📂Electrodynamics

Energy in a Magnetic Field

Explanation¹

Just like we considered the energy of the electric field created by a charge distribution in the energy of the electric field created by a charge distribution, we can think about the energy of the magnetic field created by a current distribution. When current flows through a circuit, energy is inputted. The identity of this energy is exactly the work done against electromotive force (EMF). Because of EMF, it’s hard to change the current flowing in the circuit. Therefore, to make a unit charge go around the circuit once, it must do work equal to EMF $-\mathcal{E}$. Since the definition of current is the amount of charge that passes through a conductor per unit time, the amount of work done during this unit time is

$$ \dfrac{d W}{dt}=-\mathcal{E}I=LI\dfrac{dI}{dt} $$

Integrate both sides over the time interval from when the current is $0$ to when it becomes $I$. Then, the work done during this period is

$$ \begin{equation} \int dW = \int LI dI \quad \implies \quad W=\dfrac{1}{2}LI^{2} \end{equation} $$

This value is independent of the time the current has flowed. It depends only on the geometric characteristic of the loop$(L)$ and the final current value$(I)$. Since the flux passing through the loop is proportional to its self-inductance,

$$ \Phi=LI $$

On the other hand, if we calculate the flux directly using the vector potential of the magnetic field,

$$ \Phi = \int \mathbf{B} \cdot d \mathbf{a} = \int ( \nabla \times \mathbf{A} ) \cdot d\mathbf{a} = \oint \mathbf{A} \cdot d \mathbf{l} $$

In the last equality, Stokes’ theorem was used. Therefore,

$$ LI=\oint \mathbf{A} \cdot d \mathbf{l} $$

If we substitute the above result into $(1)$,

$$ W=\dfrac{1}{2}I \oint \mathbf{A} \cdot d\mathbf{l}=\dfrac{1}{2} \oint (\mathbf{A} \cdot \mathbf{I}) dl $$

The last equality holds because both the direction of the current and the direction of the line integral are the same. Since they both follow the direction of the conductor, it is naturally the same. If this is expressed in terms of volume current,

$$ W=\dfrac{1}{2} \int ( \mathbf{A} \cdot \mathbf{J} ) d\tau $$

The ultimate goal is to express the obtained equation solely in terms of the magnetic field $\mathbf{B}$ using various methods. Using Ampère’s law $\nabla \times \mathbf{B} =\mu_{0} \mathbf{J}$,

$$ W=\dfrac{1}{2\mu_{0}} \int \mathbf{A} \cdot (\nabla \times \mathbf{B} ) d\tau $$

Partial integration of equations involving the del operator

$$ \int_{\mathcal{V}} \mathbf{A} \cdot \left( \nabla \times \mathbf{B} \right) d\tau = \int_{\mathcal{V}} \mathbf{B} \cdot \left( \nabla \times \mathbf{A} \right) d\tau + \oint_{\mathcal{S}} \left( \mathbf{B} \times \mathbf{A} \right) \cdot d \mathbf{a} $$

Using the product rule to perform partial integration,

$$ \begin{align*} W &= \dfrac{1}{2\mu_{0}} \left( \int _\mathcal{V} \mathbf{B} \cdot \mathbf{B}d\tau -\oint_\mathcal{S} (\mathbf{A} \times \mathbf{B} ) \cdot d\mathbf{a} \right) \\ &= \dfrac{1}{2\mu_{0}} \left( \int _\mathcal{V} B^{2} d\tau -\oint_\mathcal{S} (\mathbf{A} \times \mathbf{B} ) \cdot d\mathbf{a} \right) \end{align*} $$

The left-hand side is a constant quantity of work. Hence, as the value of the volume integral on the right-hand side increases, the value of the surface integral must decrease. (The same approach was used when calculating the energy of the electric field) The reason why we can freely expand the integration domain is that in areas where no current flows, $\mathbf{J}=0$ holds, and adding $0$ to the original value yields the same result. Therefore, integrating over the entire space,

$$ W=\dfrac{1}{2\mu_{0}} \int_{\mathrm{total\ space}} B^{2} d\tau $$

What this equation implies is that energy is stored in the magnetic field, and its density is $\frac{1}{2\mu_{0}}B^{2}$.