Chapter 11 Inductance and Magnetic Energy 11.1 Mutual Inductance........... 11-3 Example 11.1 Mutual Inductance of Two Concentric Co-planar Loops...............11-5 11.2 Self-Inductance.. 11-6 Example 11.2 Self-Inductance ofa Solenoid..11-6 Example 11.3 Self-Inductance of a Toroid.............. …11-7 Example 11.4 Mutual Inductance of a Coil Wrapped Around a Solenoid .11-9 11.3 Energy Stored in Magnetic Fields... .11-10 Example 11.5 Energy Stored in a Solenoid.......... 11-11 11.3.1 Creating and Destroying Magnetic Energy Animation 11-12 11.4 RL Circuits… 11-13 11.4.1 Self-Inductance and the Faraday's Law. 11-13 11.4.2 Kirchhoff's Loop Rule Modified for Inductors:a Warning 11-17 11.4.3 Example Voltmeter Readings with Time Changing Magnetic Fields...11-18 11.5 How can the Electric Field in an Inductor be Zero?... 11-21 11.6 Modified Kirchoffs Law (Misleading,see Section 11.4.2).................. 11-25 11.6.1 Rising Current............. .11-27 11.6.2 Decaying Current.... 11-29 11.7 LC Oscillations........... 11-30 11.8 The RLC Series Circuit.................. 11-34 11.9 Summary. 11-37 11.10 Appendix 1:General Solutions for the RLC Series Circuit. 11-39 11.10.1 Quality Factor.......... 11-41 11.11 Appendix 2:Stresses Transmitted by Magnetic Fields. 11-43 11.12 Problem-Solving Strategies....... 11-44 11.12.1 Calculating Self-Inductance.................. 11-44 11.12.2 Circuits containing inductors 。 11-45 11.13 Solved Problems................... 11-45 11.13.1 Energy stored in a toroid............. 11-45 11.13.2 Magnetic Energy Density 11-46 11.13.3 Mutual Inductance...... 11-47 11-1
11-1 Chapter 11 Inductance and Magnetic Energy 11.1 Mutual Inductance ........................................................................................... 11-3 Example 11.1 Mutual Inductance of Two Concentric Co-planar Loops............... 11-5 11.2 Self-Inductance ................................................................................................ 11-6 Example 11.2 Self-Inductance of a Solenoid......................................................... 11-6 Example 11.3 Self-Inductance of a Toroid............................................................ 11-7 Example 11.4 Mutual Inductance of a Coil Wrapped Around a Solenoid ............ 11-9 11.3 Energy Stored in Magnetic Fields.................................................................. 11-10 Example 11.5 Energy Stored in a Solenoid ......................................................... 11-11 11.3.1 Creating and Destroying Magnetic Energy Animation ......................... 11-12 11.4 RL Circuits..................................................................................................... 11-13 11.4.1 Self-Inductance and the Faraday’s Law................................................. 11-13 11.4.2 Kirchhoff's Loop Rule Modified for Inductors: a Warning................... 11-17 11.4.3 Example Voltmeter Readings with Time Changing Magnetic Fields... 11-18 11.5 How can the Electric Field in an Inductor be Zero? ...................................... 11-21 11.6 Modified Kirchoff’s Law (Misleading, see Section 11.4.2).......................... 11-25 11.6.1 Rising Current........................................................................................ 11-27 11.6.2 Decaying Current................................................................................... 11-29 11.7 LC Oscillations .............................................................................................. 11-30 11.8 The RLC Series Circuit .................................................................................. 11-34 11.9 Summary........................................................................................................ 11-37 11.10 Appendix 1: General Solutions for the RLC Series Circuit ........................... 11-39 11.10.1 Quality Factor ........................................................................................ 11-41 11.11 Appendix 2: Stresses Transmitted by Magnetic Fields.................................. 11-43 11.12 Problem-Solving Strategies ........................................................................... 