{L}\right) \left(\frac{B L}{\mu_0 N}\right)^2 $

$ U_B = \frac{1}{2} \frac{\mu_0 N^2 A}{L} \frac{B^2 L^2}{\mu_0^2 N^2} $

Cancel out common terms ($N^2$, $\mu_0$, one $L$):

$ U_B = \frac{1}{2} \frac{A}{L} \frac{B^2 L^2}{\mu_0} = \frac{1}{2} \frac{A L B^2}{\mu_0} $

The volume of the solenoid (where the magnetic field is present) is $V = AL$.

$ U_B = \frac{1}{2} \frac{V B^2}{\mu_0} $

The energy density ($u_B$), which is the energy per unit volume ($u_B = U_B/V$), is:

$ u_B = \frac{U_B}{V} = \frac{\frac{1}{2} \frac{V B^2}{\mu_0}}{V} $

$ u_B = \frac{B^2}{2\mu_0} $

This formula gives the energy density of the magnetic field in vacuum or air. The unit of energy density is Joules per cubic meter ($J/m^3$).

This formula for magnetic energy density is remarkably similar to the formula for electric energy density in an electric field: $u_E = \frac{1}{2}\epsilon_0 E^2$.

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Energy Density in a Material Medium

If the magnetic field is in a material medium with permeability $\mu$, the energy density is given by:

$ u_B = \frac{B^2}{2\mu} = \frac{BH}{2} $ (since $B = \mu H$)

However, the form $u_B = \frac{B^2}{2\mu}$ is more commonly used.

Although derived for a long solenoid, the formula $u_B = B^2/(2\mu_0)$ is a general result that applies to any region of space containing a magnetic field in vacuum, regardless of the source of the field. To find the total magnetic energy in a region with a non-uniform magnetic field, one must integrate the energy density over the volume: $U_B = \int u_B \, dV = \int \frac{B^2}{2\mu_0} \, dV$.

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Answer: