15. inductance - electrical, computer & energy...

15. Inductance

. A time-varying current in one wire loop induces an emf in another loop. This is called

mutual induction, or magnetic coupling. The electromagnetic parameter that enables a simpleevaluation of this emf in linear media is the mutual inductance. A single wire loop with time-varying current creates a time-varying induced electric field along itself, resulting in an inducedemf in the loop, an effect knownas self-induction.In linear media, this emf is evaluated interms of the electromagnetic parameter known as the inductance, or self-inductance, of theloop.

. The emf induced in a loop C2 by a current il(t) in a nearby loop C1 is given by

eI2(t) = 1 Elind(t). dl2!c2

(V), (15.1)

where E1ind(t) is the induced electric field of current il(t) along loop C2. In linear media themagnetic flux of current i1(t) through C2 is proportional to il(t), .

~12(t) = L12i1(t) (Wb). (15.2)

. It can be proved that L21 = L12,so that

CHAPTER 15: INDUCTANCE 149

L - 4)12(t)- L - 4)21(t)12 - il(t) - 21 - i2(t)

(henry- H), (15.3)

which is the definition of mutual inductance (also denoted by M). Mutual induction for dccurrents does not exist, but Eq. (15.3) is valid in that case also, and is used frequently for thedetermination of L12. By combining the Faraday law and Eq. (15.2), mutual inductance canalso be defined in terms of the induced emf,

(t)- d4)12(t) - L dil(t)

el2 - - - 12-dt dt (V). (15.4)

. For a single wire loop in a linear medium, the induced emf in the loop by the current i( t),in the loop is given by

e(t) = d4)seu(t)= -L di(t)dt dt (V), (15.5)

where L is the self-inductance of the loop,

L = 4)self( t )i(t)

(H). (15.6)

. For two coupled loops, the self-inductances and the mutual inductance satisfy the inequality

2Ll1L22 ~ L12, (15.7)

or

L12 = kVLl1L22, -1 < k < 1. (15.8)

The coefficient k is called the coupling coefficient.

QUESTIONS

S QI5.1. What does the expression in Eq. (15.1) for the emf induced in a wire loop actuallyrepresent? - (a) The emf of a generator concentrated at a point of the loop. (b) The emfdistributed along the loop. (c) The Thevenin equivalent to all the elemental electromotive forcesinduced around the loop.

Answer. The Thevenin equivalent to all the elemental electromotive forces induced around the loop.

QI5.2. Why does mutual (and self) inductance have no practical meaning in the dc case? -(a) Magnetic coupling in the dc case exists, but is of no interest. (b) Magnetic coupling in thedc case does not exist at all. (c) Magnetic coupling in the de case is too small to be of practicalinterest.

150 PART 3: SLOWLY TIME-VARYING ELECTROMAGNETIC FIELD

QI5.3. Explain why mutual inductance for a toroidal coil and a wire loop encircling it (e.g.,see Fig. P15.1) does not depend on the shape of the wire loop. - (a) The magnetic fl'UXdueto the current in the toroidal coil through all such loops is the same. (b) The magnetic fluxdue to the current in the loop through the toroidal coil evidently does not depend on the shapeof the loop. (c) The lines of the induced electric field of current in the toroidal coil are closedlines.

QI5.4. Explain in terms of the induced electric field why the emf induced in a coil encircling atoroidal coil and consisting of several loops (e.g., see Fig. P15.1) is proportional to the numberof turns of the loop. - Hint: follow the integration path in Eq. (15.1).

QI5.5. Can mutual inductance be negative as well as positive? Explain by consideringreference directions of the loops. - (a) It can be only positive. (b) It can have both signs. (c)It can be only negative.

S QI5.6. Mutual inductance of two simple loops is L12. We replace the two loops by twovery thin coils of the same shapes, with N1 and N2 turns of very thin wire. What is themutual inductance between the coils? Explain in terms of the induced electric field. - (a)(NdN2)L12' (b) y'N1N2L12' (c) N1N2L12'

Answer. When evaluating the induced electromotive force in one loop by the current in the other,we need to adopt a direction around that other loop for the integration of the induced electric field.This electromotive force can, therefore, be both positive and negative - it is positive if it acts in theadopted direction, and negative otherwise. Consequently, the mutual inductance can be both positiveand negative.

