Download 17. Some Examples and Applications of Time

17. Some Examples and Applications
of Time-Invariant
and Slowly Time-Varying
Magnetic Field
.. The largest permanent magnet around us is the earth itself. Its magnetic field is similar
to the field of a giant current loop with an axis declined 110 with respect to the earth's axis
of rotation. The planet's North Pole is approximately the south magnetic pole. The average
"magnitude of the earth's magnetic flux density is about 50 j.lT (at the poles it is about 60 j.lT).
The region in which the earth's magnetic field can be detected is called the magnetosphere.
. Charged particles always move when found in both electric and magnetic fields. In many
instances one of the fields is of much less influence than the other, but in some cases the effects
of both fields on a moving particle a.re of the same order of magnitude. The motion of charged
particles in electric and magnetic fields may be in a vacuum (or very rarefied gas), in gases,
and in solid or liquid conductors.
. The force on a charge Q moving in an electric and a magnetic field with a velocity v is the
Lorentz force,
F= QE+Qv x B.
(17.1)
If a charge of mass m moves in a vacuum, then this force at any instant must be equal in
magnitude and opposite in direction to the force of inertia,
dv
mill
=QE+Qv
X
(17.2)
B.
E and B in general are functions of space coordinates and of time. If a charge moves in a
material (a gas, a liquid or a solid), it collides with the particles in the material, and Eq. (17.2)
is not valid for average (drift) velocity.
.
If a charge Q (Q
arbitrary
initial
>
velocity
0) moves in a uniform, x-directed electric field of intensity E, with
Vo
= vox + 'VOy, after
QE
x(t)
= 2mt
integration
Eq.
2
+ voxt + xo,
y(t) = VOyt+ Yo,
(17.2)
yields
(17.3)
(17.4)
CHAPTER
17: SOME EXAMPLES AND APPLICATIONS OF TIME-INVARIANT ...
175
where Xo and Yo are the initial x- and y-coordinates of the charge. Consequently, the charge
will move along a parabola in the x-y plane. This is used, for example, for deflecting an
electron beam between the plates of a charged parallel-plate capacitor.
.
= Qv X B,
Since the magnetic force on the charge is Fm
it is always perpendicular to the
direction of motion. Therefore, a magnetic field cannot change the magnitude of the charge
velocity (i.e., the charge kinetic energy), but only the direction of its motion. For example, in
a uniform magnetic field of flux density B normal to the direction of motion of a particle of
charge Q and mass m, the particle describes a circle of radius
R-
mv
- QB'
(17.5)
= 21T"R = 21T"m.
(17.6)
It makes a full circle during a time
T
v
QB
The frequency of rotation of the particle,
1
QB
(17.7)
f = T = 21T"m'
is known as the cyclotron frequency, because this effect is used in cyclotrons, devices for accelerating charged particles.
.
If a strip of width d with current of density J is situated in a uniform magnetic field of flux
density B normal to the strip, there is a voltage between the strip edges. The effect is known
as the Hall effect, and the voltage as the Hall voltage, which is given by
Jd
IYl2 -- -B,
NQ
TT
(17.8 )
1
where N is the number of free charge carriers, of charge Q, per unit volume. So, if we know
the coefficient Jd/ NQ, by measuring VI2 we can measure B. Such a measuring strip is called
a Hall element.
.
The hard (and floppy) disk in every computer
is a magnetic
memory.
We write to the
disk by magnetizing a small piece of the disk surface, and we read from the disk by inducing a
voltage in a small loop ("magnetic head") which is moving in close proximity to the magnetized
disk surface element. The magnetic head is a magnetic circuit.
.
A transformer is a magnetic circuit with (usually) two windings, "primary" (e.g., with NI
turns), and "secondary" (with N2 turns), on a common ferromagnetic
core. If VI and V2 are
the primary and secondary voltages, and il and i2 the respective currents, the basic equations
of an ideal transformer (a transformer with the core of infinitely large permeability)
are
VI
v2
- N1
- N2'
(17.9)
176
PART 3: SLOWLY TIME-VARYING
i1
i2
ELECTROMAGNETIC
N2
FIELD
(17.10)
= N1'
If the secondary winding of an ideal transformer is connected to a resistor of resistance R2,
the resistance seen from the primary terminals is
Nl
Rl
= R2 (N2
2
)
.
