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CHAPTER
131
14: ELECTROMAGNETIC INDUCTION AND FARADAY'S LAW
loop. It can be expressed in several forms. If a closed conductive contour C is movmg m a
time constant magnetic field with a velocity v, it is given by
e=
t
(14.4)
(V).
(v x B) . dl
The most general expression for the induced emf is
e
= 1 Eind'
Je
dl
= - dCPthrough e
dt
in dt
(14.5)
(V).
. A qualitative law describing the sense of the induced emf around a closed loop, the Lentz
"law, states that the induced current in a loop tends to reduce the time-variation of the magnetic
flux by its own time-varying magnetic field.
. The induced currents inside conductive bodies (not wires) that are a result of the induced
electric field are called eddy currents. Due to eddy currents, the total current in a conductor
is the largest on the conductor surface. Therefore Joule's losses are increased, and the flux
inside a solid ferromagnetic core is reduced greatly for time-varying fields, so that laminated
cores must be used for alternating-c:lrrent machinery. In some instances, eddy currents are
produced on purpose, e.g., in so-called induction furnaces for melting metals.
.
Substances with zero resistivity at very low temperatures
are known as superconductors.
It is impossible to change the magnetic flux through a superconducting
loop by means of
electromagnetic
induction.
.
The most general definition of voltage between two points in an electromagnetic
field is
defined as
VAB
= iB Etotal'dl
Due to the induced electric field contained
on the path between them.
(14.6)
(V).
in Etotal, the voltage between
two points
depends
QUESTIONS
S Q14.1. An observer in a coordinate system with the source of a time-invariant magnetic field
observes a force F = Qv x B on a charge Q moving with a uniform velocity v with respect
to his coordinate system. What is the force on Q observe~\ by an observer moving with the
charge? What is his interpretation of the velocity'v? - (at-F,
due to an electric field. (b)
F, due to an electric field. (c) F, due to a magnetic field.
Answer.
The force is the same, but is interpreted
as due to an electric field, E (which is equal to
v x B). In the observer's interpretation,
v is the negative of the velocity of the first coordinate system
with respect to him/her.
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PART
3: SLOWLY TIME-VARYING ELECTROMAGNETIC
FIELD
Q14.2. Three point charges are stationary in a coordinate system of the first observer. A
second observer, in his coordinate system, moves with respect to the first with a uniform
velocity. What kinds of fields are observed by the first observer, and what by the second? (a) Both observe an electric and a magnetic field. (b) The first observes a magnetic field, the
second an electric field. (c) The first observes an electric field, the second both an electric field
and a magnetic field.
Q14.3. A small uncharged conducting sphere is moving in the field of a
Are there induced charges on the sphere surface? If they exist, how do
with the charge and an observer on the magnet explain their existence? are moving only in a magnetic field. (b) Yes (what are the explanations of
(c) No, because the magnetic field is time-constant (why?).
permanent magnet.
an observer moving
(a) No, because the'!)
the two observers?).
Q14.4. A straight metal rod moves with a constant velocity v in a uniform magnetic field of
magnetic induction B. The rod is normal to B, and v is normal to both the rod and to B.
"Sketch the distribution of the induced charges on the rod. What is the electric field of these
charges inside the rod equal to? - (a) Charges of opposite sign are induced as if the rod is
~in a uniform axial electric field. (b) There are no induced charges on the rod. (c) Charges of
opposite sign are induced on the two sides of the rod curved surface.
Q14.5. A small dielectric sphere moves in a uniform magnetic field. Is the sphere polarized?
Explain. - (a) Yes (why?). (b) No, because the field is uniform. (c) No, because the field is
time-constant.
S Q14.6. A charge Q is located close to a toroidal coil with time-varying current. Is there a
force on the charge? Explain. - (a) No, there is no electric or magnetic field outside the coil.
(b) Yes, due to the magnetic field. (c) Yes, due the electric field.
Answer. There is a force, since there is a time-varying electric field due to the time-varying current
in the coil.
Q14. 7. A wire of length l is situated in a magnetic field of flux density B parallel to the wire.
Is an electromotive force induced in the wire if it is moved (1) along the lines of B, and (2)
transverse to the lines of B? Explain. - (a) In case 1 yes, in case 2 no. (b) In case 1 no, in
case 2 yes. (c) No in both cases.
Q14.8. Strictly speaking, do currents in branches of an ac electrical circuit depend on the
circuit shape? Explain. - (a) No, because this is the circuit-theory assumption. (b) No,
because there is no induced electric field along circuit branches. (c) Yes, because there is an
induced electric field along circuit branches which depends on the circuit shape.
Q14.9. Does the shape of a dc circuit influence the currents in its branches? Explain. answers are the same as in the preceding question.
The
Q14.10. A circular metal ring carries a time-varying current, which produces a time-varying
induced electric field. If the ring is set in oscillatory motion about the axis normal to its
plane, will the induced electric field be changed? Explain your answer. - (a) Yes, because
the electrons will be additionally accelerated. (b) No, because this does not involve a motion
of charges. (c) No, because the mechanical motion includes not only free charge carriers, but
also all other charges inside matter.