Q15. 7. A two-wire line crosses another two-wire line at a distance d. The two lines are normal.Prove that the mutual inductance is zero, starting from the induced electric field. - Hint:use the expression for the induced electric field in Eq. (14.3). .

S QI5.8. The self-inductance of a toroidal coil is proportional to the square of the number ofturns of the coil. Explain this in terms of the induced electric field and induced voltage in thecoil due to the current in the coil. - Hint: follow the integration path in the evaluation of theinduced emf.

Answer. The induced electric field is proportional to the number of turns. The induced voltage in thecoil, obtained as an integral of the induced electric field along all the turns, is therefore proportionalto the square of the number of turns of the coil.

QI5.9. A thin coil is made of N turns of very thin wire pressed tightly together. If theself-inductance of a single turn of wire is L, what do you expect is the self-inductance of thecoil? Explain in terms of the induced electric field. (a) N L. (b) N2 L. (c) #L.

QI5.10. Explain in your own words what the meaning of self-inductance of a coaxial cableis. Hint: note that the lines of B are circles, and that the flux between the inner and outerconductor is that through a strip spanned between the two conductors.

QI5.11. Is it physically sound to speak about the mutual inductance between two wiresegments belonging either to two loops or to a single loop? Explain. - Hint: recall thedefinition of the induced electric field and the emf induced ina wire element.


Q15.12. Is it physically sound to speak about the self-inductance of a segment of a closedloop? Explain. Hint: the same as for the preceding question.

Q15.13. To obtain a resistive wire with the smallest self-inductance possible, the wire issharply bent in the middle and the two mutually insulated halves are pressed tightly together,as shown in Fig. Q15.13. Explain why the self-inductance is minimal in terms of the inducedelectric field and in terms of the magnetic flux through the loop. - Hint: have in mind Eq.(14.3) and note that the flux increases as the area increases (why?).

S Q15.14. Can self-inductance be negative as well as positive? Explain in terms of the flux. ---'

(a) Yes, it depends on the reference directions. (b) No, reference directions are fixed. (c) Yes,since reference directions are arbitrary.

Answer. The self inductance of a loop is defined as the flux due to the current in the loop, dividedby that current. In that, the reference directions of the flux and current are always connected by the

_,right-hand rule. Therefore the self inductance is always positive.

Q15.15. The self-inductance of two identical loops is L. What is approximately the mutual1nductance between them if they are pressed together? Explain in terms of the induced electricfield and in terms of the magnetic flux. - (a) IMI ~ L/2. (b) IMI ~ L. (c) IMI ~ 2L.

Q15.16. Two coils are connected in series. Does the total (equivalent) inductance of theconnection depend on their mutual position? Explain. - (a) It does not. (b) Only if theyhave a small number of turns. (b) It does.

h

Fig. Q15.13. A loop with small self-inductance. Fig. P15.1. A toroidal coil and wire loop.

S Q15.17. Pressed onto a thin conducing loop is an identical thin superconducting loop. Whatis the self-inductance of the conducting loop? Explain. - Hint: recall the Lentz law, which isexact for a superconducting loop.

Answer. Zero. A current is induced in the superconducting loop such that it cancels the flux due tocurrent in the other loop. So, the flux will also be zero in the other loop, and thus the self-inductanceof the other loop is zero.

Q15.18. A loop is connected to a source of voltage vet). As a consequence, a current i(t)exists in the loop. Another conducting loop with no source is brought near the first loop. Willthe current in the first loop be changed? Explain. - (a) It will not be changed, because thereis no current in the other loop. (b) It will be changed, due to the current in the other loop. (c)It will not be changed, because there is no source in the other loop.


Q15.19. Answer question Q15.I8 assuming that the source in the first loop is a dc source.Explain. - The answers are the same as for the preceding question.