11-44 11.12.1 Calculating Self-Inductance................................................................... 11-44 11.12.2 Circuits containing inductors................................................................. 11-45 11.13 Solved Problems ............................................................................................ 11-45 11.13.1 Energy stored in a toroid........................................................................ 11-45 11.13.2 Magnetic Energy Density ...................................................................... 11-46 11.13.3 Mutual Inductance ................................................................................. 11-47
11.13.4 RL Circuit.… 11-48 11.13.5 RL Circuit.11-50 11.13.6 LC Circuit......... 11-52 11.14 Conceptual Questions................. 11-53 11.15 Additional Problems.................... 11-54 11.15.1 Solenoid...... 11-54 11.15.2 Self-Inductance... 11-54 11.15.3 Coupled Inductors... 11-54 11.15.4 RL Circuit. 11-55 11.15.5 RL Circuit........... 11-56 11.15.6 Inductance of a Solenoid With and Without Iron Core.........................11-56 11.15.7 RLC Circuit.. 11-57 11.15.8 Spinning Cylinder 11-58 11.15.9 Spinning Loop... 11-59 11.15.10 LC Circuit… 11-59 11-2
11-2 11.13.4 RL Circuit............................................................................................... 11-48 11.13.5 RL Circuit............................................................................................... 11-50 11.13.6 LC Circuit .............................................................................................. 11-52 11.14 Conceptual Questions .................................................................................... 11-53 11.15 Additional Problems ...................................................................................... 11-54 11.15.1 Solenoid ................................................................................................. 11-54 11.15.2 Self-Inductance ...................................................................................... 11-54 11.15.3 Coupled Inductors.................................................................................. 11-54 11.15.4 RL Circuit............................................................................................... 11-55 11.15.5 RL Circuit............................................................................................... 11-56 11.15.6 Inductance of a Solenoid With and Without Iron Core ......................... 11-56 11.15.7 RLC Circuit............................................................................................ 11-57 11.15.8 Spinning Cylinder.................................................................................. 11-58 11.15.9 Spinning Loop........................................................................................ 11-59 11.15.10 LC Circuit .......................................................................................... 11-59
Inductance and Magnetic Energy 11.1 Mutual Inductance Suppose two coils are placed near each other,as shown in Figure 11.1.1 Coil2 N,Φ12 Coil 1 Figure 11.1.1 Changing current in coil 1 produces changing magnetic flux in coil 2. The first coil has N turns and carries a current I which gives rise to a magnetic field B.The second coil has N,turns.Because the two coils are close to each other,some of the magnetic field lines through coil 1 will also pass through coil 2.Let denote the magnetic flux through one turn of coil 2 due toI.Now,by varying I with time,there will be an induced emf associated with the changing magnetic flux in the second coil: -水9 (11.1.1) coil 2 The time rate of change of magnetic flux in coil 2 is proportional to the time rate of change of the current in coil 1: d④2=Mnd N2 dt i (11.1.2) where the proportionality constant M is called the mutual inductance.It can also be written as N2Φ2 M2= (11.1.3) 11-3