CHAPTER
14: ELECTROMAGNETIC
133
INDUCTION AND FARADAY'S LAW
S Q14.11. A device for accelerating electrons known as the betatron (Fig. Q14.11) consists of
a powerful electromagnet and an evacuated tube which is bent into a circle. Electrons are
accelerated in the tube when the magnetic flux in the core is forced to rise approximately as
a linear function of time. What is the physical mechanism of accelerating the electrons in the
betatron? - (a) By the electric field of static charges (explain). (b) By the magnetic field
(explain). (c) By the induced electric field (explain).
Answer. During the time interval in which the magnetic flux in the core rises, its magnetization and
magnetization currents increase, and an induced electric field is obtained. Due to symmetry the lines
of this field are circles centered at the electromagnet axis, and this field accelerates the electrons.. Since
they are moving in the magnetic field that exists between the magnet poles, their trajectory is bent in
approximately a circle.
Evacuated
tube
~
Fig. Q14.11.
Sketch of a betatron.
Q14.12. Is it physically sound to speak about a partial electromotive force induced in a
segment of one loop by the current in a segment of another (or even of the same) loop?
Explain. - Hint: recall the definition of the induced electric field. If a loop element is dl,
what is the physical meaning of the product Einduced . dl?
Q14.13. A vertical conducting sheet (say, of aluminum) is permitted to fall under the action of
gravity between the poles of a powerful permanent magnet. Is the motion of the sheet affected
by the presence of the magnet? Explain. - (d) It is slowed down. (b) It is not affected. (c)
It is faster than without magnet.
!
'
y.
2~
Fig.
Q14.16.
Two coupled
coils.
Fig.
Q14.17.
A permanent
magnet
approaching
a loop.
S Q14.14. A long solenoid wound on a Styrofoam core carries a time-varying current. It is
encircled by three loops, one of copper, one of a resistive alloy, and the third of a bent moist
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134
PART 3: SLOWLY TIME-VARYING
ELECTROMAGNETIC
FIELD
filament (a poor conductor). In which loop is the induced electromotive force the greatest? (a) In the copper loop. (b) In the moist filament. (c) It is the same in all three.
Answer. The electromotive force is the same in all of them, but the resulting induced current is the
greatest in the copper loop.
Q14.15. What becomes different in question Q14.14 if the solenoid is wound onto a ferromagnetic core? - (a) The answer is not changed, but the emf is greater. (b) Nothing at all is
different. (c) The answer is not changed, but the emf is smaller.
Q14.16. Assume that the current i(t) in circular loop 1 in Fig. Q14.16 in a certain time
interval increases linearly in time. Will there be a current in the closed conducting loop 2? If
you think that there will be, what is its direction? - (a) There is no current in loop 2. (b)
There is current in loop 2, in the same direction as in loop 1. (c) There is current in loop 2,
in the opposite direction to that in loop 1.
Q14.17. What is the direction of the current induced in the loop sketched in Fig. Q14.17?
-Explain. - (a) Depends on the velocity of the magnet. (b) In the clockwise direction. (c) In
the counterclockwise direction.
S Q14.18. The coil in Fig. Q14.18 consists of N densely wound turns of thin wire. What is the
voltage of the generator when compared with the case of a single turn? Explain using both the
concept of magnetic flux and that of the induced electric field. - (a) It is N times greater.
(b) It is N2 times greater. (c) It is VN times greater.
Answer. It is N times greater. In terms of the magnetic flux, the flux through N turns is N times
that through a single turn. In terms of the induced electric field, we need to integrate the product
Eind . dl along all the N turns, which amounts to N times the integral along a single turn.
B
b
n
-
;
2
1
Fig. Q14.18.
A coil rotating in magnetic field.
Fig.
Q14.20.
Three series loops.
Q14.19.
A solenoid is wound onto a long, cylindrical permanent
magnet.
A voltmeter is
connected to one end of the solenoid, and to a sliding contact that moves and makes a contact
with a larger or smaller number of turns of the solenoid. Thus the magnetic flux in the closed
loop of the voltmeter will vary in time. Does the voltmeter detect a time-varying
voltage
(assuming that it is sensitive enough to do so)? Explain. - (a) It does detect a voltage. (b)
It does not detect a voltage. (c) Depends on the velocity of the sliding contact.
Q14.20. What is the direction of the reference unit vectors normal to the three loops in Fig.
Q14.20? - (a) In all three from left to right. (b) In all three from right to left. (c) The loops
not being completely closed, it cannot be defined.
CHAPTER
14: ELECTROMAGNETIC
135
INDUCTION AND FARADAY'S LAW
Q14.21. A cylindrical permanent magnet falls without friction through a vertical evacuated
metal tube. Is the fall accelerated? If not, what determines the velocity of the magnet?
Explain. (a) It falls as if there were no tube, because there is no friction. (b) It falls as if
there were no tube, because there is no friction and no air in the tube. (c) After some time it
reaches a constant velocity, due to the magnetic force of currents induced in the tube.
v
~
(b)
(a)
Fig. Q14.23.
strip.
(a) A strip moving in a magnetic field. (b) Sketch of lines of induced currents in the
S Q14.22. A strong permanent magnet is brought near the end plate of the pendulum of a
wall timepiece. Does this influence the period of the pendulum? If it does, is the pendulum
accelerated or slowed down? Does this depend on the type of the magnetic pole of the magnet
closer to the pendulum plate? Explain. - (a) Depends on the pole closer to the pendulum.
(b) The pendulum is accelerated. (c) The pendulum is slowed down.
