14.3 Energy in a magnetic field  (Page 9/2)

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    ,

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

With the substitution of [link] , this becomes

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 $\epsilon =Ldi\text{/}dt,$ so the power absorbed by the inductor is

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:

Self-inductance of a coaxial cable

Strategy

Solution

1. We determine the magnetic field between the conductors by applying Ampère’s law to the dashed circular path shown in [link] (b). Because of the cylindrical symmetry, $\overrightarrow{B}$ is constant along the path, and
   
   ${\displaystyle \oint \overrightarrow{B}·d\overrightarrow{l}=B(2\pi r)}={\mu }_{0}I.$
   
   This gives us
   
   $B=\frac{{\mu }_{0}I}{2\pi r}.$
   
   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
   
   $u_{\text{m}}=\frac{B^2}{2{\mu }_0}=\frac{{\mu }_0{I}^2}{8{\pi }^2{r}^2},$
   
   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
   
   $u_{\text{m}}=\frac{B^2}{2{\mu }_0}=\frac{{\mu }_0{I}^2}{8{\pi }^2{r}^2}.$
   
   Thus, the total energy of the magnetic field in a length l of the cable is
   
   $U={\displaystyle {\int }_{R_1}^{R_2}dU}={\displaystyle {\int }_{R_1}^{R_2}\frac{{\mu }_0{I}^2}{8{\pi }^2{r}^2}(2\pi rl)dr}=\frac{{\mu }_0{I}^{2}l}{4\pi }\text{ln}\frac{R_2}{R_1},$
   
   and the energy per unit length is $({\mu }_0{I}^2\text{/}4\pi )\text{ln}(R_2\text{/}{R}_1)$ .
2. From [link] ,
   
   $U=\frac{1}{2}L{I}^2,$
   
   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
   
   $\frac{L}{l}=\frac{{\mu }_0}{2\pi }\text{ln}\frac{R_2}{R_1}.$

Significance

Summary

• The energy stored in an inductor U is
  
  $U=\frac{1}{2}L{I}^2.$
• The self-inductance per unit length of coaxial cable is
  
  $\frac{L}{l}=\frac{{\mu }_0}{2\pi }\text{ln}\frac{R_2}{R_1}.$

Conceptual questions

Problems