(17.11)
. Electric motors are devices which are used to transform continuously the electric energy
into mechanical energy. Of several types of electric motors, two types, the synchronous and
asynchronous motors, use rotating magnetic field. This is a magnetic field that keeps its
-magnitude constant in time, but the vector B rotates about an axis normal to itself with a
constant angular velocity CJ.).It can be obtained by two perpendicular stationary coils with
-equal currents shifted 900 in phase, or by three stationary coils at 1200 angle with symmetrical
three-phase currents. In synchronous motors, the rotating part (the rotor) is a permanent
magnet, which rotates in synchronism with the field. In asynchronous motors, the rotating
part consists of short-circuited wire loops. The rotating field induces currents in the loops, and
there is a moment on these currents, but the rotor in this case need not rotate in synchronism
with the field.
. Voltages induced in our body when we are close to power lines are much smaller than the
normal electric impulses flowing through our nerve cells, and are on the order of pV.
QUESTIONS
Q17.1. Where is the earth's south magnetic pole? - (a) Close to the earth South Pole. (b)
Close to the earth North Pole. (c) Exactly at the earth North Pole.
Q17.2. What is the order of magnitude of the earth's magnetic flux density? (b) 0.O5jlT. (c) 0.05 T.
(a) 0.05 mT.
S Q 17.3. Approximately how fast would you need to spin around your axis in the magnetic field
of the earth to induce 1 mV around the contour of your body? - (a) About three revolutions
in a second. (b) About one third of a revolution in a second. (c) About 11 revolutions in a
second.
Answer. From the answer to the preceding question, and assuming the area of the equivalent loop
to be S
1 m2, we have the equation e(t)
-d4>(t)jdt = -I.I.iBS COSl.I.it.So the amplitude of this
induced electromotive force, or voltage, would be 1 mV for an angular velocity of the body of I.I.i=
0.001/(5.10-5)
= 2 rad/s. This corresponds to about three revolutions in a second.
=
=
QI7.4. Turn your computer monitor sideways or upside down while it is on (preferably with
some brightly colored pattern on it). Do you notice changes on the screen? If yes, what and
why? - (a) Nothing happens. (b) Since the electrons are heavy, the gravitational field will
change the pattern somewhat. (c) Since the electrons are moving in the magnetic field of the
earth, the pattern will be somewhat changed.
CHAPTER
17:
SOME
EXAMPLES
AND APPLICATIONS
OF TIME-INVARIANT
...
177
Q17.5. What do you expect to happen if a magnet is placed close to a monitor? If you have
a small magnet, perform the experiment (note that the effect might remain after you remove
the magnet, but it is not permanent). Explain. - (a) The pattern on the screen is shifted and
distorted, due to the action of the magnetic field on the electron beam. (b) Nothing happens.
(c) The pattern on the screen is shifted and distorted because the magnet is heavy, and acts by
the gravitational force on the electron beam.
S Q17.6. Explain how the Hall effect can be used to measure the magnetic flux density.'- Hint: note that the Hall voltage will exist for any angle between the strip and the B-lines. So
the experiment needs to be repeated to obtain the maximal Hall voltage (why'?).
Answer. We put the Hall element into the magnetic field, and find its orientation corresponding to
the maximal Hall voltage. From this voltage we obtain the magnitude of B, and the vector B is normal
to the Hall element.
Ql 7.7. Explain how the Hall effect can be used to determine whether a semicond uctor is p- or
n- doped. - Hint: note that the Hall voltage sign depends on the sign of free charge carriers.
.
Q17 .8. What magnetic material properties are chosen for the tracks and heads in a hard disk?
- (a) Tracks have low Bremanent, head has high Bremanent. (b) Tracks have high Bremanent,
head has low Bremanent. (c) Both tracks and head have high Bremanent.
