16. Energy and Forces in the Magnetic
Field
. Almost any device that makes use of electric or magnetic forces can be made in an "electric
version" and in a "magnetic version". Realizable magnetic forces are much greater than electric
forces, and therefore devices based on magnetic forces are used much more often.
. Assume that n rigid and stationary contours, with currents il(t),i2(t),...,in(t),
and re.sistances Rl,R2,...,Rn,
are connected to generators of emf's el(t),e2(t),...,en(t),
and that
the total fluxes through the contours are ~1(t),~2(t),...
,~n(t). The elemental work done by
the generators in all the contours in a time dt satisfies the equation
dAg
= dAJ + dAm.
(16.1)
The first term on the right are Joule's losses, and the second is the work done in changing the
magnetic field.
.
The work of generators
in individual
(dAgh
contours
can be expressed
as
= Rkii(t)dt+ik(t)d~k(t), k = 1,2,...,n,
(16.2)
so that the work of all the generators is
n
dAg
n
= L Rkii(t)dt
k=1
+
L
ik(t)d~k(t).
k=1
(16.3)
The second term on the right-hand side represents the energy used to change the magnetic
field,
n
dAm
= dAg -
dAJ
=L
ik(t)d~k(t).
k=1
(16.4)
This is the law of conservation of energy for the n current contours. dAm is the work necessary
to change the fluxes through the n contours by d~l,d~2,...,d~n'
. The total work Am needed to establish de currents II, h,. .. , In in the contours, for which
the fluxes through the contours are ~1, ~2,' . . , ~n, is hence
CHAPTER 16:
159
ENERGY AND FORCES IN THE MAGNETIC FIELD
n
(Am)in
establishing
f<PIe
=L
k=l 10
0
currents
ik(t)d~k(t)
(J),
(16.5)
which for linear media becomes the energy stored in the magnetic field,
'.
n
(16.6)
(J).
Wm=~Lh~k
k=l
. In terms of the self and mutual inductances of the contours and currents in them this has
the form
n
1
Wm
which for a single contour
n
.
= ~ LLLjkljh
..j=l k=l
(J),
(16.7)
becomes
Wm
= !I~
= !LI2
2
2
(16.8)
(J).
. The density of work that needs to be done to change the magnetic flux density at a point
from Bl to B2 is given by
B2
dAm
dv
=1
H(t)dB(t)
(Jfm3).
(16.9)
B1
For linear media, this yields the density of energy at a point of the magnetic field,
dWm
dv = !
2 B2
J.l
(Jfm3).
= !J.lH2
= ~BH
2
(16.10)
The energy contained in the magnetic field in a linear medium is thus
1
Wm
= 1v 2J.lH
2
dv
(J).
(16.11)
. If a sinusoidal magnetic field exists in a piece of ferromagnetic material, there are hysteresis
losses in the material. They are proportional to the hysteresis loop area and to frequency. If the
field is uniform, the losses are also proportional to the volume of the ferromagnetic material.
. The part of inductance
as the internal inductance,
associated with the energy contained
and is defined by
inside a conductor
is known
160
PART 3: SLOWLY TIME-VARYING
Linternal
=
2(Wm)inside
conductor
ELECTROMAGNETIC
(H).
i2
FIELD
(16.12)
The external inductance is associated with the energy outside the conductor,
L
external
-
2(Wm)outside
conductor
(H).
i2
(16.13)
For a long straight wire of circular cross section the internal inductance per unit length is given
by
,
--
-
Linternal
.
If the distribution
of currents
J1.
(Him).
81r
in a magnetically
homogeneous
(16.14)
medium
is known, the Biot-
Bavart law can be used for determining the magnetic flux density. Combined with the relation
dFm = I dl
X B,
the magnetic force can then be found on any part of the current distribution.
In some important cases the magnetic force can be calculated as a derivative of the magnetic
energy. This can be done under two assumptions: (1) the fluxes through all the contours are
kept constant, or (2) the currents in all the contours are kept constant:
dWm
Fx = - (~ )
.
dWm
(
Fx = + ~
,
(16.15)
)I=constant.
(16.16)
cI>=constant
Using these formulas, the magnetic pressure on boundary surfaces between two media is obtained as
P=
B2
2(J1.2- J1.1) H;ang. + J1.1j..L2
norm.
1,
(
)
(reference
direction
into medium
1).
(16.17)
The maximal obtainable magnetic pressure is on the order of 104 times greater than the
maximal obtainable electric pressure.
QUESTIONS
Q16.1. What does Eq. (16.1) actually represent? - (a) The law of conservation of energy.
(b) An equation for evaluating the power of Joule's losses. (c) An equation for evaluating the
power of generators.
