# 5.8 Magnetic Energy

When we introduced electrostatic energy in Section 4-10, we did so by examining what happens to the energy expended in charging up a capacitor from zero voltage to some final voltage V. We introduce the concept of magnetic energy by considering an inductor with inductance L connected to a current source. Suppose that we were to increase the current i flowing through the inductor from zero to a final value I. From circuit theory, we know that the instantaneous voltage υ across the inductor is given by υ=L d i/d t. We will derive this relationship from Maxwell’s equations in Chapter 6, thereby justifying the use of the i−υ relationship for the inductor. Power p equals the product of υ and i, and the time integral of power is work, or energy. Hence, the total energy in joules (J) expended in building up a current I in the inductor is

W m=∫p d t=∫i v d t=L∫0 I i d i=1 2 L I 2(J).(5.103)

We call this the magnetic energy stored in the inductor.

To justify this association, consider the solenoid inductor. Its inductance is given by Eq. (5.95) as L=μ N 2 S/l, and the magnitude of the magnetic flux density in its interior is given by Eq. (5.90) as B=μ N I/l, implying that I=B l/(μ N). Using these expressions for L and I in Eq. (5.103), we obtain

W m=1 2 L I 2=1 2(μ N 2 l S)(B l μ N)2=1 2 B 2 μ(l S)=1 2 μ H 2 υ,(5.104)

where υ=l S is the volume of the interior of the solenoid and H=B/μ. The expression for w m suggests that the energy expended in building up the current in the inductor is stored in the magnetic field with magnetic energy density w m, defined as the magnetic energy W m per unit volume,

Furthermore, for any volume υ containing a material with permeability μ (including free space with permeability μ 0), the total magnetic energy stored in a magnetic field H is

W m=1 2∫υ μ H 2 d υ (J).(5.106)

# Example 5-10: Magnetic Energy in a Coaxial Cable

Derive an expression for the magnetic energy stored in a coaxial cable of length l and inner and outer radii a and b. The current flowing through the cable is I and its insulation material has permeability μ.

Solution: From Eq. (5.97), the magnitude of the magnetic field in the insulating material is

H=B μ=I 2 π r,

where r is the radial distance from the center of the inner conductor (Fig. 5-29). The magnetic energy stored in the coaxial cable therefore is

W m=1 2∫υ μ H 2 d υ=μ I 2 8 π 2∫υ 1 r 2 d υ.

Since H is a function of r only, we choose d υ to be a cylindrical shell of length l, radius r, and thickness d r along the radial direction. Thus, d υ=2 π r l d r and

W m=μ I 2 8 π 2∫a b 1 r 2⋅2 π r l d r=μ I 2 l 4 π ln b a=1 2 L I 2(J)

with L given by Eq. (5.99).