Q17.9. Sketch and explain the time-domain waveform of the induced emf (or current) in
the magnetic head coil in "read" mode as it passes over a piece of information recorded on a
computer disk as "110". (Assume that a "I" is a small magnet along the track with a S-to-N
orientation inside the magnet in the direction of motion of the head, and a "0" is a small
magnet magnetized in the opposite direction.) - Hint: check if the sketch of the time-domain
waveform of the induced emf as given in Fig. Q17.9 is what you expect.
v
Fig. Ql'i'.9.
Sketch of induced emf.
S Q17.10. Write Ampere's law for an ideal transformer, and derive the voltage, current and impedance (resistance) transformation ratio. The number of turns in the primary and secondary
are N1 and Nz. - Correct results are given in Eqs. (17.9)-(17.11).
Answer.
N1 i1
= Nziz, so that
iIliz
= Nz/N1'
For primary
and secondary
voltages,
The impedance is defined only in sinusoidal steady state and complex notation.
closed by an impedance Zz, the input impedance of the transformer is Zl
VIIh
=
(NlINz)z Zz.
V1/VZ = NiiNz.
If the secondary is
= (NIINz)zVz/
lz
=
178
PART 3: SLOWLY TIME-VARYING
ELECTROMAGNETIC
FIELD
QI7.I1.
What are the loss mechanisms in a real transformer, and how does each of the
contributors to loss depend on frequency? - (a) There are no losses in a real transformer.
(b) The losses exist only in the transformer windings. (c) The losses exist in the transformer
windings, and in the transformer core (hysteresis plus eddy-current losses).
S
QI7.I2.
Explain
how a synchronous
motor works.
-
(a) The rotor is a short-circuited
wire
loop, in which current is induced, and the moment of magnetic forces acts on these currents.
(b) The rotor is a permanent magnet, and the rotating magnetic field makes it rotate. (c) The
current in the stator and rotor windings are in synchronism.
Answer. It uses a rotating magnetic field to rotate a rotor in the form of a permanent magnet, with
the same angular velocity as the magnetic field (i.e., in synchronism with it).
QI7.I3. How is an asynchronous motor different from the synchronous type? the answers to the preceding question.
Hint: use
QI7.I4. Describe the two mechanisms by which ac currents can affect our body. Use formulas
~n your description. - Hint: recall that our body is conducting, and consider the influence of
.
the quasistatic and induced components of the ac electricfield produced by ac currents.
PROBLEMS
PI7.1. What is the minimum magnitude of a magnetic flux density vector that will produce
the same magnetic force on an electron moving at 100 mls as that a 10 kVIcm electric field
1 T. (b) B = 10 T. (c) B
10 kT.
produces? - (a) B
=
=
S PI7.2.
Calculate the velocity of an electron in a 10 kV CRT. The electric field is used to
accelerate the electrons, and the magnetic field to deflect them. - (a) v 3.93 .106 mls. (b)
v = 5.93 .107 mls. (c) v
1.93.107 mls.
=
=
Solution.
The velocity of an electron is given by Eq. (11.4),
me = 5.93.107mis,
v = J2ev
where e is the absolute value of electron
(me
9.108 . 10-31 kg).
=
charge
(e
= 1.602.10-19
C), and me is the electron
mass
S PI7.3. How large is the magnetic flux density vector needed for a 20 cm deflection in the CRT
1.32 mT.
in problem P17.2, ifthe length ofthe tube is 25 cm? - Refer to Fig. P17.3. (a) B
0.18T.
(b) B 0.32 mT. (c) B
=
Solution.
=
=
According to Fig. P17.3,
(R-X)2+d2
from which the radius of curvature is R
=R2,
= (x2 + d2)/(2x) = 25.62 em.
R = meV
eB
Since, at the same time,
17: SOME EXAMPLES AND APPLICATIONS OF TIME-INVARIANT
CHAPTER
(see Example 17.3), the required magnetic flux density is B
...
179
= mev/(eR)= 1.32mT.
PI7.4. A thin conductive ribbon is placed perpendicularly to the field lines of a uniform
B-field. When the current is flowing in the direction shown in Fig. P17.4, there is a measured
negative voltage V12between the two edges of the ribbon. Are the free cha.rges in the conductive
ribbon positive or negative? - (a) If V12 < 0 and B > 0, Q < O. (b) If V12 < 0 and B > 0,
Q > o.