S Q16.2. Explain why the expression dAg = e(t) i(t) dt is the work done by a generator. Hint: recall the definition of the emf of a generator, of the potential difference, and of the
current intensity.
CHAPTER
161
16: ENERGY AND FORCES IN THE MAGNETIC FIELD
Answer. The generator does the work against the electric forces of charges on its electrodes. The
work of electric forces is obtained as the potential difference between the electrodes, Vet), times the
charge dq = idt (dq > 0) transferred by the electric field from the positive to the negative electrode.
Since the electromotive force has the same value as Vet), and the generator moves the charges in the
opposite direction, e(t) i(t) dt is the work done by the generator.
Q16.3. For a simple circuit of resistance R, with an emf e(t), e(t) = Ri(t)+d~(t)/dt.
the physical meaning of the last term. - Hint: recall Faraday's law.
S
Explain
QI6.4. Why does the energy of a system of current loops not depend how the currents in the
loops attained their final values? - (a) Because Joule's losses are the same. (b) Because the
induced electromotive forces along the loops are always the same. (c) Because otherwise the
law of conservation of energy would be violated.
Answer.
This would violate the law of conservation of energy. We could then always make the
..currents attain their final value in a way using less energy and then make them go to zero in a way that
requires more energy to build up the currents, and therefore returns more energy when the currents
are reduced to zero. In this way we would make a surplus
of energy with no source of energy.
.
Q16.5. Is Eq. (16.7) valid for nonlinear magnetic media? Explain. - (a) Yes, because the
derivation does not depend on the medium. (b) No, because for nonlinear media the flux is
not proportional to the current. (c) No, because for nonlinear media the starting equation, Eq.
{16.1}, is not valid.
QI6.6. The current in a thin loop 1 is increased from zero to a constant value I. A thin
resistive loop 2 has no generators in it, but is in the magnetic field of the current in loop 1.
Both loops are made of a linear magnetic material. Are the power pg(t) of the generator in
loop 1 and the final value Wm of the energy stored in the system affected by the presence of
loop 2? - (a) pg(t) is affected, Wm not. (b) Pg(t) is not affected, Wm is. (c) Neither is
affected.
QI6.7. Repeat question Q16.6 with loop 2 open-circuited.
for the preceding questions.
-
The answers are the same as
Q16.8. A body of a linear magnetic material is placed in the vicinity of loop 1 of question
Q16.6. Is some energy associated with the magnetization of the body? - (a) None. (b) Yes,
some. (c) Depends on the size of the body.
Q16.9. Eq.(16.6) was derived by assuming that the currents were increased inside stationary
conductors. Using the law of conservation of energy as an argument, prove that this expression
must be valid for the magnetic energy of the system considered, irrespective of the process by
which the current system is obtained. - Hint: note that, if this were true, different processes
would require different energies to obtain the same system.
Q16.10. Using a sound physical argument, explain why the work in Eq. (16.5) done by the
generators in establishing a given time-constant magnetic field is a function of the process by
which the system of currents is established when ferromagnetic materials are present in the
field. - Hint: recall hysteresis losses.
S Q16.1l. Will the magnetic energy of a system of fixed quasi-filamentary dc current loops be
changed if a closed conducting loop with no current is introduced into the system? Explain.
162
PART 3: SLOWLYTIME-VARYING ELECTROMAGNETIC FIELD
- (a) It will not be changed. (b) It will be changed. (c) It will be changed only if the loop is
superconductive.
Answer. Only if the loop is superconductive, because then a current will be induced (and remain) in
it, which would change the magnetic field, and its energy. If the loop has resistance, however small,
a current will be induced in the loop while it is introduced in the field, but it will soon become zero
once the loop is in its final position.
Q16.12. Is it possible to determine theoretically the self-inductances and mutual inductances
in a system of current loops by starting from Eq. (16.7) if Wm is known? Explain. - (a)
Yes, because the process is reciprocal. (b) No, because this is just one equation. (c) Yes, if the
number of loops is less than three.
Q16.13. Two equal thin loops of self-inductance L are pressed onto each other, so that
IL121 ~ L. If the currents in the loops are hand
h, what is the magnetic energy of the
system? Answer the question if the two currents are (1) in the same direction and (2) in
2£12 = Wm2. (b) Wm1 = 2L12, Wm1 = -2£12. (c)
opposite directions. - (a) Wm1
:
2
=
Wm1 = 2LI , Wm1 = O.
S
Q16.14. How would you make an electric version of a generator of sinusoidal emf? - Hint:
e.g., start from a rectangular metal plate cut into two mutually insulated halves. Let it rotate in
a uniform electric field, the two plate halves being connected by sliding contacts to a resistor.