Q15.20. A thin, flat loop of self-inductance L is placed over a flat surface of very highpermeability. What is the new self-inductance of the loop? - Hint: replace the surface by theimage of the loop.

PROBLEMS

P15.1. Find the mutual inductance between an arbitrary loop and the toroidal coil iri Fig.PI5.I. There are N turns around the torus, and the permeability of the core is p. - Assumethe coil and the loop are oriented in the same way. (a) L12 = (pNh/,lr)ln(bja). (b) L12 =(2IlNh/,rr)ln(b/a). (c) L12 = (IlNh/211")ln(b/a).

II

(a) (b)

Fig. P15.2. (a) Two parallel two-wire lines, and (b) the surface for determining the magnetic flux.

S P15.2. Find the mutual inductance of two two-wire lines running parallel to each other. Thecross section of the lines is shown in Fig. PI5.2a. - Hint: let a current II flow in line 1. Usethe sUTfacein Fig. 15.2b to evaluate the magnetic flux per unit length of line II due to currentII in conductor 1, and similarly for conductor 2. (a) L{ II = (po/211")In[(r14T23)/(r13r24)]. (b)

L{ II = (Ilo /211")In[( T24r23) / (TI3TI4)] . (c) L{ II = (po /2~) In[(r13T23)/ (T14T24)]', ,

Solution. Suppose that a current II flows through line I, in the directions indicated in the figure.Let us first evaluate the magnetic flux per unit length of line II due to current II in conductor 1. Thesimplest surface for determining the flux is that shown in Fig. 15.2b in dashed line. So we can write

/ jT14 I

jT14

«PI,II )due to current in conductor 1 = Bl (r) dr = /-lo I dr = /-lolJ In r14 .T13 211" T13 r 211" r13

By analogy, the magnetic flux per unit length of line II due to current II in conductor 2 is

/ /-loll r24«PI,II)due to current in conductor 2 =-- 2

In-.11" r23

Hence the total flux per unit length of line II is

/ / / /-loll r14 r23<PI,II = «PI,II)due to current in conductor 1 + «PI,II)due to current in conductor 2 = -2

In-.11" r13r24


The mutual inductance per unit length of the two lines is hence

, - <l>LII = /l0 In r14 r23 .LI,II - h 211" r13 r24

What is the order of magnitude of this inductance per one meter? What happens if r14 r23 <Q3r24, or if r14r23 =r13r24? Explain. . "

PI5.3. A cable-car track runs parallel to a two-wire phone line, as in Fig. PI5.3. Thecable-car power line and track form a two-wire line. The amplitude of the sinusoidal currentthrough the cable-car wire is 1m and its angular frequency is 1.1.).All conductors are very thincompared to the distances between them. Find the amplitude of the induced emf in the sectionof the phone line b long. - Hint: use the expression for mutual inductance from the precedingproblem.

D

Powerline

Phoneline

H ~ ~II d I.. . I

Power Telephoneline line

Fig. P15.3. Cable car parallel to phone line. Fig. P15.4. Parallel power and phone lines.

PI5.4. Parallel to a two-wire symmetrical power line along a distance h is a two-wire telephoneline, as shown in Fig. PI5.4. (1) Find the mutual inductance between the two lines. (2) Findthe amplitude of the emf induced in the telephone line when there is a sinusoidal current ofamplitude 1m and frequency f in the power line. As a numerical example, assume the following:f = 100Hz, Im=100A, h=50m, d=10m, a=50cm, and b=25cm. Hint: use the results ofproblem P15.2. (a) Lab = 18.5nH, (eind)m = 0.985mV. (b) Lab = 32.5nH, (eind)m =1.785 m V. (c) Lab = 12.5 nH, (eind)m = 0.785 m V.

S PI5.5. Two coaxial thin circular loops of radii a and b are in the same plane. Assuming thata » b and that the medium is air, determine approximately the mutual inductance of theloops. As a numerical example, evaluate the mutual inductance if a = 10 cm, and b = 1 em.- (a) L12 = ~2.97nH. (b) L12 = ~1.97nH. (c) L12 = ~3.97nH.