Answer. Eddy currents are induced in the pendulum, and the magnetic force on them tends to slow
down the pendulum. This effect does not depend on the magnetic pole of the magnet, only on the
intensity of the magnetic field.
QI4.23. Fig. Q14.23a shows a sketch of a flat strip moving between the poles of a permanent
magnet. Sketch the lines of the induced current in the strip. - Hint: check if the lines sketched
in Fig. Q14.2b are as you expected.
Q14.24. The strip in question ,Q14.23 has in case (1) longitudinal, and in case (2) transverse
slots with respect to the direction of motion. In which case are induced currents greater?
Explain. - (a) In case 1. (b) The same in both cases. (c) In case 2.
QI4.25.
Explain in detail how the right-hand side in the Faraday law for a rigid contour
moving in a magnetic field,
e = Ie Eind . dl
= Ie (v X B)
. dl
= Ie (~;
is obtained from the middle expression in it. out of the integral.
X
B) . dl = :t Ie (ds
X B)
. dl,
Hint: discuss why the factor djdt can be taken
I
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PART 3: SLOWLY TIME-VARYING
ELECTROMAGNETIC
FIELD
S Q14.26. The magnetic flux through a contour C at time t is ~1, and at time t + f:::1t
it is ~2.
Is the time increment of the flux through C (~2 - ~d, or (~1 - ~2)? Explain. - Hint: recall
the definition of the increment of a function of one variable.
Answer. It is (cI>2- cI>1)' Recall that the increment of a function is always defined as the difference
of its value for an increased independent variable, and its value before this increase.
Q14.27. Is the distribution of the induced electromotive force around a contour seen from the
right-hand side in Eq. (14.5)? Is it seen from the middle expression in that equation?
Is it
seen from the right-hand side in the Faraday law in the text of the question Q14.25? - Hint:
trace where in these equations induced elemental electromotive forces are integrated along the
contour.
S Q14.28. Imagine an electric circuit with several loops situated in a slowly time-varying magnetic (and induced electric) field. Can you analyze such a circuit by circuit-theory methods? If
you think you can, explain in detail how you would do it. - Hint: there is a distributed electromotive force in all such loops. How it can be incorporated in the circuit-theory equations?
Answer.
There is a distributed
electromotive
force in all such loops.
in the circuit-theory
solution as equivalent Thevenin generators.
They need to be incorporated
Q14.29.
Does it make any sense at all to speak about the electromotive force induced in an
open loop? If you think that this makes sense, explain what happens. Hint: recall that even
for a closed loop the induced emf is, in fact, a sum of elemental emf's. Will there be an induced
current in an open loop?
Q14.30. Explain in detail what a positive and what a negative electromotive force in Eq.
(14.5) mean. - (a) A negative emf acts in the direction of the contour C. (b) A positive emf
acts in the direction opposite to that of the contour. (c) A positive emf acts in the direction of
the contour.
S Q14.31. Why is the reduction of eddy current losses possible only if the vector B is parallel to
a thin ferromagnetic sheet? - (a) Because of boundary conditions for vector B. (b) Because
only then the flux through a contour in the sheet is small. (c) Because then the magnetic field
in the sheet is smaller.
Answer. If vector B is parallel to a thin ferromagnetic sheet, possible current loops have small areas,
the enclosed magnetic flux is small, and the induced electromotive force in them is also small. If it
is normal to the sheet, large current loops (as large as the sheet itse,lf) are possible, enclosing a large
flux.
0
Fig.
Q14.32.
B(t)
A loop in the form of an 8.
CHAPTER
14:
ELECTROMAGNETIC
INDUCTION AND FARADAY'S LAW
137
Q14.32. What is the induced electromotive force in the loop shown in Fig. Q14.32, if the
magnetic field is time-varying? If the right half of the loop is turned about the x axis by
180 degrees, what is then the induced electromotive force? Explain both in terms of the
-2SdB(t)jdt
in both cases. (b)
magnetic flux and of the induced electric field. - (a) e(t)
2SdB(t)jdt in both cases. (c) e(t)
-2SdB(t)jdt
in the first case, zero in the second.
e(t)
=
=
=
Q14.33. A solid conducting body is placed near a loop with time-varying current. Are any
forces acting on free charges inside the body? Explain. - (a) No, because there is only a
magnetic field inside the body. (b) Yes, because there is an induced electric field inside the
body. (c) Yes, because inside the body there is both a magnetic and an electric field.
Q14.34. Is there a magnetic force between the body and the loop from. the preceding example?
Explain. - (a) Yes, there is an electric force. (b) Yes, there is a magnetic force on the induced
currents. (c) No, because there is no magnetic field inside the body.
S Q14.35. A planar insulated loop with time-varying current is placed on the surface of a plane
conducting sheet. What happens in the sheet? Is the power required to drive the current in
the loop different when it resides on the sheet than when it is isolated in space? Explain. Hint: there is an induced electric field in the sheet due to the loop current, but also an induced
electric field along the loop due to the currents induced in the sheet.
Answer. A current will be induced in the sheet. It is accompanied by both losses and change in the
loop inductance, and so the power required to drive the current in the loop is different when the sheet
is present.
Q14.36. Of two closed conducting loops, C1 and C2, C1 is connected to a generator of timevarying voltage. Is there a current in C2? Explain. - Hint: think in terms of the induced
electric field. Can you think of an example in which there is no induced current in C2 ?
