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 simple evaluation of this emf in linear media is the mutual inductance. A single wire loop with timevarying current creates a time-varying induced electric field along itself, resulting in an induced emf in the loop, an effect knownas self-induction. In linear media, this emf is evaluated in terms of the electromagnetic parameter known as the inductance, or self-inductance, of the loop.
.
The emf induced in a loop C2 by a current il(t) in a nearby loop C1 is given by eI2(t)
=1
!c2
Elind(t).
dl2
(V),
(15.1)
where E1ind(t) is the induced electric field of current il(t) along loop C2. In linear media the magnetic flux of current i1(t) through C2 is proportional to il(t),
.
~12(t)
.
It can be proved that L21
= L12i1(t)
= L12, so
that
(Wb).
(15.2)
CHAPTER
15:
149
INDUCTANCE
L
- 4)21(t) 21 - i2(t)
- 4)12(t)- L
12 -
il(t)
-
(henry- H),
(15.3)
which is the definition of mutual inductance (also denoted by M). Mutual induction for dc currents does not exist, but Eq. (15.3) is valid in that case also, and is used frequently for the determination of L12. By combining the Faraday law and Eq. (15.2), mutual inductance can also be defined in terms of the induced emf,
el2 (t) --
d4)12(t) -- - L 12-dil(t) 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 )
(H).
i(t) .
(15.6)
For two coupled loops, the self-inductances and the mutual inductance satisfy the inequality 2
Ll1L22
~ L12,
(15.7)
or
(15.8)
-1 < k < 1.
L12 = kVLl1L22, 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 actually represent? - (a) The emf of a generator concentrated at a point of the loop. (b) The emf distributed along the loop. (c) The Thevenin equivalent to all the elemental electromotive forces induced around the loop. Answer. QI5.2.
The Thevenin equivalent to all the elemental electromotive forces induced around the loop. 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 the dc case does not exist at all. (c) Magnetic coupling in the de case is too small to be of practical interest.
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., - (a) The magnetic fl'UXdue to the current in the toroidal coil through all such loops is the same. (b) The magnetic flux due to the current in the loop through the toroidal coil evidently does not depend on the shape of the loop. (c) The lines of the induced electric field of current in the toroidal coil are closed lines.
see Fig. P15.1) does not depend on the shape of the wire loop.
QI5.4. Explain in terms of the induced electric field why the emf induced in a coil encircling a toroidal coil and consisting of several loops (e.g., see Fig. P15.1) is proportional to the number of 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 considering reference 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 two very thin coils of the same shapes, with N1 and N2 turns of very thin wire. What is the mutual 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 the
adopted direction, and negative otherwise. Consequently, the mutual inductance can be both positive and 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 of turns of the coil. Explain this in terms of the induced electric field and induced voltage in the coil due to the current in the coil. - Hint: follow the integration path in the evaluation of the induced emf. Answer. The induced electric field is proportional to the number of turns. The induced voltage in the coil, obtained as an integral of the induced electric field along all the turns, is therefore proportional to 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 the self-inductance of a single turn of wire is L, what do you expect is the self-inductance of the coil? 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 cable is. Hint: note that the lines of B are circles, and that the flux between the inner and outer conductor is that through a strip spanned between the two conductors. QI5.11. Is it physically sound to speak about the mutual inductance between two wire segments belonging either to two loops or to a single loop? Explain. - Hint: recall the definition of the induced electric field and the emf induced ina wire element.
CHAPTER 15: INDUCTANCE
151
Q15.12. Is it physically sound to speak about the self-inductance of a segment of a closed loop? Explain. Hint: the same as for the preceding question. Q15.13. To obtain a resistive wire with the smallest self-inductance possible, the wire is sharply 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 induced electric 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. since reference directions are arbitrary.
(b) No, reference directions
are fixed.
