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== 14.3#h(0.6em)Energy in a Magnetic Field


  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 #strong[magnetic energy density],


  #math.equation(block: true, alt: "u sub m equals the fraction B squared over 2 μ sub 0")[$u_("m") = frac(B^(2), 2 μ_(0))$]
  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 #math.equation(block: false, alt: "B equals μ sub 0 n I")[$B = μ_(0) n I$] everywhere inside the solenoid. Thus, the energy stored in a solenoid or the magnetic energy density times volume is equivalent to


  #math.equation(block: true, alt: "U equals u sub m open parenthesis V close parenthesis equals the fraction open parenthesis μ sub 0 n I close parenthesis squared over 2 μ sub 0 open parenthesis A l close parenthesis equals the fraction 1 over 2 open parenthesis μ sub 0 n squared A l close parenthesis I squared .")[$U = u_("m") ( V ) = frac(attach(( μ_(0) n I ), t: 2), 2 μ_(0)) ( A l ) = frac(1, 2) ( μ_(0) n^(2) A l ) I^(2) .$]
  With the substitution of , this becomes


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "U equals the fraction 1 over 2 L I squared .")[$U = frac(1, 2) L I^(2) .$]
]


  Although derived for a special case, this equation gives the energy stored in the magnetic field of #emph[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 #math.equation(block: false, alt: "ε equals L d i / d t ,")[$ε = L d i "/" d t ,$] where #math.equation(block: false, alt: "i")[$i$] is the induced current at that instance. Therefore, the power absorbed by the inductor is


  #math.equation(block: true, alt: "P equals ε i equals L the fraction d i over d t i .")[$P = ε i = L frac(d i, d t) i .$]
  The total energy stored in the magnetic field when the current increases from 0 to #emph[I] in a time interval from 0 to #emph[t] can be determined by integrating this expression:


  #math.equation(block: true, alt: "U equals ∫ 0 t P d t prime equals ∫ 0 t L the fraction d i over d t prime i d t prime equals L ∫ 0 I i d i equals the fraction 1 over 2 L I squared .")[$U = display(∫_(0)^(t) P d t^(′) =) display(∫_(0)^(t) L frac(d i, d t^(′)) i d t^(′) =) L display(∫_(0)^(I) i d i =) frac(1, 2) L I^(2) .$]
  #examplebox("Example 1")[Self-Inductance of a Coaxial Cable][
shows two long, concentric cylindrical shells of radii #math.equation(block: false, alt: "R sub 1")[$R_(1)$] and #math.equation(block: false, alt: "R sub 2 .")[$R_(2) .$] As discussed in Capacitance on capacitance, this configuration is a simplified representation of a #strong[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(figph[Figure a shows two concentrically arranged hollow cylinders. The radius of the inner one is R1 and that of the outer one is R2. Figure 2 shows a dotted circle with radius r in between the two cylinders. Figure c shows a cylinder of length and radius r in between the two cylinders. Its thickness is dr.], alt: "Figure a shows two concentrically arranged hollow cylinders. The radius of the inner one is R1 and that of the outer one is R2. Figure 2 shows a dotted circle with radius r in between the two cylinders. Figure c shows a cylinder of length and radius r in between the two cylinders. Its thickness is dr.", caption: [(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 #emph[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  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 .


    Solution


    + We determine the magnetic field between the conductors by applying Ampère’s law to the dashed circular path shown in (b). Because of the cylindrical symmetry, #math.equation(block: false, alt: "B →")[$arrow(B)$] is constant along the path, and
#math.equation(block: true, alt: "∮ B → times d l → equals B open parenthesis 2 π r close parenthesis equals μ sub 0 I .")[$display(∮ arrow(B) · d arrow(l) = B ( 2 π r )) = μ_(0) I .$]
This gives us
#math.equation(block: true, alt: "B equals the fraction μ sub 0 I over 2 π r .")[$B = frac(μ_(0) I, 2 π r) .$]
In the region outside the cable, a similar application of Ampère’s law shows that #math.equation(block: false, alt: "B equals 0")[$B = 0$], since no net current crosses the area bounded by a circular path where #math.equation(block: false, alt: "r greater than R sub 2 .")[$r > R_(2) .$] This argument also holds when #math.equation(block: false, alt: "r less than R sub 1 ;")[$r < R_(1) ;$] 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
#math.equation(block: true, alt: "u sub m equals the fraction B squared over 2 μ sub 0")[$u_("m") = frac(B^(2), 2 μ_(0))$]
the energy stored in a cylindrical shell of inner radius #emph[r], outer radius #math.equation(block: false, alt: "r plus d r ,")[$r + d r ,$] and length #emph[l] (see part (c) of the figure) is
#math.equation(block: true, alt: "u sub m equals the fraction μ sub 0 I squared over 8 π squared r squared .")[$u_("m") = frac(μ_(0) I^(2), 8 π^(2) r^(2)) .$]
Thus, the total energy of the magnetic field in a length #emph[l] of the cable is
#math.equation(block: true, alt: "U equals ∫ R sub 1 R sub 2 d U equals ∫ R sub 1 R sub 2 the fraction μ sub 0 I squared over 8 π squared r squared open parenthesis 2 π r l close parenthesis d r equals the fraction μ sub 0 I squared l over 4 π ln the fraction R sub 2 over R sub 1 ,")[$U = display(∫_(R_(1))^(R_(2)) d U) = display(∫_(R_(1))^(R_(2)) frac(μ_(0) I^(2), 8 π^(2) r^(2)) ( 2 π r l ) d r) = frac(μ_(0) I^(2) l, 4 π) #h(0.2em) "ln" #h(0.2em) frac(R_(2), R_(1)) ,$]
and the energy per unit length is #math.equation(block: false, alt: "open parenthesis μ sub 0 I squared / 4 π close parenthesis ln open parenthesis R sub 2 / R sub 1 close parenthesis")[$( μ_(0) I^(2) "/" 4 π ) "ln" ( R_(2) "/" R_(1) )$].
+ From ,
#math.equation(block: true, alt: "U equals the fraction 1 over 2 L I squared ,")[$U = frac(1, 2) L I^(2) ,$]
where #emph[L] is the self-inductance of a length #emph[l] of the coaxial cable. Equating the previous two equations, we find that the self-inductance per unit length of the cable is
#math.equation(block: true, alt: "the fraction L over l equals the fraction μ sub 0 over 2 π ln the fraction R sub 2 over R sub 1 .")[$frac(L, l) = frac(μ_(0), 2 π) #h(0.2em) "ln" #h(0.2em) frac(R_(2), R_(1)) .$]