11-3 Inductance and Magnetic Energy 11.1 Mutual Inductance Suppose two coils are placed near each other, as shown in Figure 11.1.1 Figure 11.1.1 Changing current in coil 1 produces changing magnetic flux in coil 2. The first coil has N1 turns and carries a current I 1 which gives rise to a magnetic field ! B1 . The second coil has N2 turns. Because the two coils are close to each other, some of the magnetic field lines through coil 1 will also pass through coil 2. Let !12 denote the magnetic flux through one turn of coil 2 due to I 1 . Now, by varying I 1 with time, there will be an induced emf associated with the changing magnetic flux in the second coil: ! 12 = "N2 d#12 dt = " d dt ! B1 $ d ! A2 coil 2 %% . (11.1.1) The time rate of change of magnetic flux !12 in coil 2 is proportional to the time rate of change of the current in coil 1: N2 d!12 dt = M12 dI1 dt , (11.1.2) where the proportionality constant M12 is called the mutual inductance. It can also be written as M12 = N2 !12 I 1 . (11.1.3)
The SI unit for inductance is the henry [H]: 1 henry =1H=1T.m/A. (11.1.4) We shall see that the mutual inductance M,depends only on the geometrical properties of the two coils such as the number of turns and the radii of the two coils. In a similar manner,suppose instead there is a current 1,in the second coil and it is varying with time (Figure 11.1.2).Then the induced emf in coil 1 becomes d=-dfB,d瓜 =-N dt di (11.1.5) coil I and a current is induced in coil 1. B2 Coil2 Coil 1 NΦ2 Figure 11.1.2 Changing current in coil 2 produces changing magnetic flux in coil 1. This changing flux in coil I is proportional to the changing current in coil 2, d=Ma dt N、dt (11.1.6) where the proportionality constant M is another mutual inductance and can be written as NΦL M= (11.1.7) 12 The mutual inductance reciprocity theorem states that the constants are equal M2=M≡M. (111.8) 11-4
11-4 The SI unit for inductance is the henry [H]: 1 henry = 1 H = 1 T!m2 /A . (11.1.4) We shall see that the mutual inductance M12 depends only on the geometrical properties of the two coils such as the number of turns and the radii of the two coils. In a similar manner, suppose instead there is a current I2 in the second coil and it is varying with time (Figure 11.1.2). Then the induced emf in coil 1 becomes ! 21 = "N1 d#21 dt = " d dt ! B2 $ d ! A1 coil 1 %% , (11.1.5) and a current is induced in coil 1. Figure 11.1.2 Changing current in coil 2 produces changing magnetic flux in coil 1. This changing flux in coil 1 is proportional to the changing current in coil 2, N1 d!21 dt = M21 dI2 dt , (11.1.6) where the proportionality constant M21 is another mutual inductance and can be written as M21 = N1 !21 I2 . (11.1.7) The mutual inductance reciprocity theorem states that the constants are equal M12 = M21 ! M . (11.1.8)
We shall not prove this theorem.It's left as a difficult exercise for the reader to prove this result using Ampere's law and the Biot-Savart law. Example 11.1 Mutual Inductance of Two Concentric Co-planar Loops Consider two single-turn co-planar,concentric coils of radii R and R,,with R>>R, as shown in Figure 11.1.3.What is the mutual inductance between the two loops? Figure 11.1.3 Two concentric current loops Solution:The mutual inductance can be computed as follows.Using Eq.(9.1.15)of Chapter 9,we see that the magnitude of the magnetic field at the center of the ring due to I in the outer coil is given by B= Lo (11.1.9) 2R Because R>>R,,we approximate the magnetic field through the entire inner coil by B.Hence,the flux through the second (inner)coil is =B4=π=LR 2R (11.1.10) 2R Thus,the mutual inductance is given by M= 2=4π及 11.1.11) 112R The result shows that M depends only on the geometrical factors,R and R,and is independent of the current I,in the coil. 11-5