(c) Yes,
Answer. The self inductance of a loop is defined as the flux due to the current in the loop, divided by 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 mutual 1nductance between them if they are pressed together? Explain in terms of the induced electric
field 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 the connection depend on their mutual position? Explain. - (a) It does not. (b) Only if they have a small number of turns. (b) It does.
h
Fig.
S
Q15.13.
A loop with small self-inductance.
Fig. P15.1.
A toroidal coil and wire loop.
Q15.17. Pressed onto a thin conducing loop is an identical thin superconducting loop. What is the self-inductance of the conducting loop? Explain. - Hint: recall the Lentz law, which is exact for a superconducting loop. Answer. Zero. A current is induced in the superconducting loop such that it cancels the flux due to current in the other loop. So, the flux will also be zero in the other loop, and thus the self-inductance of 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. Will the current in the first loop be changed? Explain. - (a) It will not be changed, because there is 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.
152
PART
3: SLOWLY TIME-VARYING ELECTROMAGNETIC
FIELD
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 high permeability. What is the new self-inductance of the loop? - Hint: replace the surface by the image 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. - Assume (pNh/,lr)ln(bja). (b) L12 the coil and the loop are oriented in the same way. (a) L12 (2IlNh/,rr)ln(b/a). (c) L12 (IlNh/211")ln(b/a).
=
=
=
II
(b)
(a)
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. The cross section of the lines is shown in Fig. PI5.2a. - Hint: let a current II flow in line 1. Use the sUTfacein Fig. 15.2b to evaluate the magnetic flux per unit length of line II due to current II 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. The simplest surface for determining the flux is that shown in Fig. 15.2b in dashed line. So we can write
/
«PI,II )due to current in conductor
j
=
1
I
T14
Bl (r) dr
T13
= /-lo I 211"
j
T14
dr
T13
r
= /-lolJ In r14 . 211"
r13
By analogy, the magnetic flux per unit length of line II due to current II in conductor 2 is / «PI,II)due
/-loll
to current
in conductor
2
r24
= -- 211" In-.r23
Hence the total flux per unit length of line II is /
/
=-
conductors
II 211'
The external
= (1(211'){J-lln[(a+2d)ja]+J-lo In[bj(a+d)]}. (c) L = (lj211'){J-lln[(a+d)ja]+J-loln[bj(a+
(
l
J.L
self-inductance
is given
a+d dr
-+J.LO
a
r
l
,
=-=II
b
dr
-
a+d r
per unit length
L
by
1
)
=-
II 211'
(
a
+d
J.Lln-+J.Lolna
a + d
of the cable is thus
(
a+d
b
b
)
J.Lln-+J.Loln. 211' a a+d
).
15:
CHAPTER
155
INDUCTANCE
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 is d (d ~ a). Both conductors are coated with a thin layer of ferrite b thick (b ~ d) and of permeability p. The ferrite is an insulator. Calculate the external self-inductance per (1/1I"){pIn[(a + b)ja] + Po In[dj(a + b)]}. (b) L' unit length of the line. - (a) L'
=
(1/211"){p In[( a + b)/ a] + Po In[d/( a + b)]}. (c) L'
=
= (1/211"){p In[(2a
+ b)/ a] + PO In[dj(2a + b)]}.
b
b c
a
J..I-,
h/2
J..I-2
h/2
a
h
J..I-l
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 permeabilities PI and P2, as in each part of Fig. P15.12. Find the self-inductance of the toroidal coil 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 are of 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)].
=
N2
-
J;
.
h I
N1
ill'I" i
h
I
-I: '---'-" I N3
.a
I
I
b c
a
Fig. P15.13.
Three toroidal coils.
Fig. P15.14.
Pol
A strip line with a two-layer dielectric.