Answer. Imagine the page you are reading to be cut in half, parallel to the printed lines, with the
two halves insulated by a thin dielectric rod representing the axis of rotation. Let the two plates be
connected by sliding contacts to a resistor. If this structure rotates in a uniform electric field, the
plates will be charged according to the sinusoidal time variation, and there will be a current of this
time variation in the resistor. Efficiency of such a generator, however, is extremely small, which the
reader can readily prove himself/herself.
Q16.15.
How would you make an electric version of a generator of "rectified" sinusoidal
emf? - Hint: having in mind the answer to the preceding question, recall the commutator in
generators based on electromagnetic induction.
Q16.16. Imagine somebody came to you with a piece of a ferromagnetic material he developed
and stated that the working point moves along the hysteresis loop in the clockwise direction.
Would you believe him? Explain. - (a) Yes, if we are certain that he is a reliable person. (b)
Yes, because the motion of the working point can be either way. (c) No, because such a piece
of material would produce energy, instead of dissipating it.
Q16.17. Is it possible to derive Eq. (16.9) from Eq. (16.10)? Explain. of the two is more general.
Hint: recall which
Q16.18. If a hysteresis loop was obtained by a sinusoidally varying H(t), will the hysteresis
losses be exactly equal to the area of this loop if H(t) varies as a triangular function of time
(that is, varies linearly from -Hm to Hm, then back to-Hm,
and so on)? Explain. - (a)
They will be the same, provided the areas of the triangle and the sinusoid are the same. (b)
They do depend on the time variation of the field, and will not be the same. (c) They will be
the same, because the shape of the hysteresis loop does not depend on H(t), as long as it is
periodic and of the same amplitude.
"
CHAPTER
16: ENERGY AND FORCES IN THE MAGNETIC FIELD
163
Q16.19. If the frequency of the alternating current producing a magnetic field is f (cycles
per second), what is the power per unit volume necessary to maintain the field in a piece
of ferromagnetic substance ? (a) It equals f2 times the integral around the hysteresis loop of
H(t)dB(t). (b) f times that integral. (c).J! times that integral.
S Q16.20. According to the expression in Eq. (16.10), the volume density of magnetic energy is
always greater in a vacuum than in a paramagnetic or idealized linear ferromagnetic material
for the same B. Using a sound physical argumeQt, explain this result. - Hint: recall what is
done when such materials are magnetized.
Answer. A paramagnetic or ferromagnetic material is magnetized so that it enhances the magnetic
field. Less energy will be needed to obtain the same B inside such a material, than in a vacuum.
Q16.21. The magnetization curve of a real ferromagnetic material is approximated by a
. nonlinear, but single-valued, function B( H) (not by a hysteresis loop). Is it possible to speak
about the energy density of the magnetic field inside the material? If you think it is, what is
the energy density equal to? - (a) No, it is not possible, because the material is not linear.
.:(b) No, it is not possible, because such an approximation is not possible to make. (c) Yes, it
is equal to the integral of H(t)dB(t) from B
0 to the value of B at the point considered.
=
S Q16.22. A thin toroidal ferromagnetic core is magnetized to saturation and the current in
the excitation coil is reduced to zero, so that the operating point in the H - B plane is H = 0,
B = Br. Is it possible to speak in that case about the energy of the magnetic field stored in
the core? Is it possible to speak about the energy used to create the field? Explain. - Hint:
recall that when we say "energy stored in the core", we mean how much energy we would get
if the field goes to zero.
Answer. When we say "energy stored in the core" , we mean how much energy we get if the field goes
to zero. In this case we cannot make the field go to zero except by introducing more energy in the core.
So we cannot define any magnetic energy stored in the core. The energy used to create this field is
defined, however, by the integral of H(t)dB(t) from B
0 to B
Br along the magnetization curve,
multiplied by the volume of the core. We know that one part of this energy was lost to hysteresis
losses.
=
=
Q16.23. The first part of the magnetization curve can be approximated as B( H) = C H2,
where C is a constant. How can you evaluate in that case the energy density necessary for the
magnetization of the material? Is that also the energy density of the magnetic field? - Hint:
use Eg. (16.9).
Q16.24. Is it possible for the initial magnetization curve to be partly decreasing in B as H
increases? What would that mean? - (a) Yes, it depends on the material. (b) No, because
this means a flow of energy from the material back to the sources during the magnetization
process. (b) Yes, for any material, with appropriate time variation of current in the coil.
Q16.25. Evaluate approximately the density of hysteresis losses per cycle for the hysteresis
200A/m and Bm
0.5T. - (a) 30Jjm3. (b) 80Jjm3. (c)
loop in Fig. Q16.25 if Hm
=
=
150Jlm3.
Q16.26. Why are hysteresis\losses linearly proportional to frequency? as an argument.