11-5 We shall not prove this theorem. It’s left as a difficult exercise for the reader to prove this result using Ampere’s law and the Biot-Savart law. Example 11.1 Mutual Inductance of Two Concentric Co-planar Loops Consider two single-turn co-planar, concentric coils of radii R1 and R2 , with R1 >> R2 , as shown in Figure 11.1.3. What is the mutual inductance between the two loops? Figure 11.1.3 Two concentric current loops Solution: The mutual inductance can be computed as follows. Using Eq. (9.1.15) of Chapter 9, we see that the magnitude of the magnetic field at the center of the ring due to I 1 in the outer coil is given by B1 = µ0 I 1 2R1 . (11.1.9) Because R1 >> R2 , we approximate the magnetic field through the entire inner coil by B1 . Hence, the flux through the second (inner) coil is !12 = B1A2 = µ0 I 1 2R1 " R2 2 = µ0 " I 1R2 2 2R1 . (11.1.10) Thus, the mutual inductance is given by M = !12 I 1 = µ0 " R2 2 2R1 (11.1.11) The result shows that M depends only on the geometrical factors, R1 and R2 , and is independent of the current I 1 in the coil
11.2 Self-Inductance Consider again a coil consisting of N turns and carrying current I in the counterclockwise direction,as shown in Figure 11.2.1.If the current is steady,then the magnetic flux through the loop will remain constant.However,suppose the current I changes with time,then according to Faraday's law,an induced emf will arise to oppose the change.The induced current will flow clockwise if dl/dt >0,and counterclockwise if dl/dt <0.The property of the loop in which its own magnetic field opposes any change in current is called self-inductance,and the emf generated is called the self- induced emf or back emf,which we denote as g,.The self-inductance may arise from a coil and the rest of the circuit,especially the connecting wires. Figure 11.2.1 Magnetic flux through a coil Mathematically,the self-induced emf can be written as dΦ e,=-N B.tum =N- dBd瓜 (11.2.1) d Because the flux is proportional to the current I,we can also express this relationship by dl e,=-L (11.2.2) di where the constant L is called the self-inductance.The two expressions can be combined to yield L= NOB (11.2.3) Physically,the self-inductance L is a measure of an inductor's"resistance"to the change of current;the larger the value of L,the lower the rate of change of current. Example 11.2 Self-Inductance of a Solenoid 11-6
11-6 11.2 Self-Inductance Consider again a coil consisting of N turns and carrying current I in the counterclockwise direction, as shown in Figure 11.2.1. If the current is steady, then the magnetic flux through the loop will remain constant. However, suppose the current I changes with time, then according to Faraday’s law, an induced emf will arise to oppose the change. The induced current will flow clockwise if dI / dt > 0 , and counterclockwise if dI / dt < 0 . The property of the loop in which its own magnetic field opposes any change in current is called self-inductance, and the emf generated is called the selfinduced emf or back emf, which we denote as ! L . The self-inductance may arise from a coil and the rest of the circuit, especially the connecting wires. Figure 11.2.1 Magnetic flux through a coil Mathematically, the self-induced emf can be written as ! L = "N d#B,turn dt = "N d dt ! B$d ! A turn %% . (11.2.1) Because the flux is proportional to the current I , we can also express this relationship by ! L = "L dI dt . (11.2.2) where the constant L is called the self-inductance. The two expressions can be combined to yield L = N!B I . (11.2.3) Physically, the self-inductance L is a measure of an inductor’s “resistance” to the change of current; the larger the value of L , the lower the rate of change of current. Example 11.2 Self-Inductance of a Solenoid
Compute the self-inductance of a solenoid with N turns,length 1,and radius R with a current I flowing through each turn,as shown in Figure 11.2.2. N turns Figure 11.2.2 Solenoid Solution:Ignoring edge effects and applying Ampere's law,the magnetic field inside a solenoid is given by Eq.(9.4.3) B=,M求=4lk, (11.2.4) where n=N//is the number of turns per unit length.The magnetic flux through each turn is Φg=BA=nl(πR2)=4nIπR (11.2.5) Thus,the self-inductance is L-Nou-HnTRI. (11.2.6) We see that L depends only on the geometrical factors(n,R and 1)and is independent of the current / Example 11.3 Self-Inductance of a Toroid (a) (b) Figure 11.2.3 A toroid with N turns 11-7