Q14.37. An elastic metal circular ring carrying a steady current I is periodically deformed to
a flat ellipse, and then released to retain its original circular shape. Is an electromotive force
induced in the loop? Explain. - (a) No, because the current is not time-varying. (b) No,
because there is no time variation of flux through the ring. (c) Yes, because the segments of
the loop are moving the magnetic field of the current in the loop.
Q14.38. In the introduction to this Chapter, it was mentioned that the flux through a superconducting loop cannot be changed. Does this mean that the flux through a superconducting
loop cannot be changed by any means? - (a) It cannot be changed by any means. (b) It
cannot be changed by the induced electric field. (c) It can be changed by the induced electric
field which varies in time very slowly.
Q14.39. Why does the shape of voltmeter leads influence the time-varying voltage it measures?
Why does this influence increase with frequency? - Hint: have, in mind the induced electric
field and Eq. (14.3).
PROBLEMS
S P14.1. Starting from Eq. (14.3), prove that the lines of the induced electric field vector of a
circular current loop with a time-varying current are circles centered at the loop axis. - Hint:
refer to Fig. P 14.1 and have in mind the symmetry of the structure.
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138
PART 3: SLOWLY TIME-VARYING
ELECTROMAGNETIC
FIELD
Solution. Let the loop be in the plane of the paper (Fig. P14.1). Consider an arbitrary point P in
space (not necessarily in the plane of the loop). Let dEind and dE:nd be the induced electric field
vectors at P due to two symmetrical current elements, i dl and i dl' , shown in the figure. In accordance
to the expression in Eq. (14.3), these vectors are such that the vector dEind + dEfnd is tangential to
the circular contour C centered at the loop axis and indicated in dashed line in the figure. The circular
contour C is thus a line of the vector Eind'
/'
,....--
-........
/
.
/
'
/
dE'
/
"
;{ t
md '
I
~idl'
\
Eind
" dEind+"dE.'
,
(\
',dE,.,
~dl
\
Fig. P14.1.
",
/
i
"
md
--- ---C
/
/
/
,..../'
A circular loop with a time-varying current.
P14.2.
Assume that you know the induced electric field Eind(t) along a circular line C
of radius a in problem P14.1. A wire loop of radius a coincides with C. Evaluate the total
electromotive force induced in the loop. Prove that this is, actually, the voltage of the Thevenin
generator equivalent to the distributed infinitesimal generators around the loop. - (a) e(t)
Eind(t)27ra. (b) e(t) = -Eind(t) 27ra. (c) e(t) = Eind(t)7ra,
=
=
S P14.3. Two coaxial solenoids shown in Fig. P14.3 are connected in series. A current i(t)
1.5 sin 1O00tA, where time is in seconds, flows through the solenoids. The dimensions are
a = 1cm, b = 2 cm, and L = 50cm. The number of turns in both is the same, NI = N2 = 1000.
Find the approximate induced electric field at points Al (surface of the inner solenoid), A2
(halfway between the two solenoids), and A3 (right outside of the outer solenoid). - Hint: the
points AI, A2 and A3 are far from the solenoid ends, so the solenoids can be considered to be
infinitely long. Therefore, for NI = N2, B = J.LoNIi/L (a < r < b), and B = 0 (r < a, r> b).
Noting that the lines of the electric field are circles, use Faraday's law to find Eind(r).
Solution.
The points AI, A2 and A3 are far from the solenoid ends, so we can approximately
find
the required fields by assuming that the solenoids are infinitely long. Therefore, since NI = N2, the
magnetic flux density is given by
B
= J-tONIi
L
(a<r<b),
B =0
(r < a, r > b),
where r is the distance from the axis of the solenoids (what is the reference direction of B?).
Faraday's law tells us that
def>
Eind27rr= - dt '
CHAPTER
14:
ELECTROMAGNETIC
<I>being the magnetic flux through the surface spanned
the reference direction of Eind ?). Hence we obtain
Eind
=0
r2
(r < a) ,
It is left to the reader
Eind
139
INDUCTION AND FARADAY'S LAW
=-
to find numerical
- a2
dB
2r
dt
over the circular
of radius
b2 - a2 dB
(a < r < b),
values of Eind at points
contour
Eind
=-
dt
2r
r (what
is
(r>b).
A b A2 and A3.
--
ti3
i(t)
Q
i(t}
J
~'"
2a
00
2b
Fig. P14.3.
Fig. P14.4.
Two coaxial solenoids.
Rotating current loop and charge.
P14.4. A circular loop of radius a, with a current I, rotates about the axis normal to its
plane with an angular frequency UJ.A small charge Q is fastened to a loop radius and rofates
with the loop, as in Fig. P14.4. Is there a force on the charge? If it exists, determine its
direction. - (a) There is no force on the charge. (b) There is a force, radial. (c) There is a
force, parallel to the axis of rotation.
c
d
:
08
I
I
:.
I
I
fL > > /-La
I
I
:
+::.-t_-j--1a
Fig.
Pl4.5.
v
A partly
shielded
2a
b
wire loop.
Fig.
P14.6.
A solenoid
and a coil.
P14.5. A side of a rectangular wire loop is partly shielded from the magnetic field normal to
the loop plane with a hollow ferromagnetic cylinder, as in Fig. P14.5. Therefore the sides ad
and bc are situated in a different magnetic field. If the loop, together with the cylinder, moves
in the indicated direction with a velocity v, will there be a current in the loop? If the answer
is yes, this could serve for measuring
the veiocity with respect to the earth's
magnetic
field.