Hint: use Eq. (16.9)
164
PART 3: SLOWLY TIME-VARYING
ELECTROMAGNETIC
FIELD
S Q16.27. Is the volume density of hysteresis losses in a thick toroidal ferromagnetic core with
a coil carrying a sinusoidal current the same at all points of the core? Is the answer the same
if the current intensity is such that at all points of the core saturation is attained, and if it is
not? - Hint: recall that H is not of the same value in the entire core.
Answer. It is not the same, since the hysteresis loops have different sizes for different distances from
the toroid axis. If the saturation is attained at all points, this argument does not hold any more, and
the density of hysteresis losses is then the same at all points of the core.
Q16.28. Why can the self-inductance of a thick conductor not be defined naturally in terms
of the induced emf or flux through the conductor? - (a) There is no possibility to define a
unique closed loop. (b) The closed loop must not be inside a conductor. (c) The axis of a thick
loop is not defined.
Q16.29. Why is it very difficult to obtain the internal inductance using the flux definition
of self-inductance? To answer the question, consider two wires, one thin and the other thick,
with the same current I flowing through them. - Hint: discuss the possibility of defining a
~closed contour for defining the total flux in such cases.
Q16.30.
Direct current due to a lightning stroke on a three-phase line propagates along the
three conductors.
Will the force repel or attract
the conductors?
-
(a) There will be no force
on the conductors. (b) The force will be repulsive. (c) The force will be attractive.
h
Hm
a
Fig.
Q16.25.
A hysteresis
loop.
Fig. P16.6.
A thick toroidal coil.
Q16.31. If a lightning stroke hits a transformer, in some cases it may be noticed that the
transformer "swells". Explain why. - (a) It is heated up and explodes. (b) The magnetic
forces -always tend to increase the size of a current loop. (c) It is due to chemical changes in
the transformer insulation.
PROBLEMS
S P16.1. Write the explicit expression for the magnetic energy of three current loops with
currents It, h, and h. Assume that the self-inductances and mutual inductances of the loops
are known. - Hint: use Eq. (16.7) with n = 3. Note that the double sum in that case results
in a total of nine terms.
CHAPTER
16: ENERGY AND FORCES IN THE MAGNETIC FIELD
Solution.
The energy of such a system is
3
Wm
1
21
165
3
= ~ :L:LLkj1k1j
k=l j=l
21
...
2'
= -L1111
+ -L2212 + -L3313 + L12hlz + L13hl3 + L23lzl3.
222
(This is Eq. (16.11) for n
=3.)
PI6.2. Find the magnetic energy per unit length in the dielectric of a coaxial cable of inner
conductor radius a and outer conductor radius b, carrying a current I. The permeability of
L~xternaII2/2.
- (a) W~
(}.loI2/2rr)ln(b/a). (b)
the dielectric is }.l0. Show that W~
"W~ = (}.loI2/rr) In(b/a). (c) W~ = (}.loI2 /4rr) In(b/a).
=
=
'PI6.3. Find the total inductance per unit length of a coaxial cable that has an inner conductor
of radius a and an outer conductor with inner radius b and outer radius c. The permeability
of the conductors and the dielectric is }.lo,and current is distributed uniformly over the cross
sections of the two conductors. - Hint: add the two internal inductances to the external
inductance.
PI6.4.
A thin ferromagnetic toroidal core is made of a material that can be characterized
by a constant permeability }.l 4000}.l0. The mean radius of the core is R=10 cm
and the core cross-sectional area is 8=1 cm2. A current of 1=0.1 A is flowing through N =500
turns wound around the core. Find the energy spent on magnetizing the core. Is this equal to
the energy contained in the magnetic field in the core? - (a) Wm 1 mJ. (b) Wm 2.1 mJ.
(c) Wm = 3.2 mJ.
=
approximately
=
=
PI6.5. In the toroidal core of the preceding problem, a small part of length 10=2 mm is cut
out so that now there is a small air gap in the core. The current in the coil is kept constant
while the piece is being cut out. Find the energy contained in the magnetic field in this case.
- (a) 38}.lJ. (b) 56}.lJ. (c) 73}.lJ.
PI6.6. Show that the same expression for the self-inductance ofthe toroidal coil in Fig. P16.6
is obtained from the expression ~ = LI, and the expression 2Wm
LI2. - Hint: the first
expression requires the determination of the magnetic flux through all the coil turns, and the
second expression requires the evaluation of the magnetic energy in the toroid core using Eq.
(16.11).
=
S PI6.7.
On a thin ferromagnetic toroidal core of cross-sectional area 8
turns of thin wire are tightly
wound.
The mean radius of the core is R
= 1cm2, N
= 16 em.