Solution. Assume a current h in the larger loop. The magnetic flux through the smaller loop can beapproximately evaluated as <1>12=Bl11"b2, where Bl =/loh /(2a) is the magnetic flux density of thelarger loop in its center (see problem P12.7). Thus, the mutual inductance of the loops if oriented inthe same direction is L12 = <l>2/h =/l01l"b2 /(2a) =1.97nH.

PI5.B. Two coaxial thin circular loops of radii a and b are in air a distance d (d» a, b)apart. Determine approximately the mutual inductance of the loops. As a numerical example,evaluate the mutual inductance if a = b = 1cm and d = 10cm. - (a) L12 = 19.7pH. (b)L12 = 29.7 pH. (c) L12 = 39.7 pH.


PIS.7. Inside a very long solenoid wound with N' turns per unit length is a small flat loop ofsurface area S. The plane of the loop makes an angle 0 with the solenoid axis. Determine andplot the mutual inductance between the solenoid and the loop as a function of O. The mediumis air. - (a) L12 = I.J1-oN'ScosO. (b) LI2 = I.2J1-oN'SsinO. (c) L12 =I.J1-oN'SsinO.

il(tJ!ll

Ji-r"" '"2

R

Fig. P15.B. Two coupled circuits.

PIS.B. Assume that within a certain time interval the current in circuit 1 in Fig. P15.8 growslinearly, i1(t) = 10+ltjtl' Will there be any current in circuit 2 during this time? If yes, whatis the direction and magnitude of the current? The number of turns of the two coils is thesame. - Adopt the reference direction 'of current in the resistor, i2(t), to be from left to theright and note that J-lr -+ 00 (a perfect transformer). (a) i2(t) = 5il(t)j6. (b) i2(t) = iI(t).(c) i2(t) = 6 i1(t)j5.

PI5.9. Three coupled closed circuits have self-inductances equal to LI, L2, and L3, resistancesR1, R2, and R3, and mutual inductances LI2, L13, and L23. Write the equations for thecurrents in all three circuits if a voltage VI(t) is connected to circuit 1 only. Then write theequations for the case when three sources, of voltages VI(t), V2(t), and V3(t) are connected tocircuits 1, 2, and 3, respectively. - Hint: write circuit-theory equations, noting that in allthree circuits there are two additional emf's due to the currents in the other two circuits.

S PlS.10. A coaxial cable has conductors of radii a and b. The inner conductor is coated with alayer of ferrite d thick (d < b- a) and of permeability J-l. The rest of the cable is air-filled. Findthe external self-inductance per unit length of the cable. What should your expression reduce to

when (1) d =0, and (2) when d = b-a? - (a) L' = (1(211'){J-lln[(a+2d)ja]+J-lo In[bj(a+d)]}.(b) L' = (lj1l'){J-lln[(a+d)ja]+J-loln[bj(a+d)]}. (c) L = (lj211'){J-lln[(a+d)ja]+J-loln[bj(a+d)]}.

Solution. Assume a current I in the cable. The flux through the flat surface of length 1 and widthb - a between the cable conductors is given by

II

( la+d dr lb dr

)II

(a + d b

)4>=- J.L -+J.LO - =- J.Lln-+J.Loln- .211' a r a+d r 211' a a + d

The external self-inductance per unit length of the cable is thus

, <I> 1 (a+d b

)L =-=- J.Lln-+J.Loln- .II 211' a a + d


See also Example 15.5.

When (1) d = 0 and (2) d = b - a, the expression for L' reduces to that of (1) an air-filled, and(2) a ferrite-filled coaxial cable. (Convince yourself that this is true.)