-
Hint: use superposition. (a) There will be no current in ~he loop. (b) There will be a current
in the loop. (c) Current in the loop exists, but its magnitude is too small to be measured.
I
140
P14.6.
PART 3: SLOWLY TIME-VARYING
A current i(t)
= Imsin(27rft) = 2.5sin314tA
ELECTROMAGNETIC
FIELD
is flowing through the solenoid in Fig.
P14.6, where frequency is in hertz and time is in seconds. The solenoid has Nl
=
= 50 turns
of wire, and the coil K shown in the figure has N2
3 turns. Calculate the emf induced
in the coil, as well as the amplitude of the induced electric field along the coil turns. The
dimensions indicated in the figure are: a 0.5cm, b = 1cm, and L = 10cm. Plot the induced
emf as a function
of N1, N2, and
(b) e = -O.116cos314tmV,
O.616mVjm.
f. -
=
(a) e
= 0.116cos314tmV, Eindm = O.616mVjm.
Eindm = O.313mVjm.
(c) e = -O.116cos314tmV,
Eindm '-
I-AA'
-1
a
A'
v
d
A
Fig. P14.8.
Fig. P 14. 7. A rotating conductor.
P14.7.
A conductor and two-wire line.
If the conductor AA' in Fig. P14.7 is rotating at a constant angular velocity and
makes n turns per second, find the voltage VAA' as a function
of time.
Assume
that at t
=0
the conductor was in the position shown in the figure. - (a) VAA,(t) = 27rnBadcos(27rnt).
(b) VAA,(t) = 27rnBadsin(27rnt). (c) VAA,(t) = -27rnBadsin(27rnt).
P14.8. A two-wire line is parallel to a long straight conductor with a dc current 1 (Fig.
P14.8). The two-wire line is open at both ends, and a conductive bar is sliding along it with a
uniform velocity v, as shown in the figure. Find the potential difference between the two line
conductors. - (a) VA'A(t) = J-lolvln[(d+a)ja]j(27r). (b)VA'A(t) = J-lolvln[(d+a)ja]j7r. (c)
VA'A(t) = J-lolvln[(d + 2a)ja]j(27r).
S P14.9. A rectangular wire loop with sides of length a and b is moving away from a
wire with a current 1 (Fig. P14.9). The velocity of the loop, v, is constant. Find the
emf in the loop. The reference direction of the loop is shown in the figure. Assume
t = 0 the position of the loop is defined by x = a. - For the reference direction
P14.9, (a) e = J-lolabvj[27r(a+ vt)(2a + vt)]. (b) e
e
= -J-lolabvj[7r(a + vt)(2a + vt)].
Solution.
given by
in Fig.
= -J-lolabvj[27r(a+ vt)(2a + vt)].
(c)
The magnetic flux through the surface S of the rectangle, oriented towards the reader, is
cI> ::::
-
1
5
where x
straight
induced
that at
= a + vi.
The induced
B dS
=-
x+a
l
x
1-'01b dr
= _l-'olb
21TT
emf in the loop, for the reference
In x + a
21T'
x'
direction
in Fig. PH.9,
is
CHAPTER
14: ELECTROMAGNETIC
141
INDUCTION AND FARADAY'S LAW
e = --del> = dt
J.Lolabv
21r(a+ vt)(2a + vt)
.
d
B
{y-
I)
x=o
I
I
I
a
x
'-.~
x
Fig. P14.9.
v
Fig. P14.11.
A moving frame in magnetic field.
Measurement of fluid velocity.
P14.10. The current flowing through the straight wire from the preceding problem is now i(t)
(a function of time). Find the induced emf in the loop, which is moving away from the wire as
in the preceding problem. What happens to your expression for the emf when (1) the frame
stops moving, or (2) when i(t) becomes a dc current, I? - Hint: there are two components
of the emf, one due to the motion of the frame, and the other to the magnetic field varying in
time.
.. .
Fig.
P14.12.
Cross section
of a solenoid.
Fig. P14.13.
A tubular coil.
P14.11. A liquid with a small but finite conductivity is flowing through a flat insulating
pipe with an unknown velocity v. The velocity of the fluid is roughly uniform over the cross
section of the pipe. To measure the fluid velocity, the pipe is in a magnetic field with a flux
density vector B normal to the pipe, as shown in Fig. P14.11. Two small electrodes are in
contact with the fluid at the two ends of the pipe. A voltmeter with large input impedance
shows a voltage V when connected to the electrodes. Find the velocity of the fluid. - (a)
v = Vj(2Bd). (b) v = 2Vj(Bd). (c) v = Vj(Bd).
P14.12.
Shown in Fig. P14.12 is the cross sec~ion of a very long solenoid of radius a
= 1 cm,
with N' = 2000 turnsjm. In the time interval O-:St :S 1 s, a current i(t) = 50t A flows through
the solenoid. Determine the acceleration of an electron at points AI, A2, and A3 indicated
in the figure. (Note: the acceleration, a, is found from the relation F = ma, where F is
I
142
PART 3: SLOWLY TIME-VARYING
(a) al
al = 0, a2
5.5. 107 mjs2 and a3
a3
4.3 .107 mjs2.
the force on the electron.)