= 1000
It may be
assumed that the magnetic field is uniform over the cross section of the toroid. The idealized
initial magnetization curve is shown in Fig. P16.7. Determine the work Am done in establishing
the magnetic field inside the toroid if the intensity
-
of the current
(a) 0.021J. (b) 0.050J. (c) 0.075J. Repeat the problem if (1) I
=
Solution.
The magnetic field intensity in the core, H
N1/(21rR)
1000 A/m, so the density of work done in establishing the field is
through
the coil is I
= 2 A.
= 1 A, and (2) I = 0.5A.
=:: 2000A/m,
is larger than
166
PART 3: SLOWLY TIME-VARYING
-dAm
=10
dv
1T
1
H dB
A
ELECTROMAGNETIC
J
= -1000
- . 1T = 500 3
2
m
m
(the density of work is proportional to the area of the triangle between the magnetization
the B axis in Fig. P16.7). The total work is
Am
FIELD
dAm
= -2-rrRS
dv
=0.05J.
curve and
.
Repeat the problem if (1) 1= 1 A, and (2) 1= 0.5 A.
B [T]
B [T]
1000
Fig.
Pl6.7.
1000
H [Aim]
An idealized
magnetization
curve.
Fig. Pl6.8.
H[A/m]
An idealized magnetization
curve.
P16.8. Repeat problem P16.7 if the idealized initial magnetization curve is as shown in Fig.
P16.8. - (a) 0.031 J. (b) 0.07 J. (c) 0.10 J.
B
B
""
Bm
H
H
Hm
Hm
-Hm
""
-Bm
Fig. Pl6.l2.
An idealized hysteresis loop.
Fig. Pl6.l3.
An idealized hysteresis loop.
P16.9.
On the toroidal core shown in Fig. P16.6, N = 650 turns of thin wire are tightly
wound.
The intensity
of the time-constant
current
in the coil is I
=
2 A and the idealized
initial magnetization curve of the core is as shown in Fig. P16.7. Determine the work done
5 cm, b = 15cm, and h = 10cm. - (a)
in establishing the magnetic field in the core if a
5.41J. (b) 8.23J. (c) 3.14J.
=
P16.10. Repeat the preceding problem for intensities of current through the coil of (1) 0.5A,
and (2) 1A. - The result in case 2: (a) 2.54J. (b) 6.38J. (c) 8.22J.
CHAPTER
16: ENERGY AND FORCES IN THE MAGNETIC FIELD
167
PI6.11. The initial magnetization curve of a ferromagnetic material can be approximated by
BoH/(Ho + H), where Bo and Ho are constants. Determine the work done per unit
B(H)
volume in magnetizing this material from zero to a magnetic field intensity H. - Hint: first
determine dB, and then the work done per unit volume using Eq. (16.9).
=
s
PI6.12. The idealized hysteresis loops of the ferromagnetic core in Fig. P16.6 are as in Fig.
P16.12. Determine the power of hysteresis losses in the core if it is wound with N turns of
wire with sinusoidal current of amplitude 1m and frequency f. Assume that saturation is not
reached at any point, and neglect the field of eddy currents. - Hint: the energy lost per unit
volume of the core in one cycle is numerically equal to the area of the hysteresis loop at the
considered point of the core. Note that, in the present case, the hysteresis loops are different
for different points. Physteresislosses= f Whysteresislosses, where Whysteresislosses =
(b) JLnormal
N2 l~ h In -,
411"
ab
(a) JLnormal
N2 l~ h In-,ab
211"
_and JLnormal is the normal permeability,
(c) 2JLnormai
N2 l~ h In -,
11"
ab
JLnormal = Bm/ Hm.
Solution. The energy lost per unit volume of the core in one cycle is numerically equal to the area
of the hysteresis loop at the considered point of the core. In the present case, the hysteresis loops are
different for different points, since the amplitude of the magnetic field intensity is
Hm(r)
=
Nlm
21!'r .
Consequently, we have to determine the elemental losses in thin circular rings of radius r and thickness
dr (a < r < b), the total losses being given as their sum (integral). Note that (Fig. P16.12)
Bm (r) - Bm
Hm(r)
-
Hm
= const = JLnormal,
where JLnormal is the normal permeability of the core.
Losses in one such ring in one cycle are numerically equal to the area of the loop in Fig. P16.12
multiplied by the volume of the ring:
dWhysteresis losses = [4Bm (r)Hm (r)] (21!'rhdr)
= 8JLnormaIH~(r)l!'rhdr.
Losses in the entire core per cycle are
W hysteresis losses =
Finally, noting that
given by
l
b
b
2JLnormal N212m h dr = 2JLnormai N 2 1m
2 h I
l!'r
n-.
a
11'
a
f is the number
of cycles per unit time, the power of the hysteresis
Physteresis losses = f Whysteresis losses'
losses is
168
PART 3: SLOWLY TIME-VARYING
ELECTROMAGNETIC
FIELD
P16.13. Repeat problem P16.12 for idealized hysteresis loops shown in Fig. P16.13, assuming
Bm/ Hm for all the loops is the same and that saturation is not reached at any point. Neglect
the field of eddy currents. - Hint: the losses are one half of those in the preceding problem
(why?).