    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 #math.equation(block: false, alt: "open parenthesis R sub 2 close parenthesis")[$( R_(2) )$] or decrease the inner radius #math.equation(block: false, alt: "open parenthesis R sub 1 close parenthesis")[$( R_(1) )$]. 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.
]


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
How much energy is stored in the inductor of  after the current reaches its maximum value?


    
    #solutionbox[
0.50 J
]
]


  
    
=== Summary


    - The energy stored in an inductor #emph[U] is
#math.equation(block: true, alt: "U equals the fraction 1 over 2 L I squared .")[$U = frac(1, 2) L I^(2) .$]
- The self-inductance per unit length of coaxial cable is 
#math.equation(block: true, alt: "the fraction L over l equals the fraction μ sub 0 over 2 π ln the fraction R sub 2 over R sub 1 .")[$frac(L, l) = frac(μ_(0), 2 π) #h(0.2em) "ln" #h(0.2em) frac(R_(2), R_(1)) .$]


  
  
    
=== Conceptual Questions


    
      Show that #math.equation(block: false, alt: "L I squared / 2")[$L I^(2) "/" 2$] has units of energy.


    
  
  
    
=== Problems


    
      At the instant a current of 0.20 A is flowing through a coil of wire, the energy stored in its magnetic field is #math.equation(block: false, alt: "6.0 times 10 to the power −3 J .")[$6.0 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "J" .$] What is the self-inductance of the coil?


    
    
      Suppose that a rectangular toroid has 2000 windings and a self-inductance of 0.040 H. If #math.equation(block: false, alt: "h equals 0.10 m")[$h = 0.10 #h(0.2em) "m"$], what is the current flowing through a rectangular toroid when the energy in its magnetic field is #math.equation(block: false, alt: "2.0 times 10 to the power −6 J ?")[$2.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "J" ?$]


    
    #solutionbox[
0.01 A
]


    
      Solenoid #emph[A] is tightly wound while solenoid #emph[B] has windings that are evenly spaced with a gap equal to the diameter of the wire. The solenoids are otherwise identical. Determine the ratio of the energies stored per unit length of these solenoids when the same current flows through each.


    
    
      A 10-H inductor carries a current of 20 A. How much ice at #math.equation(block: false, alt: "0 ° C")[$0 "°" #h(0.2em) "C"$] could be melted by the energy stored in the magnetic field of the inductor? (#emph[Hint]: Use the value #math.equation(block: false, alt: "L sub f equals 334 J/g")[$L_("f") = 334 #h(0.2em) "J/g"$] for ice.)


    
    #solutionbox[
6.0 g
]


    
      A coil with a self-inductance of 3.0 H and a resistance of #math.equation(block: false, alt: "100 Ω")[$100 #h(0.2em) "Ω"$] carries a steady current of 2.0 A. (a) What is the energy stored in the magnetic field of the coil? (b) What is the energy per second dissipated in the resistance of the coil?


    
    
      A current of 1.2 A is flowing in a coaxial cable whose outer radius is five times its inner radius. What is the magnetic field energy stored in a 3.0-m length of the cable?


    
    #solutionbox[
#math.equation(block: true, alt: "U sub m equals 7.0 times 10 to the power −7 J")[$U_("m") = 7.0 #h(0.2em) × #h(0.2em) 10^(−7) #h(0.2em) "J"$]
]