=
=
S P14.13.
sketched
-
= 0, a2
= 3.5.107
= 7.3.107
mjs2.
ELECTROMAGNETIC
mjs2
and a3
(c) al = 0, a2
= 5.3.
FIELD
107mjs2. (b)
= 2.5.107
mjs2 and
Determine approximately the induced electric field strength inside the tubular coil
in Fig. P14.13.
The current
intensity
in the coil is I
= 1cm, b =
= 0.02 cos 106t A,
the number
1.5cm, and L = 10cm. Eind= 0.162sin106tVim. (b) Eind = 1.12sin 106t Vim. (c) Bind = -6.31 sin106tVim.
of turns is N = 100, and the coil dimensions are a
(a)
Solution. Starting from Eq. (14.3), we conclude that the induced electric field in the tube (for r < a)
is axial and approximately uniform. The magnetic flux density vector in the tube is circular (the lines
of B are circles centered at the tube axis), of intensity
B
= p.oNi
211"r
-~r is the distance from the tube axis). By neglecting the induced electric field outside the tube (this
can be done since L ~ b), we can write
EindL = - d4>
dt'
4>= p.oNiL In-b
211"
a '
so that Eind = O.162sin106tV/m.
Sj
S2
.
Sn
B
Fig.
P14.14.
A test of electromagnetic
induction.
Fig. P14.15.
~
B
An idealized electric motor.
P14.14. Sketched in Fig. P14.14 is an experimental setup for the analysis of electromagnetic
induction. By closing sequentially the switches 51"", 5n, it is possible to change the magnetic
flux through the closed contour shown from zero to a maximal value. Will the voltmeter
indicate an emf induced in the circuit? Explain. - Hint: note that there is neither a motion
of the contour in the magnetic field, nor a time variation of the magnetic field in which the
contour is situated.
P14.15. Find the angular velocity of the rotor of an idealized electric motor shown in Fig.
P14.15 for the case when no load is connected to it. Does the value of R influence the angular
velocity? The rotor is in the form of a metal wheel with four spokes, situated in a uniform
magnetic field of magnetic flux density B, as shown in the figure. What is the direction of
rotation of the rotor? - (a) IN = Ej(Ba2), clockwise. (b) IN= 2Ej(Ba2), clockwise. (c)
IN= E j ( B a2), counterclockwise.
CHAPTER
14:
ELECTROMAGNETIC
INDUCTION
AND FARADAY'S
143
LAW
P14.16. A circular loop of radius a rotates with an angular velocity w about the axis lying
in its plane and containing the center of the loop. It is situated in a uniform magnetic field of
flux density B(t) normal to theaxis of rotation. Determine the induced emf in the loop. At
t = 0, the position of the loop is such that B is normal to its plane. - (a) e = 2B1I"a2w sin wt.
(b) e
= -B1I"a2wsinwt.
(c) e
= B1I"a2wsinwt.
=
=
P14.17. A winding of N
1000 turns of wire with sinusoidal current of amplitude 1m
200 mA is wound on a thin toroida.l ferromagnetic core of mean radius a 10 em. Fig. P14.17a
shows the idealized hysteresis loop of the core corresponding to the sinusoidal magnetization
of the core to saturation in both directions. Also wound on the toroid are several turns of wire
over the first winding. Plot the emf induced in the second winding during one period of the
sinusoidal current in the first winding. - Compare your results with those in Fig. P14.17b.
=
r.Jt
B [T]
-318.3
1
r.Jt
H [Aim]
-1
r.Jt
(b)
(a)
Fig. P14,17.
(a) An idealized hysteresis loop. (b) Time dependence of H, B, and eind.
S P14.18. Shown in Fig. P14.18 is a rectangular loop encircling a very long solenoid of radius
R and with N' turns of wire per unit length. The amplitude of current in the winding is 1m,
and its angular frequency is w. Determine the emf induced in the entire rectangular loop, as
well as in its sides a and b separately. - Hint: note that the induced electricfield is normal to
the loop diagonals, and determine the flux through the four triangles obtained by dividing the
loop by its diagonals.
Solution. The magnetic flux through the loop is
4>
= /-LoN'lm
cos",-,t.
Consider the triangle with a side a and a vertex coinciding with the center of the loop (there are two
such triangles in Fig. Pl4.I8), and also the triangle with a side b. The fluxes through these triangles
are
1I"-a
4>a= -
211"
4>,
a
4>b= -211"4>,
I
PART 3: SLOWLY TIME-VARYING
144
where 0:/2
=tan-l
ELECTROMAGNETIC
FIELD
(b/a) (see the figure).
There is no emf induced in the sides of triangles indicated in dashed line in the figure (vector
Eind is normal to these sides). So, the induced electromotive forces in the entire loop, in side a, and
in side b, are respectively given by
d<p
e = -d't'
.
ea
---,
-
d<Pa
eb=
dt.
-
d<Pb
dt
The same results can be obtained as line integrals of the induced electric field, but it is not so simple.
,
, ,
,
"
R
,..
/
/
/ /
/
//////
b
0<
,..
i(t;",
)
,
a
Fig.
P14.18.
A rectangular
loop encircling
a solenoid.