P16.14. A ferromagnetic core of a solenoid is made of thin, mutually insulated sheets. To
estimate the eddy current and hysteresis losses, the total power losses were measured at two
frequencies, II and h, for the same amplitude of the magnetic flux density. The total power
losses were found to be PI and P2, respectively. Determine the power of hysteresis losses.
and of eddy current losses at both frequencies. Hint: note that Pl/2 = Physteresis losses 1/2 +
=
Peddy current losses 1/2
C1 fl/2 + C2f;/2' and solve the two equations obtained at the two
frequencies for the constants C1 and C2'
+
s
-Hs
dx
/ /
/ /
Fig.
P16.15
12
Tube of flux of a current
Fig. P16.17.
loop.
An electromagnet.
*P16.15. Prove that Eq. (16.9) is valid for any magnetic field, not necessarily uniform. Hint: consider an arbitrary current contour, .divide the field into elemental tubes of magnetic
flux, as in Fig. P16.15, and use Eq. (16.5).
S P16.16. Two coaxial solenoids of radii a and b, lengths It and 12, and number of turns N1 and
N2 have the same c1J.rrentI flowing through them. Find the axial force that the solenoids exert
on each other if the thinner solenoid is pulled by a length x (x < It, 12) into the other solenoid.
Neglect edge effects and assume that the medium is air. Hint: assume that b < a, that the
turns of both solenoids (and the currents in them) are oriented in the same direction, and find
the mutual inductance of the solenoids as a function of x. Then use Eq. (16.16). (a) Fx
JloN[N'i7rb212/(l112)' (b) Fx
= JloN1N27rb212/(l112)'
(c) Fx
= 4JloN1N27rb212/(l112)'
=
Solution.
Assume that b < a, and that the turns of both solenoids (and the currents in them) are
oriented in the same direction. The mutual inductance of the solenoids is then
L12
The total magnetic
energy of the system
= J.tONIN27rb2X
/1/2
is
1
Wm
= (2"Ll +
1
)
2
2"L2+ L12 I.
169
CHAPTER 16: ENERGY AND FORCES IN THE MAGNETIC FIELD
Note that only the inductance L12 depends on x. The force on the inner solenoid is given by Eq.
(16.21),
Fx
= dWm
dx
I
1= constant
= dL12
= J10N1N21rb2 [2.
[2
dx
/1/2
Since Fx > 0, we conclude that the thicker solenoid tends to pull in the thinner one, i.e., the centers
of the two solenoids tend to be at the same point.
Does the result remain the same when we change the direction of current in one of the solenoids?
fi!)!
a
x
F'
F'
2b
y
x
dx
I
i
H
*, Y
Fig. P16.18.
x
A two-conductor line.
Fig. P16.19
A two-conductor line.
P16.17. An electromagnet and the weight it is supposed to lift are shown in Fig. P16.17. The
dimensions are S
100cm2, 11= 50 em, 12= 20em. Find the current through the winding
=
of the electromagnet and the number of turns in the winding so that it can lift a load that
is W
= 300 kiloponds
magnetization
(a kp is 9.81 N) heavy. The electromagnet is made of a material whose
by B(H) = 2H/(400 + H), where B is in T and
Note: the solution is not unique. (a) For N = 200, I = 2.431A. (b) For
curve can be approximated
H is in A/m. -
N = 100, 1= 1.223A. (c) For N = 50, 1= 10.38A.
,,
a
r
y
,
b/2
",,01
// [ "
/
'
2
,
Y
-b/> 1 I .=Pl
,,
,,
a
Fig. P16.20.
dY
A two-conductor line.
al
.L
[0
I
.. J I..
.x
Fig. P16.21.
y
y
dy
Cross section of a stripline.
P16.18. One of the conductors of a two-conductor line is in the form of one half of a thin
circular cylinder. The other conductor is a thin wire running along the axis of the first
(Fig. P16.18). If a current I flows through the two conductors in opposite directions, determine the force per unit length on the conductors. - (a) F~n conductor 2 = J.LoI2/(7r2a).(b)
F~n conductor 2 = J.LoI2/(47r2a). (c) }~n conductor 2 = 2J.LOI2/(7r2a).
.
110
PART 3: SLOWLY TIME-VARYING
ELECTROMAGNETIC
FIELD
P16.19. A thin conductor 2 runs parallel to a thin metal strip 1 (Fig. P16.19). Both a
and b are much larger than the thickness of the strip. Determine the force per unit length
on the two conductors for a current I flowing through them in opposite directions.
