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