11-7 Compute the self-inductance of a solenoid with N turns, length l , and radius R with a current I flowing through each turn, as shown in Figure 11.2.2. Figure 11.2.2 Solenoid Solution: Ignoring edge effects and applying Ampere’s law, the magnetic field inside a solenoid is given by Eq. (9.4.3) ! B = µ0NI l kˆ = µ0nI kˆ , (11.2.4) where n = N / l is the number of turns per unit length. The magnetic flux through each turn is !B = BA = µ0nI "(# R2 ) = µ0nI# R2 . (11.2.5) Thus, the self-inductance is L = N!B I = µ0n 2 " R2 l . (11.2.6) We see that L depends only on the geometrical factors ( n , R and l ) and is independent of the current I . Example 11.3 Self-Inductance of a Toroid (a) (b) Figure 11.2.3 A toroid with N turns
Calculate the self-inductance L of a toroid,which consists of N turns and has a rectangular cross section,with inner radius a,outer radius b,and height h,as shown in Figure 11.2.3(a). Solution:According to Ampere's law (Example 9.5), ∮B.ds=∮Bds=Bdds=B(2πr)=4,NI. (11.2.7) The magnitude of the magnetic field inside the torus is given by B=4oNI (11.2.8) 2πr The magnetic flux through one turn of the toroid may be obtained by integrating over the rectangular cross section,with d4=hdr as the differential area element (Figure 11.2.3b), (11.2.9) The total flux is N.Therefore,the self-inductance is L= b -In (11.2.10) 2π Again,the self-inductance L depends only on the geometrical factors.Let's consider the situation where a>>b-a.In this limit,the logarithmic term in the equation above may be expanded as (11.2.11) and the self-inductance becomes L-MoN'h.b-aNA BoN'A (11.2.12) 2π a2πa 19 where 4=h(b-a)is the cross-sectional area,and 1=2xa.We see that the self- inductance of the toroid in this limit has the same form as that of a solenoid. 11-8
11-8 Calculate the self-inductance L of a toroid, which consists of N turns and has a rectangular cross section, with inner radius a , outer radius b, and height h , as shown in Figure 11.2.3(a). Solution: According to Ampere’s law (Example 9.5), ! B! d ! s "" = Bds "" = B ds = "" B(2#r) = µ0NI . (11.2.7) The magnitude of the magnetic field inside the torus is given by B = µ0NI 2!r . (11.2.8) The magnetic flux through one turn of the toroid may be obtained by integrating over the rectangular cross section, with dA = hdr as the differential area element (Figure 11.2.3b), !B = ! B" d ! ## A = µ0NI 2$r % & ' ( ) * hdr a b # = µ0NIh 2$ ln b a % & ' ( ) * . (11.2.9) The total flux is N!B . Therefore, the self-inductance is L = N!B I = µ0N2 h 2" ln b a # $ % & ' ( . (11.2.10) Again, the self-inductance L depends only on the geometrical factors. Let’s consider the situation where a >> b ! a . In this limit, the logarithmic term in the equation above may be expanded as ln b a ! " # $ % & = ln 1+ b ' a a ! " # $ % & ( b ' a a , (11.2.11) and the self-inductance becomes L ! µ0N2 h 2" # b $ a a = µ0N2 A 2"a = µ0N2 A l , (11.2.12) where A = h(b ! a) is the cross-sectional area, and l = 2!a . We see that the selfinductance of the toroid in this limit has the same form as that of a solenoid
Example 11.4 Mutual Inductance of a Coil Wrapped Around a Solenoid A long solenoid with length and a cross-sectional area 4 consists of N,turns of wire. An insulated coil of N,turns is wrapped around it,as shown in Figure 11.2.4. (a)Calculate the mutual inductance M,assuming that all the flux from the solenoid passes through the outer coil. (b)Relate the mutual inductance M to the self-inductances L and L,of the solenoid and the coil. Figure 11.2.4 A coil wrapped around a solenoid Solutions: (a)The magnetic flux through each turn of the outer coil due to the solenoid is ①=BA=44A (11.2.13) 1 where B=uoNI/I is the uniform magnetic field inside the solenoid.Thus,the mutual inductance is M-N LN N,4 (11.2.14) (b)From Example 11.2,we see that the self-inductance of the solenoid with N,turns is given by Φ=44 L= 1 1, (11.2.15) where is the magnetic flux through one turn of the inner solenoid due to the magnetic field produced by I.Similarly,we have L=uN2A/I for the outer coil.In terms of L and L,,the mutual inductance can be written as 11-9