- (a)
F' = [(JLoI2)j(7rb)].tan-1(bja).
(b) p' = [(JLoI2)j(27rb)]In(bja). (c) p' = [(JLoI2)j(27rb)]
tan-1 (bja).
P16.20. Determine the force per unit length on the conductors of the line with a cross section
as shown in Fig. P16.20. The current in the conductors is I and the medium is air. - (a)
F{2 = (JLoI2V2)j(47rb). (b) F{2 = (JLoI2V2)j(4b). (c) F{2 = (JLoI27rV2)j(2b).
-
I
I
I
-
I
d
-
I
Fig. P16.22.
J
B
I&i
dF
2a.12b
----Short-circuited two-wire line.
Fig. P16.23.
Short-circuited
coaxial cable.
*P16.21. Determine the force per unit length on the conductors of the stripline with a cross
section as shown in Fig. P16.21. The current in the conductors is I, in opposite directions. The solution is relatively complex, and we give only the correct answer:
F ' --xF ' - JLoI2
2
2b
7r
(2bt
an
-I!!. a
an1
a2 + b2
a2
).
P16.22. A thin two-wire line has conductors of circular cross section of radius a and the
distance between their axes d and is short-circuited by a straight conducting bar, as shown in
Fig. P16.22. If a current I flows through the line, what is the force on the bar? Assume that
the section of the line to the left of the bar is very long, and that the medium is air. - (a)
F ~ [(JLoI2)j(47r)]1n(dja). (b) F ~ [(JLoI2)j(47r)]1n(2dja). (c) F ~ [(JLoI2)j(27r)]ln(dja).
S P16.23. A long air-filled coaxial cable is short-circuited at its end by a thin conducting plate,
as shown in Fig. P16.23. Determine the force on the end plate corresponding to a current
[(2JLoI2)j7r]In(bja). (b)
of intensity I through the cable. - The force is axial. (a) Fx
Fx
= [(JLoI2)j7r] In(bja).
((:) Fx
= [(JLoI2)j(47r)]
=
In(bja).
Solution.
The force on the plate can be obtained by integrating the elemental forces acting on the
plate elements. Here we solve the problem in a different way, starting from energy considerations.
Let us introduce the z axis along the cable axis, directed towards the plate. Imagine that a
magnetic force on the plate, which we need to find, has moved the plate by dz along the z axis, and
that the current I in the cable has been kept constant during such an experiment. It is easy to conclude
that during the experiment the magnetic energy contained in the cable has been increased by
1 I
2
dWm = - L dz I ,
2
CHAPTER
111
16: ENERGY AND FORCES IN THE MAGNETIC FIELD
where L' is the external self inductance per unit length of the coaxial cable,
L'
= JJO In ~
211"
a'
The force on the plate is hence
"
Fx
= dWm
dz
I
1= constant
= "!..L' [2 = JJo[2
2
411"
ln~.
a
P16.24. Determine approximately the force between two parallel coaxial circular loops with
currents It and 12. The radii of the loops are a and b, respectively, with a ~ b, and the
..distance between their centers z. - The force is axial. If the cu.rrents in the loops are in the
same direction, it is attractive.
(a) Fz = 3J..lo7ra2b2Ithz
2(z2 + a2)3/2 '
(b) Fz = J..lo7ra2b21112Z
(z2 + a2)3/2 '
(c) Fz = J..lo7ra2b2Ithz
2(z2 + a2)3/2 .
b
t (vxB)xB
N]
~
2"[8
~
v
!v
Fig. P16.25.
9"
A strip moving in magnetic field.
Fig. P16.26.
A plate falling in magnetic field.
S PI6.25. A metal strip of conductivity 0"and of small thickness b moves with a uniform velocity
v between the round poles of a permanent magnet. The radius of the poles of the magnet is
a and the width of the strip is much larger than a (Fig. P16.25). The flux density B is very
nearly constant over the circle shown hatched and practically zero outside it. Assuming that
o"v X B/2, determine the force on
the induced current density in that circle is given by J
=
the strip. -
(a) F
= 7rO"a2bvB2/4.
(b) F
= 7rO"a2bvB2/2.
(c) F
= 7rO"a2bvB2/3.
= Jdv
Solution.
The magnetic force on the current element Jdv is given by dF
magnetic force per unit volume acting on the currents in the hatched circle is
dF
-dv
=J
(J'
x B
= -2 (v x B)
x B.
x B, so that the
172
PART
3: SLOWLY TIME-VARYING ELECTROMAGNETIC
FIELD
The vector dFjdv is constant in the circle. It is directed in the direction opposite to that of v, that is,
it opposes the motion of the strip. The total force is in the same direction, of intensity
F=
P16.26.