PI5.1t. The conductor radii of a two-wire line are a and the distance between them isd (d ~ a). Both conductors are coated with a thin layer of ferrite b thick (b ~ d) andof permeability p. The ferrite is an insulator. Calculate the external self-inductance perunit length of the line. - (a) L' = (1/1I"){pIn[(a + b)ja] + Po In[dj(a + b)]}. (b) L' =(1/211"){p In[( a + b)/ a] + Po In[d/( a + b)]}. (c) L' = (1/211"){p In[(2a + b)/ a] + POIn[dj(2a + b)]}.

b b

a c a

J..I-,

J..I-2

h/2

h/2

J..I-l h

J..I-2

(a) (b)

Fig. P15.12. Two toroidal~oils with inhomogeneous core.

PI5.12. The core of a toroidal coil of N turns consists of two materials, of respective perme-abilities PI and P2, as in each part of Fig. P15.12. Find the self-inductance of the toroidalcoil and the mutual inductance between the coil and the loop positioned as in Fig. P15.1 if (1)the ferrite layers are of equal thicknesses, hj2, in Fig. P15.12a, and (2) the ferrite layers areof equal height h and the radius of the surface between them is c (a < c < b), in Fig. P15.12b.- Hint: note that in both cases in the core H = N I j (211"r),where I is the current in the coil,and r is the radial distance from the torus axis. Note also that, in both cases, L12 = L/ N(why?). (a) L(I) = {[(PI + IL2)N2h]/(271')}In(b/a), L(2) = (N2h/1I")[plln(c/a) + p2In(b/c)].(b) L(1) = {[(PI + p2)N2h]/(411")}In(bja), L(2) = (N2hj211")[Plln(cja) + P2In(bjc)]. (c)L(1) = {[(ILl + IL2)N2h]j(1I")}In(bja); L(2) = (N2hj1l")[Plln(cja) + p2In(bjc)].

N1

hN2

h

a Pol

Fig. P15.13. Three toroidal coils. Fig. P15.14. A strip line with a two-layer dielectric.

J;

I

- .

ill i'I"

-I: I"'---'--I .a-

N3 I bI c


PI5.IS. Three toroidal coils are wound in such a way that the coils 2 and 3 are inside coill, asin the cross section shown in Fig. Pl5.l3. The medium is air. Find the self-inductances L1, L2,and L3, and mutual inductances L12, L13, and L23. What are the different values of inductancethat can be obtained by connecting the three windings in series in different ways? - We givethe results for self-inductances only. (a) L1 = (J1.oNfh/27r)ln(c/a), L2 = (J1.oN?h/7r)ln(c/a),L3 = (J1.oNlh/7r)ln(b/a). (b) Ll = (J1.oNfh/47r)ln(c/a),L2 = (J1.oN?h/47r)ln(c/a),L3 =(J1.oNlh/47r) In(b/a). (c) L1 = (J1.oNfh/7r)ln(c/a), L2 = (J1.oN?h/27r)ln(c/a), and L3 =(J1.oNlh/27r)ln(b/a). .

PI5.14. The width of the strips of a long, straight strip line is a and their distance is d (Fig.Pl5.l4 for d2 = 0). Between the strips is a ferrite of permeability J1.. Neglecting edge effects,find the inductance of the line per unit length. - The correct solution is given in the answersto the next problem, for d2 = 0 and J1.1= J1..

S.P15.15. The width of the strips of a strip line is a and their distance is d. Between thestrips there are two ferrite layers of permeabilities J1.1and J1.2,and the latter is d2 thick,:as in Fig. Pl5.l4. Neglecting edge effects, find the inductance of the line per unit length.- Hint: note that between the strips H = l/a. (a) L' = [2J1.1(d - d2)/a] + J1.2d2/a. (b)L' = [J1.1(d - d2)/a] + 2J1.2d2/a. (c) L' = [J1.1(d - d2)/a] + J1.2d2/a.

Solution. See problem P13.13. The magnetic field intensity vector between the strips is parallel tothem and lies in the transversal cross section of the line. Its magnitude is equal to H = 1/ a. Themagnetic flux through the surface spanned between the strips, which we adopt to be normal to H, andof length I, is given by

q, = J.lIH(d - d2)l + J.l2Hd21,

so that the inductance per unit length of the line is

L' = q, = J.llCd - d2) + J.l2d2II a .

