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Electronic Devices and Circuits
According to JNTUK syllabus
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Index of first unit
Fundamentals of atomic theory
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Introduction and EssentialFundamentals of
atomic theory
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HISTORY OF THE ATOM
460 BC Democritus develops the idea of atoms
he pounded up materials in his pestle and
mortar until he had reduced them to smaller
and smaller particles which he called
ATOMA
(greek for indivisible)
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HISTORY OF THE ATOM
1808 John Dalton
suggested that all matter was made up of
tiny spheres that were able to bounce around
with perfect elasticity and called them
ATOMS
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HISTORY OF THE ATOM
1898 Joseph John Thompson
found that atoms could sometimes eject a far
smaller negative particle which he called an
ELECTRON
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HISTORY OF THE ATOM
Thompson develops the idea that an atom was made up of
electrons scattered unevenly within an elastic sphere surrounded
by a soup of positive charge to balance the electron's charge
1904
like plums surrounded by pudding.
PLUM PUDDINGMODEL
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HISTORY OF THE ATOM
In 1910 Ernest Rutherford and his team
Geiger and Marsden conducted a famousexperiment .
they fired Helium nuclei at a piece of gold foil
which was only a few atoms thick.
they found that although most of them
passed through. About 1 in 10,000 hit
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HISTORY OF THE ATOM
gold foil
helium nuclei
They found that while most of the helium nuclei passed
through the foil, a small number were deflected and, to their
surprise, some helium nuclei bounced straight back.
helium nuclei
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HISTORY OF THE ATOM
Rutherfords new evidence allowed him to propose a more
detailed model with a central nucleus.
He suggested that the positive chargewas all in a central
nucleus. With this holding the electrons in place by electrical
attraction
However, this was not the end of the story.
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HISTORY OF THE ATOM
1913 Niels Bohr
studied under Rutherford at the Victoria
University in Manchester.
Bohr refined Rutherford's idea by adding
that the electrons were in orbits. Rather
like planets orbiting the sun. With each
orbit only able to contain a set number of
electrons.
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HELIUM ATOM
+N
N+
--
proton
electron neutron
Shell
What do these particles consist of?
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ATOMIC STRUCTURE
Particle Charge Mass
Proton +ve Charge1.60210-19 C
1.67262158 10-27kg
Or
1.0086649156 amu
Neutron No charge 1,6749 x 10-27kgOr
1.00727638 amu
Electron -ve charge-1.60210-19C
9.10938188 10-31
kgOr0.0005446623 amu
An atomic mass unit (symbolized AMU or amu) is defined as precisely 1/12 the mass of an
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2. In a transition from one stationary state correspondingto a definite energy W2to another stationary state, with
an associated energy W1, radiation will be emitted. The
frequency of this radiant energy is given by
f= (W2-W1)/hwhere h is Plancks constant in joule-seconds, the Ws are
expressed in joules, andf is in cycles per second, or hertz.
3. A stationary state is determined by the conditionthat the angular momentum of the electron in this
state is quantized and must be an integral multiple
of h/2Thus
mvr = n h/2where n is an integer.
1. Not all energies as given by classical mechanics arepossible, but the atom can possess only certain discrete
energies. While in states corresponding to these discrete
energies, the electron does not emit radiation, and the
electron is said to be in a stationary, or non radiating,state.
the energy level in joules of each
state is found to be
The Bohr Atom : Bohr in 1913 postulated the following
three fundamental laws:
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wavelength in angstroms
E energy value of the
stationary states in electron volts
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According to Schrodinger four quantum numbers are
required to define the wave function.
1. Theprincipal quantum number n is an integer 1, 2, 3, . . .and determines the total energy associated with a particular
state. This number may be considered to define the size of
the classical elliptical orbit, and it corresponds to the
quantum number n of the Bohr atom.
2. The orbital angular momentum quantum number ltakeson the values 0, 1, 2, . . . , (n 1). This number indicates
the shape of the classical orbit. The magnitude of this
angular momentum is
3. The orbital magnetic number mlmay have the values0, 1, 2, ... l. This number gives the orientation of the
classical orbit with respect to an applied magnetic field. The
magnitude of the component of angular momentum along
the direction of the magnetic field is ml(h/2)
4. Electron spin. In order to explain certain spectroscopic andmagnetic phenomena, Uhlenbeck and Goudsmit, in 1925,
found it necessary to assume that, in addition to traversing
its orbit around the nucleus, the electron must also rotate
about its own axis. This intrinsic electronic angularmomentum Is called electron spin. When an electron system
is subjected to a magnetic field, the spin axis will orient itself
either parallel or anti parallel to the direction of the field.