-dF
dv
1m2b
1
= -7rua2
2
bvB2.
A thin metal plate is falling between the poles of a permanent magnet (Fig. P16.26)
=
under the action of the gravitational
field. The pole radius is a
2 em and the flux density
in the gap is B = 1 T. Determine approximately
the velocity of the plate, if its thickness is
b = 0.5mm, its surface area S = 100cm2, its conductivity (7= 36.106 8/m (aluminum), and
its mass density pm = 2.7g/cm3 (aluminum). - (a) v ~ 2.34cm/s. (b) v ~ 3.58cm/s. (c)
v ~ 1.17 cm/s.
",""
'e'//
r-
I
I
I
--
=t
L-
i=Imcos c.Jt
J.Lo
---~
H
=:~~
I
I
I
h
__I
).L
-u
_u
Fig. P16.27.
A ring in a magnetic field.
Fig. P16.28.
u
A V-shaped tube in a magnetic field.
P16.27. A metal ring K of negligible resistance is placed above a short cylindrical electromagnet, as shown in Fig. P16.27. Determine qualitatively the time dependence of the total
force on the ring if the current in the electromagnet coil is of the form i( t) = 1m cos 1.J.)t. With the reference direction of the force upward, (a) Pr = Frm COSl.J.)t.(b) Fr = Prm COS3I.J.)t.
(c) Fr = Frm cos2l.J.)t.
P16.28. A U-shaped glass tube is filled with a paramagnetic liquid of unknown magnetic
susceptibility Xm. A part of the tube inside the dashed square in Fig. P16.28 is exposed to a
uniform magnetic field of intensity H. Under the influence of magnetic forces, a difference h of
the levels of the liquid in the two sections of the tube is observed. Given that the mass density
of the liquid is pm and that the medium above the liquid is air, determine Xm. - Hint: use Eq.
2pmgh/(J1.oH2). (c) Xm
2pmgh/(7rJ1.oH2).
(16.17). (a) Xm = Pmgh/(2J1.oH2). (b) Xm
=
=
(g ~ 9.81 m/s2 ).
P16.29. Plot the "scale calibration curve" Ftot(l) for the ammeter
Ftod 1) is the total force acting on the iron nail for a given current
a = 1mm, I = 5 em, N' = 10 turns/em (you need to look up the
iron and its mass density). - Hint: the downward magnetic force,
sketched in Fig. P16.29.
1 in the coil. Given are
relative permeability for
Ftot(I),
can be evaluated
CHAPTER
16: ENERGY AND FORCES IN THE MAGNETIC FIELD
173
by means of Eq. (16.16) once the inductance of the coil with the nail a distance x inside it is
known. (a) F:: pSN'2]2. (b) F:: !pSN'2]2.
(c) F :: 2pSN'2]2.
S P16.80. Derive the general expression for pressure of magnetic forces, Eq. (16.17). - Hint:
use boundary conditions combined with Eqs. (16.15) and (16.16) for the force on a patch ~S of
the boundary, assuming it ,to move by ~x into medium 1 under the pressure at the interface.
~
~
Scale
I [A]
.
Solution.
The normal component of force (per unit area) acting on the interface between two media
J-Lland J-L2depends on the direction of Band H in both media. All other cases can
be deduced if we analyze just the following two: (1) when Band H are normal, and (2) when they
are parallel to the interface.
of permeabilities
In case (1), during a virtual displacement dx of the interface, B, and hence also the magnetic
flux <1>,remain constant, so that Eq. (16.20) applies. Assume that the reference direction of the
displacement is into the medium of permeability J-Ll' According to the boundary condition in Eq.
(13.18), the change in energy located in the magnetic field is
1
1
dWm
1
2
= -2 (-J-L2 - -J-Ll) Bnormal~Sdx,
where AS is the area of a patch at the interface.
into medium of permeability
J-Llis
So the pressure
p(l) - - -2- dWm - ! 2. - 2.
-
AS
dx
-
2
(
J-Ll
In case (2), H (i.e. the currents) remains constant,
condition in Eq. (13.17) tells us that
1
dWm
on the sur.face for a reference direction
)
B2
J-L2
normal'
and Eq. (16.21) applies.
2
= 2' (J-L2- J-Ll)HtangASdx,
from which
p (2)
= -AS1 -dWm
dx
--
1
H2
2' (J-L2 - J-Ll) tang'
The boundary
174
PART 3: SLOWLY TIME-VARYING
ELECTROMAGNETIC
FIELD
Noting that
J1.H2
=,..IlH2tang
B2
+
normal
J1.
,
from the expressions for p(1) and p(2) above it may be deduced that
surface in the general case is given by the expression in Eq. (16.17).
the pressure
on the boundary