PI5.16. A long thin solenoid of length b and cross-sectional area S is situated in air and hasN tightly wound turns of thin wire. Neglecting edge effects, determine the inductance of thesolenoid. - (a) L = 2J1.oN2S/b. (b) L = J1.oN2S/(47rb). (c) L = J1.oN2S/b.

P15.l7. A thin toroidal core of permeability J1.,mean radius R and cross-sectional area S isdensely wound with two coils of thin wire with Nl and N2 turns, respectively. The windingsare wound one over the other. Determine the self- and mutual inductances of the coils, and thecoefficient of coupling between them. - (a) Ll = J1.NfS/(27rR), L2 = J1.N?S/(27rR),L12 =J1.N1N2S/(27rR), k = 1. (b) L1 = J1.NfS/(47rR), L2 = J1.N?S/(47rR), L12 = j-lN1N2S/(47rR),k = 1. (c) L1 = J1.N'fS/R, L2 = J1.Ni.S/R, L12 = J1.N1N2S/R, k = 1.

P15.lB. A thin toroidal ferromagnetic core, of mean radius R and cross-sectional area S isdensely wound with N turns of thin wire. A current i(t) = 10 + 1mcos r..;t,where 10 and1m are constants, and 10 :;}>1m, is flowing through the coil. Which permeability would youadopt in approximately determining the coil self-inductance? Assuming that this permeabilityis J1.,determine the self-inductance of the coil. Does it depend on 10? - (a) Differential


permeability. L = J.LN2Sj(21rR). (b) Normal permeability. L = J.LN2Sj(lrR). (c) Initialpermeability. L = J.LN2 Sj( 411'R).

S P15.19. A thin solenoid is made of a large number of turns of very thin wire tightly wound inseveral layers. The radius of the innermost layer is a, of the outermost layer b, and the solenoidlength is d (d ~ a, b). The total number of turns is N, and the solenoid core is made outof cardboard. Neglecting edge effects, determine approximately the solenoid self-inductance.Note that the magnetic flux through the turns differs from one layer to the next. Plot this fluxas a function of radius, assuming the layers of wire are very thin. - (a) L = J.L01rN2(3a2+2ab + b2)j(3d). (b) L = J.L0.lrN2(3a2 + 6ab + b2)jd. (c) L = J.L01rN2(3a2 + 2ab + b2)j(6d). .

Solution. There is a total of dN =Ndr/(b - a) turns in a layer of radius r and thickness dr. Sucha layer can be considered as a very long solenoid, so it produces a uniform magnetic field inside itself.The magnetic flux density is given by

dB = /JodNId

Outside the layer, dB =O. By integrating the above expression, we get the magnetic flux density dueto all the turns to be

B(r) = /JONId (r < a), B(r) = /JoNI(b- r)(b- a)d

(a < r < b), B(r) =0 (r > b).

The magnetic flux through a single turn, of radius r, is

<1>0= /J01l'NI(3br2 - 2r3 - a3)3(b - a)d

Finally, the total flux we obtain as

<J>= J~. <J>odN,

so that the self-inductance of the solenoid is

L = <I> = /J01l'N2(3a2 + 2ab + b2) .I 6d

Check the result by considering the case with (b - a) -- O.

P15.20. Repeat problem P15.19 for a thin toroidal core. Assume that the mean toroid radiusis R, the total number of turns N, the radius of the innermost layer a, and that of the outermostlayer b, with R ~ a, b. - Hint: the solution is practically the same as that of problem P15.19;just set d = 211'R.


P15.21. The current intensity in a circuit of self-inductance L and negligible resistance waskept constant during a period of time at a level 10' Then during a short time interval ilt, thecurrent was linearly reduced to zero. Determine the emf induced in the circuit during this timeinterval. Does this have any connection with a spark you have probably seen inside a switchyou turned off in the dark? Explain. - Adopt the reference direction of the circuit to be thesame as that of the current. (a) e = -2L 10/ ilt. (b) e = -L 10/ilt. (c) e = -L 10/(2ilt).

15. inductance - electrical, computer & energy...

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