The spin is thus quantized to one of two possible values. The
electronic angular momentum is given by ms(h/2), where
the spin quantum number msmay assume only two values,
+1/2 or 1/2 .
1. Theprincipal quantum number n2. The orbital angular momentum quantum number l
3. The orbital magnetic number ml
4. Electron spin ms
A l bi l h
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1S
2S 2P
3S 3P
Represents Requirement of
electrons to get argon
configuration
Actual orbital shape
1S2S3S
P(14)+N(14)
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Electron atomic and molecular orbitals. The chart of orbitals (left) is arranged by
increasing energy Note that atomic orbits are functions of three variables (two
angles, and the distance from the nucleus, r). These images are faithful to the
angular component of the orbital, but not entirely representative of the orbital as a
whole.
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Basic Information
Name:Silicon
Symbol:Si
Atomic Number:14Atomic Mass:28.0855 amu
Melting Point:1410.0 C (1683.15 K, 2570.0 F)
Boiling Point:2355.0 C (2628.15 K, 4271.0 F)
Number of Protons/Electrons:14
Number of Neutrons:14
Classification:MetalloidCrystal Structure:Cubic
Density @ 293 K:2.329 g/cm3
Color:grey
Date of Discovery:1823
Discoverer:Jons Berzelius
Name Origin:From the Latin word silex(flint)
Uses:glass, semiconductors
Obtained From:Second most abundant element. Found in clay,
granite, quartz, sand
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Basic Information
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Basic Information
Name:Germanium
Symbol:Ge
Atomic Number:32
Atomic Mass:72.61 amuMelting Point:937.4 C (1210.55 K, 1719.3201 F)
Boiling Point:2830.0 C (3103.15 K, 5126.0 F)
Number of Protons/Electrons:32
Number of Neutrons:41
Classification:Metalloid
Crystal Structure:CubicDensity @ 293 K:5.323 g/cm3
Color:grayish
Date of Discovery:1886
Discoverer:Clemens Winkler
Name Origin:From the Latin word Germania, meaning Germany
Uses:semiconductorsObtained From:refining of copper, zinc, lead
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I UNITSyllabusElectron Ballistics and Applications:1.1Force on Charged Particles in Electric field,
1.2 Constant Electric Field,1.3 Potential, Relationship between Field Intensity and Potential,
1.4Two Dimensional Motion,
1.5 Electrostatic Deflection in Cathode ray Tube, CRO,
1.6 Force in Magnetic Field,
1.7 Motion in Magnetic Field,
1.8 Magnetic Deflection in CRT,
1.9 Magnetic Focusing,
1.10 Parallel Electric and Magnetic fields
1.11 Perpendicular Electric and Magnetic Fields.
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Force on a charged particle in an Electric Field
The Electric field intensity at any point is defined as The Force on unit Positive charge at that point.
There fore the force (f) on unit positive charge q in an electric field
is f = qThe resulting force fis in Newton and is in the direction
of electric field , q is in coulombs and
is in Volts/meters.
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The Newton second law of motion says force (f)= m a
where mis mass of electron and ais acceleration of electron.
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Index
By relating the force of electric field on the electron
with Newtons second law of motion we can write that
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V
+
++
+
+
+
+
+
-
-
-
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-
-
-
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constant electrified field
Suppose that an electron is
situated between the two
plates of parallel plate
capacitor which are
contained in an evacuated
envelope as shown in fig.
A difference of potential
is applied between the
two plates .
The direction of electric
field is from positive tonegative plate.
If the distance between The
plates is very Smallcompared with dimension of
the plates .
the electric field maybe
considered to be uniform .
The direction of electric
field is along the ve x
direction. That is the onlyforce on the electron.
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V
+
++
+
+
+
+
+
-
-
-
-
-
-
-
-
constant electrified field
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V
+
++
+
+
+
+
+
-
-
-
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-
-
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constant electrified field
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t t l t ifi d fi ld
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V
+
++
+
+
+
+
+
-
-
-
-
-
-
-
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constant electrified field
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Two Dimensional electronic
motion in a uniform electric field
When
Vd=0 Volts
When
Vd is +Ve
Electric Field
When an electron moving in a straight line with a uniform velocity (Vox) it willcontinue its motion in a straight line path .