P14.19. The cross section of a thick coil with a large number, N, of turns of thin wire, is
shown in Fig. P14.19. The coil is situated in atime-varying magnetic field of flux density B(t),
in the indicated direction. Determine the emf induced in the coil. - Hint: note that different
layers of turns have different fluxes through them. The induced emf in the coil is e -d q>/ dt,
where: (a) q>= 71"
B(t)N( a2 + ab + b2) cos a. (b) q> -(1/6)71"B(t)N( a2 + ab + b2)cos a. (c)
q> = -(1/3)71"B(t)N( a2 + ab + b2) cos a.
=
=
b
.I.I.I.I.I.T.I.
.T.'.'.'.'.""
,'.T.'.T.'.'."
Fig. P14.19.
S
v
Fig. P14.20.
A coil of rectangular cross section.
A conductor encircled by a coil.
P14.20.
The conductor whose cross section is shown shaded in Fig. P14.20 carries a sinusoidal
current of amplitude 1m and angular frequency VJ. The conductor is encircled by a flexible
thin rubber strip of cross-sectional area S, densely wound along its length with N' turns of
wire per unit length. The measured amplitude of the voltage between the terminals of the
strip winding is Vm. Determine 1m. - (a) 1m Vm/(J..LoN'S). (b) 1m 2Vm/(J..LoN'S). (c)
=
1m
=
= v'm/(2J..LoN'S).
=
Solution.
There are dN
N'dl turns of wire on a length dl of the strip. The magnetic flux through
a single turn is <1>0 B . S, and that through dN turns is
=
CHAPTER
14:
ELECTROMAGNETIC
145
INDUCTION AND FARADAY'S LAW
dcp= cpodN = N'dl. BS,
since Sand dl are colinear. The total flux through all the turns of the flexible solenoid is thus
cp=
i
dcp=N'S
i
B.dl=JJON'Si(t),
according to the Ampere law applied to the contour C along the strip. The induced emf in the winding.
is e = -dcp/dt,so that, finally, the expression for the amplitude of i(t) reads 1m
Vm/(JJoN'SIN).
=
~(t)
R
A
x
LI
D
Fig. P14.21.
A test of electromagnetic induction.
Fig. P14.22.
K
A coil with a sliding contact.
P14.21. Wire is being wound from a drum Dl onto a drum D, at a rate of N' turns per unit
time (Fig. P14.21). The end of the wire on drum D is fastened to the ring R, which has a
sliding contact F. A voltmeter is connected between the contact F and another contact G.
Through the drum D there is a constant flux (>, as indicated. What is the electromotive force
measured by the voltmeter? - Hint: recall that the emf can be induced in only two ways: by
moving sections of the contour in the magnetic field, or by varying the magnetic field in time.
S P14.22.
magnetic
A cylindrical coil is tightly wound around a ferromagnetic core with time-varying
flux (>(t)
= (>m cos f.AJt,as
shown in Fig.
P14.22.
The length
of the coil is L, and
the number of turns in the coil is N. If a sliding contact K moves along the coil according
to the law x = L(1 + cos f.AJ1t) /2, what is the time dependence of the electromotive force
between contacts A and K? Plot your result. - (a) e
Nf.AJ(>msinf.AJt(1
+ COSf.AJlt).(b)
=
e = (Nf.AJ(>m/2) sin f.AJt(1 + COSf.AJlt). (c) e = (3Nf.AJ(>m/2) sin f.AJt(1 + sin f.AJlt).
Solution. According to Faraday's law of electromagnetic induction, an electromotive force is induced
in each turn of the coil, and its value is
el = --
dcp
.
= INCPm
sin INt.
dt
Since between the contacts A and J( there are at the same instant
n-
-
Nz
L
= N2 (1 +coslNlt)
146
turns,
PART 3: SLOWLY TIME-VARYING
the electromotive
force induced
ELECTROMAGNETIC
FIELD
in all of them is
e = nel =
N"",«I>m.
2
Slll"",t(1 + cOS""'l
t).
Note that in this case Faraday's law cannot be written in the form
e
=-
d«l>total=
dt
- d(n«l»
dt
= -n d«l>- «I>dn.
dt
dt
The second term on the right represents the emf that would be induced as the result of an increase of
the number of turns. By the same arguments as in the preceding problem, it is clear that such an emf
does not exist.
S=-'~
N
G
.
"
s
Fig. P14.23.
.
A rotating magnet with sliding contacts.
P14.23. A cylindrical conducting magnet of circular cross section rotates about its axis with
a uniform angular velocity. A galvanometer G is connected to the equator of the magnet
and to the center of one of its bases by means of sliding contacts, as shown in Fig. P14.23.
If such an experiment is performed; the galvanometer indicates a certain current through the
circuit (Faraday, 1832). Where is the electromotive force induced: in the stationary conductors
connecting the sliding contacts with the galvanometer, or in the magnet itself? - Hint: note
that the magnetic field is time-constant in spite of the magnet rotation, and the material of the
magnet rotates in this field.
-+~
(a)
"
.
D~
(b)
(c)
"
Fig. P14.24. (a) A moving elastic loop encircles a magnetized toroid. (b) The loop is moving, but
closed by conducting toroid body. (c) The loop does not move and does not encircle the toroid.