11-9 Example 11.4 Mutual Inductance of a Coil Wrapped Around a Solenoid A long solenoid with length l and a cross-sectional area A consists of N1 turns of wire. An insulated coil of N2 turns is wrapped around it, as shown in Figure 11.2.4. (a) Calculate the mutual inductance M , assuming that all the flux from the solenoid passes through the outer coil. (b) Relate the mutual inductance M to the self-inductances L1 and L2 of the solenoid and the coil. Figure 11.2.4 A coil wrapped around a solenoid Solutions: (a) The magnetic flux through each turn of the outer coil due to the solenoid is !12 = BA = µ0N1I 1 l A. (11.2.13) where B = µ0N1I 1 / l is the uniform magnetic field inside the solenoid. Thus, the mutual inductance is M = N2 !12 I 1 = µ0N1N2A l . (11.2.14) (b) From Example 11.2, we see that the self-inductance of the solenoid with N1 turns is given by L1 = N1 !11 I 1 = µ0N1 2 A l , (11.2.15) where !11 is the magnetic flux through one turn of the inner solenoid due to the magnetic field produced by I 1 . Similarly, we have L2 = µ0N2 2 A / l for the outer coil. In terms of L1 and L2 , the mutual inductance can be written as
M=VL凸 (11.2.16) More generally the mutual inductance is given by M=kWLL2,0≤k≤1， (11.2.17) where k is the coupling coefficient.In our example,we have k =1,which means that all of the magnetic flux produced by the solenoid passes through the outer coil,and vice versa,in this idealization. 11.3 Energy Stored in Magnetic Fields Because an inductor in a circuit serves to oppose any change in the current through it, work must be done by an external source such as a battery in order to establish a current in the inductor.From the work-energy theorem,we conclude that energy can be stored in an inductor.The role played by an inductor in the magnetic case is analogous to that of a capacitor in the electric case. The power,or rate at which an external emf works to overcome the self-induced emf E and pass current I in the inductor is dw PL=dt est =IEext (11.3.1) If only the external emf and the inductor are present,then =which implies that P dm=-1e,=+L l (11.3.2) d If the current is increasing with dl/dt>0,then P>0,which means that the external source is doing positive work to transfer energy to the inductor.Thus,the internal energy U of the inductor is increased.On the other hand,if the current is decreasing with dl/dt <0,we then have P<0.In this case,the external source takes energy away from the inductor,causing its internal energy to decrease.The total work done by the external source to increase the current form zero to is then w.-Sd-irdr-i (11.3.3) This is equal to the magnetic energy stored in the inductor, 11-10
11-10 M = L1L2 . (11.2.16) More generally the mutual inductance is given by M = k L1L2 , 0 ! k ! 1, (11.2.17) where k is the coupling coefficient. In our example, we have k = 1, which means that all of the magnetic flux produced by the solenoid passes through the outer coil, and vice versa, in this idealization. 11.3 Energy Stored in Magnetic Fields Because an inductor in a circuit serves to oppose any change in the current through it, work must be done by an external source such as a battery in order to establish a current in the inductor. From the work-energy theorem, we conclude that energy can be stored in an inductor. The role played by an inductor in the magnetic case is analogous to that of a capacitor in the electric case. The power, or rate at which an external emf ! ext works to overcome the self-induced emf ! L and pass current I in the inductor is PL = dWext dt = I! ext . (11.3.1) If only the external emf and the inductor are present, then ! ext = "! L which implies that PL = dWext dt = !I" L = +IL dI dt . (11.3.2) If the current is increasing with dI / dt > 0 , then P > 0 , which means that the external source is doing positive work to transfer energy to the inductor. Thus, the internal energy UB of the inductor is increased. On the other hand, if the current is decreasing with dI / dt < 0 , we then have P < 0 . In this case, the external source takes energy away from the inductor, causing its internal energy to decrease. The total work done by the external source to increase the current form zero to I is then Wext = dWext = ! LI 'dI ' 0 I ! = 1 2 LI 2 . (11.3.3) This is equal to the magnetic energy stored in the inductor