If an uniform Electric field ()is introduced in its straight line path in a
perpendicular direction . It has Two forces now and the resultant Is path of
parabola.
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ELECTROSTATIC DEFLECTION IN A
CATHODE-RAY TUBE
The essentials of a
cathode-ray tube for electrostatic
deflection are illustrated in Fig. The hot
cathode K emits electrons which are
accelerated towards the anode by the
potential Va.
Those electrons which are not collected by
the anode pass through the tiny anode bole
and strike the end of the glass envelope.
This has been coated with a material that
fluoresces when bomb boarded by
electrons.
Thus the positions where the electrons strike
the screen are made visible to the eye. The
displacement D of the electrons is determinedby the potential Vd(assumed constant) applied
between the deflecting plates, as shown. The
velocity voxwith which the electrons emerge
from the anode hole is given by
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on the assumption that the initial velocities of
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on the assumption that the initial velocities of
emission of the electrons from the cathode are
negligible.
Since no field is supposed to exist in the region
from the anode to the point 0, the electronswill move with a constant velocity in a straight-
line path. In the region between the plates the
electrons will move in the parabolic path given
by
The path is a straight line from the point of emergence M at the edge of the
plates to the point P on the screen, since this region is field-free.
The straight-line path in the region from the deflecting plates to the screen is, of
course, tangent to the parabola at the point. M. The slope of the line at this
point, and so at every point between M and P, is
From the geometry of the figure,
the equation of the straight line MP is
found to be
When y = 0, x = 1/2, which indicates thatwhen the straight line MP is extended
backward, it will intersect the tube axis
at the point 0, the center point of the
plates. This result means that 0 is, in
effect, a virtual cathode, and regardless
of the applied potentials Vaand Vd, theelectrons appear to emerge from this
cathode and move in a straight line to
the point P.
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At th i t P D d L l/2
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By inserting the known values of
ay(= eVd/dm) and voxthis becomes
The idealization made in connection with the foregoingdevelopment, viz, that the electric field between the deflecting
plates is uniform and does not extend beyond the edges of the
plates, is never met in practice.
Consequently, the effect of fringing of the electric field may be
enough to necessitate corrections amounting to as much as 40
percent in the results obtained from an application of Eq
Typical measured values of sensitivity are
1.0 to 0.1 mm/V, corresponding to a voltage requirement of 10
to 100 V to give a deflection of 1 cm.
At the point P, y = D, and x = L + l/2.
This result shows that the deflection on the screen of a cathode-ray tube
is directly proportional to the deflecting voltage Vd
applied between the plates.
Consequently, a cathode-ray tube may be used as a
linear-voltage indicating device.
the deflection (in meters) on the
screen per volt of deflecting voltage.
Is defined as deflection sensitivity
Thus
An inspection of above Eq. shows that the
sensitivity is independent of both the deflecting
voltage Vdand
the ratio e/m.
Furthermore, the sensitivity varies inversely
with the accelerating potential V.
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Vd=0
Ha
t
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- -
+ Vd +
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+ +
-Vd
-
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Horizontal sweep voltage applied on Horizontal Deflection plates
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Horizontal sweep voltage applied on Horizontal Deflection plates.
And a sinusoidal voltage applied for vertical deflection plates.
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Index
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FORCE IN A MAGNETIC FIELD
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To investigate the force on a moving charge in a magnetic
field, the well- known motor law is recalled.
It has been verified by experimentally that, if a conductor of
length L, carrying a current of I, is situated in a magnetic field
of intensity B,
the forcef, acting on this conductor is f = BIL sinwherefmis in newtons, B is in webers per square meter
(Wb/m),Iis in amperes, L is in meters and is the angle
between I and B.
If is 900 then f= BIL
FORCE IN A MAGNETIC FIELD
One weber per square meter is called tesla and equals to 104G
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Force on an electron
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If N electrons are contained in a
length L of conductor and if it takes an electron
a time T sec to travel a distance of L meters in
the conductor, the total number of electronspassing through any cross section of wire in unit
time is N/T.