I
14: ELECTROMAGNETIC
CHAPTER
141
INDUCTION AND FARADAY'S LAW
P14.24. A ferromagnetic toroid with no air gap is magnetized so that no magnetic field exists
outside it. The toroid is encircled by an elastic metal loop, as in Fig. P14.24a. The loop is now
taken out of the toroid in such a way that during the process, the loop is always electrically
closed through the conducting material of the toroid, as in Figs. P14.24b and c. The magnetic
flux through the contour was obviously changed from a value cI>,the flux through the toroid,
to zero. A formal application of Faraday's law leads to the conclusion that a certain charge
will flow through the circuit during the process, but in this case it is not possible to detect any
current (Herring, 1908). Explain the negative 'j'esult of the experiment. - Hint: recalrthat
the emf can be induced in only two ways: by moving sections of the contour in the magnetic
.
field, or by varying the magnetic field in time.
P14.25. Discuss the possibility of constructing a generator of electromotive force constant
in time, operating on the basis of electromagnetic induction. - Hint: solve the equation
~~ = -E = constant, which gives the flux necessary to obtain an emf constant in time.
S -'P14.26.
= 1 mm
In a straight copper wire of radius a
there is a sinusoidal current i(t)
=
1 cos I.J.)t
A. A voltmeter is connected between points 1 and 2, with leads of the shape shown
=
=
in Fig. P14.26. If b 50 cm and c 20 cm, evaluate the voltage measured by the voltmeter
for (1) I.J.) 314 rad/s, (2) I.J.)
= 104 rad/s, and (3) I.J.)
= 106 rad/s. Assume that the resistance
of the copper conductor per unit length, R', is approximately that for a dc current (which
actually is not the case, due to the so-called skin effect), and evaluate for the three cases
R'bi(t) and the voltage induced in
the difference between the potential difference VI - V2
1.98 m V.
the leads of the voltmeter. - The rms value of the potential difference VI - V2
The difference between this potential difference and the voltage indicated by the voltmeter for
the three specified frequencies is: (a) (1) 217.4ILV, (2) 5.75 mV, and (3) 5.75 V. (b) (a) (1)
117.4ILV, (2) 3.75mV, and (3) 3.75 V. (c) (a) (1) 87.4ILV, (2) 2.75mV, and (3) 2.75 V.
=
=
Solution.
The voltage between the terminals of the voltmeter is
Vvoltmeter
where R'
=
=
= (VI -
V2)
-
e
= R'bi - e,
=
1/(crcu 1I"a2) (crCu
5.7 . 107 S/m), and e is the induced emf in the rectangular contour
containing the voltmeter and the wire segment between points 1 and 2. This emf is given by (see
solution to problem PH.10)
e
The rms value of the potential
= 211"dt
b di
difference
J.LO
VI
-
In c + a
a
.
V2 amounts
to 1.98 m V, and does not depend
on
frequency. The difference between this potential difference and the voltage indicated by the voltmeter
for the three specified frequencies is: (1) 117.4J.LV, (2) 3.75mV, and (3) 3.75 V. This difference
represents an error in measuring the potential difference using the voltmeter with such leads. We see
that
in case (2) the relative
error is as large as 189
%, and that in case (3) such a measurement
is
meaningless.
P14.27. A circular metal loop of radius R, conductivity 0' and cross-sectional area S encircles
a long solenoid with a time-varying current i(t) (Fig. P14.27). The solenoid has N' turns of
wire per unit length, and its radius is r. Determine the current in the loop, and the voltage
between points A and B of the loop along paths a, b, c, and d. Neglect the induced electric field
in the loop due to the loop current itself. Determine also the voltage between points A and
I
148
PART 3: SLOWLY TIME-VARYING
C, along the path AcC, and along the path AbBC. -
ELECTROMAGNETIC
FIELD
Let the loop be oriented clockwise, and
let the emf inducedin it bee. (a) VAaB= VAbB= -VAcB = -VAdB = e/3, VAcC= -3e/4,
VAbBC= e/4. (b) VAaB= VAbB= -VAcB = -VAdB = -e/2, VAcC= e/4, VAbBC= -e/4.
(c) VAaB= VAbB= -VAcB = -VAdB = e/2, VAcC= -e/4, VAbBC= 3e/4.
b
v
d
c
c
2
2a
i(t)
Fig.
P14.26.
Measurement
of ac voltage.
Fig. P14.27.
A solenoid encircled by a circular loop.
P14.28.
Repeat problem P14.27 assuming that the two halves of the loop have different
conductivities, al and az, and they meet at points A and B. - Let the conductivity of
the semic'ircleAaB be a}, and that of AcB az. (a) VAaB = VAbB
= eaz/(al + az), VAcB=
VAdB= -eaI/(al +az), VAcC= VAcB/2,VAbBC= VAaB/2. (b)VAaB= VAbB= eaz/(2al +
az), VAcB = VAdB = -eaI/(2al + az), VAcC= VAcB/2, VAbBC= VAaB/2. (c) VAaB =
VAbB = eaz/(al + 2az), VAcB = VAdB = -eaI/(al + 2az), VAcC = VAcB/2, VAbBC=
VAaB/2.
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 known as 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 Cz by a current il(t) in a nearby loop C1 is given by
elz(t)
=
1
Elind(t).
dlz
(V),
(15.1)
JC2
where Elind(t) is the induced electric field of current il(t) along loop Cz. In linear media the
magnetic flux of current il(t) through Cz is proportional to il(t),
cJ>12(t) = LIZil(t)
.
It can be proved that LZI
= LIZ, so that
(Wb).
(15.2 )