$$\require{cancel}$$

# 14.4: Energy in a Magnetic Field

• • Contributed by OpenStax
• General Physics at OpenStax CNX

Learning Objectives

By the end of this section, you will be able to:

• Explain how energy can be stored in a magnetic field
• Derive the equation for energy stored in a coaxial cable given the magnetic energy density

The energy of a capacitor is stored in the electric field between its plates. Similarly, an inductor has the capability to store energy, but in its magnetic field. This energy can be found by integrating the magnetic energy density,

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

over the appropriate volume. To understand where this formula comes from, let’s consider the long, cylindrical solenoid of the previous section. Again using the infinite solenoid approximation, we can assume that the magnetic field is essentially constant and given by $$B = \mu_0 nI$$ everywhere inside the solenoid. Thus, the energy stored in a solenoid or the magnetic energy density times volume is equivalent to

$U = u_m(V) = \dfrac{(\mu_0nI)^2}{2\mu_0}(Al) = \dfrac{1}{2}(\mu_0n^2Al)I^2. \label{14.19}$

With the substitution of Equation 14.3.12, this becomes

$\boxed{U = \dfrac{1}{2}LI^2.}$

Although derived for a special case, this equation gives the energy stored in the magnetic field of any inductor. We can see this by considering an arbitrary inductor through which a changing current is passing. At any instant, the magnitude of the induced emf is $$\epsilon = Ldi/dt$$, where i is the induced current at that instance. Therefore, the power absorbed by the inductor is

$P = \epsilon i = L\dfrac{di}{dt}i.$

The total energy stored in the magnetic field when the current increases from 0 to I in a time interval from 0 to t can be determined by integrating this expression:

$U = \int_0^t Pdt' = \int_0^t L\dfrac{di}{dt'}idt' = L\int_0^l idi = \dfrac{1}{2}LI^2. \label{14.22}$

Example $$\PageIndex{1}$$: Self-Inductance of a Coaxial Cable

Figure $$\PageIndex{1}$$ shows two long, concentric cylindrical shells of radii $$R_1$$ and $$R_2$$. As discussed in Capacitance on capacitance, this configuration is a simplified representation of a coaxial cable. The capacitance per unit length of the cable has already been calculated. Now (a) determine the magnetic energy stored per unit length of the coaxial cable and (b) use this result to find the self-inductance per unit length of the cable. Figure $$\PageIndex{1}$$: (a) A coaxial cable is represented here by two hollow, concentric cylindrical conductors along which electric current flows in opposite directions. (b) The magnetic field between the conductors can be found by applying Ampère’s law to the dashed path. (c) The cylindrical shell is used to find the magnetic energy stored in a length l of the cable.

Strategy

The magnetic field both inside and outside the coaxial cable is determined by Ampère’s law. Based on this magnetic field, we can use Equation \ref{14.22} to calculate the energy density of the magnetic field. The magnetic energy is calculated by an integral of the magnetic energy density times the differential volume over the cylindrical shell. After the integration is carried out, we have a closed-form solution for part (a). The self-inductance per unit length is determined based on this result and Equation \ref{14.22}.

Solution

1. We determine the magnetic field between the conductors by applying Ampère’s law to the dashed circular path shown in Figure $$\PageIndex{1b}$$. Because of the cylindrical symmetry, $$\vec{B}$$ is constant along the path, and $\oint \vec{B} \cdot d\vec{l} = B(2\pi r) = \mu_0 I.$ This gives us $B = \dfrac{\mu_0I}{2\pi r}. \nonumber$ In the region outside the cable, a similar application of Ampère’s law shows that $$B = 0$$, since no net current crosses the area bounded by a circular path where $$r > R_2$$. This argument also holds when $$r < R_1$$; that is, in the region within the inner cylinder. All the magnetic energy of the cable is therefore stored between the two conductors. Since the energy density of the magnetic field is $u_m = \dfrac{B^2}{2\mu_0}\nonumber$ the energy stored in a cylindrical shell of inner radius r, outer radius $$r + dr$$ and length l (see part (c) of the figure) is $u_m = \dfrac{\mu_0I^2}{8\pi^2 r^2}.\nonumber$ Thus, the total energy of the magnetic field in a length l of the cable is $U = \int_{R_1}^{R_2} dU = \int_{R_1}^{R_2} \dfrac{\mu_0I^2}{8\pi^2r^2}(2\pi rl)dr = \dfrac{\mu_0I^2l}{4\pi} ln \dfrac{R_2}{R_1},\nonumber$ and the energy per unit length is $$(\mu_0I^2 / 4\pi)ln(R_2/R_1).$$
2. From Equation \ref{14.22}, $U = \dfrac{1}{2}LI^2,$ where L is the self-inductance of a length l of the coaxial cable. Equating the previous two equations, we find that the self-inductance per unit length of the cable is $\dfrac{L}{l} = \dfrac{\mu_0}{2\pi}ln\dfrac{R_2}{R_1}.\nonumber$

Significance

The inductance per unit length depends only on the inner and outer radii as seen in the result. To increase the inductance, we could either increase the outer radius ($$R_2$$) or decrease the inner radius ($$R_1$$). In the limit as the two radii become equal, the inductance goes to zero. In this limit, there is no coaxial cable. Also, the magnetic energy per unit length from part (a) is proportional to the square of the current.

Exercise $$\PageIndex{1}$$

How much energy is stored in the inductor of Example 14.3.1 after the current reaches its maximum value?

Solution

0.50 J