(c) If V12
>
0 and B
> 0, Q > o.
d
x
R-x
d
Fig. P17.3.
PI7.5.
A stream of electrons in a CRT.
A current strip in a magnetic field.
What is the voltage V12equal to in problem P17.4 if B
and the concentration
-1.10-7
Fig. P17.4.
of free carriers
V. (b) V12~ -5.10-7
in the ribbon
is N
V. (c) V12~ -8.10-7
= 0.8T,t = 0.5mm,1= 0.8A
= 8 . 1028 m -3
? -
(a) Vi2
'"
V.
PI7.6.
The magnetic head in the form of a toroidal coil wound on a thin ferromagnetic
toroidal core with a narrow air gap is in write mode. Calculate the magnitude of the current I
in the winding that would be needed to produce a Bo
1pT field in the gap. There are N = 5
turns on the core, the core can approximately be considered as linear, of relative permeability
of pr = 1000, the gap is Lo =20 pm wide, the cross-sectional area of the core is S
10-9 m2,
and the mean radius of the core is r =0.1 mm. Assume that the fringing field in the gap makes
3.76pA.
the gap cross-sectional area effectively 10% larger than that of the core. - (a) I
=
(b) I
= 5.08pA. (c) 1=
6.34pA.
=
=
PI7. 7. The head and the tracks in magnetic hard disks are made of different magnetic
materials. Sketch and explain the preferred hysteresis curves for the two materials, indicating
the differences. Which has higher loss in an ac regine? - (a) The hysteresis loopfor the head
material should be wide, and that for the disk material should be narrow. (b) Opposite to the
preceding answer. (c) Both materials should have narrow hysteresis loops.
PI7.S. A CRT needs 10 kV to produce an electric field for electron acceleration. Design a
wall-plug transformer to convert from llOV in the U.S. and Canada or 220V in Europe and
Asia. Assume you have a core made of a magnetic material that has a very high permeability.
- Assume the number of turns in the primary winding to be ten. The number of the turns in
the secondary winding in the two cases is then: (a) 110 and 220. (b) 400 and 200. (c) 450
and 900.
S PI7.9. Assume that for obtaining a rotating magnetic field you use three coils. The axes of
the coils are at 60 degrees, not 90 degrees with respect to each other. What is the relative
180
PART
3: SLOWLY TIME-VARYING ELECTROMAGNETIC
phasing of three sinusoidal currents (of equal amplitude)
magnetic field? Plot the current waveforms as a function
FIELD
in the coils that will give a rotating
of time.
Solution.
Assume that the three coils are in the zy plane. Let the first coil be along the z axis,
producing a magnetic flux density vector Bl (t) in the z direction. Let the second coil produce a
magnetic flux density vector B2(t) at an angle of 120 degrees with the z axis, and the third coil
produce a vector B3 (t) at an angle of 240 degrees (or -120 degrees) with the z axis. Assume that
the three vectors are of the same amplitude and have initial phases 0 degrees, -120 degrees and -240
degrees, respectively, i.e., that
Bl (t) = Bm cos,,-,t,
B2(t) = Bm cos(,,-,t- 21r/3),
It is necessary to consider the instantaneous
flux density separately:
Bx(t)
= Bl (t) - B2(t)
cos1r/3
By(t)
- 41r/3).
values of the z- and y-components of the magnetic
- B3(t)
= [B2(t)
Bl (t) = Bm cos(,,-,t
cos 1r/3
= Bl (t) -
1
-[B2
(t) + B3 (t)],
2
- B3(t)]cos1r/B.
- B3(t) are obtained by representing the cosines
- = cos a cos b + sin a sin b. The final result is
The sum B2(t) + B3(t) and the difference B2(t)
the expessions for B2 (t) and B3 (t) as cos( a b)
in
3
Bx(t)
= -Bm
2
cos,,-,t.
By(t) = ~Bm
sin,,-,t.
2
These are the same expressions as for the two B-field components in a two-phase system, and the
B-vector rotates with an angular velocity "-', as in the two-phase case.