Energy Stored in a Magnetic Field

Why energy is stored

• A changing magnetic flux in a coil causes an induced emf. The induced emf so produced opposes the change, and hence energy has to be spent to overcome it in order to build up the magnetic field.
• This energy may be recovered as heat in a resistance of the circuit. That fact gives the logical concept of the energy being stored in the magnetic field.
• The electrostatic parallel is already familiar: the total electrostatic energy stored in the medium between the plates of a capacitor of capacitance $C$ carrying charge $q$ at potential $V$ is $U_E = \dfrac{q^2}{2C} = \dfrac{1}{2}CV^2$.

Derivation of the stored energy

• The induced emf in an inductor is

$$e = -L\dfrac{di}{dt}$$

• The work done in moving a charge $dq$ against this emf is

$$dW = -e\,dq = L\dfrac{di}{dt}\,dq = L\cdot\dfrac{di\cdot dq}{dt}$$

$$\therefore\ dW = L\,i\,di \qquad\left[\because \dfrac{dq}{dt} = i\right]$$

• Therefore the total work done in raising the current from zero to $i$ is

$$W = \int dW = \int_0^{i} L\,i\,di = \dfrac{1}{2}Li^2 = U_B$$

• This is the energy stored ($U_B$) in the magnetic field, in joule. It is analogous to the energy $U_E$ stored in the electric field of a capacitor, and to the kinetic energy $\dfrac{1}{2}mv^2$ of a moving mass — which is why $L$ is called the electrical inertia of a circuit.

Energy density of a magnetic field (12.13)

• Consider a long solenoid of length $l$ near its middle, cross-sectional area $A$, carrying a current $i$. The volume associated with this length is $Al$.
• The energy $U_B$ stored by this length must lie entirely within the volume $Al$, because the magnetic field outside a long solenoid is almost zero. Moreover the energy is uniformly distributed within that volume, because $\vec{B}$ is uniform everywhere inside.
• Diagram in words: a long solenoid drawn in section, with closely spaced turns, straight uniform field lines running along the axis inside it and no field lines outside; a length $l$ of the interior is marked as the volume $Al$.
• Hence the energy stored per unit volume in the magnetic field is

$$u_B = \dfrac{U_B}{Al} = \dfrac{1}{2}\dfrac{Li^2}{Al} = \dfrac{1}{2}\left(\dfrac{L}{l}\right)\dfrac{i^2}{A}$$

• For a long solenoid the inductance per unit length is $\dfrac{L}{l} = \mu_0 n^2A$, where $n$ is the number of turns per unit length. Substituting,

$$u_B = \dfrac{1}{2}\dfrac{\mu_0 n^2 A i^2}{A} = \dfrac{1}{2}\mu_0 n^2i^2$$

• The field at interior points of a solenoid is $B = \mu_0 n i$, so $\mu_0 n^2 i^2 = \dfrac{B^2}{\mu_0}$, giving

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

• This gives the density of stored energy at any point where the magnetic field is $B$, in $\text{J m}^{-3}$. It holds good for all magnetic fields, no matter how they are produced.
• The parallel with electrostatics is exact: $u_E = \dfrac{1}{2}\varepsilon_0E^2$ for the electric field, $u_B = \dfrac{B^2}{2\mu_0}$ for the magnetic field, with $\mu_0$ and $\varepsilon_0$ the permeability and permittivity of free space.

Worked pattern: inductance from stored energy (coaxial cable)

• For a coaxial cable of length $l$ carrying current $I$ down an inner cylinder of radius $a$ and back along an outer cylinder of radius $b$, Ampere's law gives the field between the cylinders as $B = \dfrac{\mu_0 I}{2\pi r}$, and zero elsewhere.
• The energy density is $u_B = \dfrac{B^2}{2\mu_0} = \dfrac{\mu_0 I^2}{8\pi^2r^2}$.
• The energy in a cylindrical shell of length $l$, radius $r$ and thickness $dr$ is $u_B(2\pi r l\,dr) = \dfrac{\mu_0 I^2l}{4\pi}\dfrac{dr}{r}$.
• Integrating from $a$ to $b$, $W = \dfrac{\mu_0 I^2 l}{4\pi}\ln\left(\dfrac{b}{a}\right)$.
• Comparing with $W = \dfrac{1}{2}LI^2$ gives the inductance of the coaxial cable,

$$L = \dfrac{\mu_0 l}{2\pi}\ln\left(\dfrac{b}{a}\right)$$

Standard results

• $U_B = \dfrac{1}{2}Li^2$ joule — total energy stored in an inductor.
• $u_B = \dfrac{B^2}{2\mu_0}$ joule per cubic metre — energy per unit volume at a point where the field is $B$; valid for every magnetic field.
• $u_B = \dfrac{1}{2}\mu_0n^2i^2$ inside a long solenoid.
• Comparisons: $U_E = \dfrac{1}{2}CV^2$, $u_E = \dfrac{1}{2}\varepsilon_0E^2$, $KE = \dfrac{1}{2}mv^2$.