156
PART 3: SLOWLY TIME-VARYING
ELECTROMAGNETIC
FIELD
PI5.IS. Three toroidal coils are wound in such a way that the coils 2 and 3 are inside coill, as in 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 inductance that can be obtained by connecting the three windings in series in different ways? - We give the results for self-inductances
L3 = (J1.oNlh/7r)ln(b/a). (J1.oNlh/47r)
In(b/a).
only. (a) L1
= (J1.oNfh/27r)ln(c/a), L2 = (J1.oN?h/7r)ln(c/a),
(b) Ll = (J1.oNfh/47r)ln(c/a),L2 = (J1.oN?h/47r)ln(c/a),L3 =
(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. Pl5.l4
.
The width of the strips of a long, straight strip line is a and their distance is d (Fig. for d2
= 0).
Between the strips is a ferrite of permeability
find the inductance of the line per unit length. to the next problem, for d2 = 0 and J1.1= J1..
J1.. Neglecting
edge effects,
The correct solution is given in the answers
S.P15.15. The width of the strips of a strip line is a and their distance is d. Between the strips 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. l/a. (a) L' [2J1.1(d - d2)/a] + J1.2d2/a. (b) - Hint: note that between the strips H 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 to them and lies in the transversal cross section of the line. Its magnitude is equal to H 1/ a. The magnetic flux through the surface spanned between the strips, which we adopt to be normal to H, and of 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'
= IIq, =
J.llCd - d2) + J.l2d2
a
.
PI5.16. A long thin solenoid of length b and cross-sectional area S is situated in air and has N tightly wound turns of thin wire. Neglecting edge effects, determine the inductance of the solenoid. - (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 is densely wound with two coils of thin wire with Nl and N2 turns, respectively. The windings are wound one over the other. Determine the self- and mutual inductances of the coils, and the coefficient 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 is = 10 + 1m cos r..;t,where 10 and 1m are constants, and 10 :;}>1m, is flowing through the coil. Which permeability would you adopt in approximately determining the coil self-inductance? Assuming that this permeability is J1.,determine the self-inductance of the coil. Does it depend on 10? - (a) Differential
densely wound with N turns of thin wire. A current i(t)
157
CHAPTER 15: INDUCTANCE
permeability. permeability.
L L
= J.LN2Sj(21rR).
= J.LN2
Sj(
(b) Normal permeability.
L
= J.LN2Sj(lrR).
(c) Initial
411'R).
S P15.19. A thin solenoid is made of a large number of turns of very thin wire tightly wound in several layers. The radius of the innermost layer is a, of the outermost layer b, and the solenoid length is d (d ~ a, b). The total number of turns is N, and the solenoid core is made out of 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 flux
=
as 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. Such a 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 Outside the layer, dB to all the turns to be
B(r)
= O. By integrating
= /JONId
= /JodNI d
the above expression, we get the magnetic flux density due
B(r) = /JoNI(b- r) (b- a)d
(r < a),
(a < r < b),
B(r)
=0
(r > b).
The magnetic flux through a single turn, of radius r, is
0 = /J01l'NI(3br2 - 2r3 - a3) 3(b - a)d Finally, the total flux we obtain as
= J~. so that the self-inductance
of the solenoid
L
=
I
odN,
is
= /J01l'N2(3a26d+
Check the result by considering the case with (b
2ab + b2) .
- a) -- O.
P15.20. Repeat problem P15.19 for a thin toroidal core. Assume that the mean toroid radius is R, the total number of turns N, the radius of the innermost layer a, and that of the outermost layer b, with R ~ a, b. - Hint: the solution is practically the same as that of problem P15.19; just set d = 211' R.
158
PART 3: SLOWLY TIME-VARYING
ELECTROMAGNETIC
FIELD
P15.21. The current intensity in a circuit of self-inductance L and negligible resistance was kept constant during a period of time at a level 10' Then during a short time interval ilt, the current was linearly reduced to zero. Determine the emf induced in the circuit during this time interval. Does this have any connection with a spark you have probably seen inside a switch you turned off in the dark? Explain. - Adopt the reference direction of the circuit to be the same as that of the current. (a) e -2L 10/ ilt. (b) e -L 10/ilt. (c) e -L 10/(2ilt).
=
=
=