Thus the total charge per second passing any point, which, by definition, is
the current in amperes, isI = Ne/T
The force in newtons on a lengthL m (or the force on theN conduction
charges contained therein) is
f = BIL = BL(Ne/T)
sinceL/T is the average, or drift, speed rn/sec of the electrons, the force per
electron is
fm= eBv
The subscript m indicates that the force is of magnetic origin.
Force on an electron
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Current density
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Current densitythe current density, denoted by the symbolJ,
is the current per unit area of the conducting medium.
That is, assuming a uniform current distribution,
J =I/Awhere J is in amperes per square meter, and
A is the cross-sectional area (in meters) of the conductor.
This becomes,
J=Ne/ TA
But T =L/v. ThenTherefore J= Nev/LA
it is evident thatLA is simply the volume containing theN electrons,
and soN/LA is the electron concentration n (in electrons per cubic meter).
Thus n = N/LA
and J=nev=v
where =ne is the charge density, in coulombs per cubic meter,
and is vin meters per second.
This derivation is independent of the form of the conductingmedium. Consequently, it does not necessarily represent a wire
conductor. It may represent equally well a portion of a gaseous-
discharge tube or a volume element in the space-charge cloud of a
vacuum tube or a semiconductor. Furthermore, neither nor v
need be constant, but may vary from point to point in space or
may vary with time.
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MOTION IN A MAGNETIC FIELD
Consider an electron to be placed in the region of
the magnetic field. If the particle is at rest, fm= 0and the particle remains at rest. If the initial velocity
of the particle is along the lines of the magnetic flux,
there is no force acting on the particle, particle
whose initial velocity has no component normal to
a uni form magnetic field will continue to move with
constant speed along the lines of flux.
Now consider an electron moving with a speed v. to enter a
constant uniform magnetic field normally, as shown in Fig
Since the forcefmis perpendicular to vand so to the motion at
every instant, no work is done on the electron.
This means that its kinetic energy is not increased, and so its speed
remains unchanged. Further, since vand B are each constant in
magnitude, thenfmis constant in magnitude and perpendicular to
the direction of motion of the particle. This type of force results in
motion in a circular path with constant speed.
It is analogous to the problem of a mass tied to a rope and twirled
around with constant speed. The force (which is the tension in the
rope) remains constant in magnitude and is always directed toward
the center of the circle, and so is normal to the motion.
To find the radius of the circle, it is recalled that a particle moving in
a circular path with a constant speed vhas an acceleration toward
the center of the circle of magnitude v /R where R is the radius of
the path in meters.
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DEFLECTION IN A CATHODE-RAY TUBE MAGNETIC
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Hence the electron moves in a straight line
from the cathode to the boundary 0 of the
magnetic field. In the region of the uniformmagnetic field the electron experiences a force
of magnitude eBv, where v is the speed.
The path OM will be the arc of a circle whose
center is at Q. The speed of the particles will
remain constant and equal to
DEFLECTION IN A CATHODE-RAY TUBE MAGNETIC
the a cathode- ray tube may employ a
magnetic as well as an electric field in order to
accomplish the deflection of the electronbeam. However, since it is not feasible to use a
field extending over the entire length of the
tube, a short coil furnishing a transverse field
in a limited region is employed, as shown in
Fig.
The magnetic field is taken as pointing out of
the paper, and the beam is deflected upward.
It is assumed that the magnetic field intensity
B is uniform in the restricted region shown and
is zero outside of this area
The angle is, by definition of radian measure, equal to the length of the arc
OM divided by R, the radius of the circle, if we assume a small angle of
deflection, thenWhere
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S=D/B is called the magnetic-deflection
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In most practical cases, L is very much larger than
l, so that little error will be made in assuming
that the straight line MP, if projected backward,
will pass through the center 0 of the region of
the magnetic field. ThenSo, we can write
The deflection per unit magnetic fieldintensity,D/B, given by
S=D/B is called the magnetic deflection
sensitivity of the tube. It is observed that this
quantity is independent of B. This condition is
analogous to the electric case for which the
electrostatic sensitivity is independent of the
deflecting potential.
However, in the electric case, the sensitivity
varies inversely with the anode voltage,
whereas it here varies inversely with the
square root of the anode voltage.
Another important difference is in the
appearance of elm in the expression for the
magnetic sensitivity, whereas this ratio did not
enter into the final expression for the electric
case.
Because the sensitivity increases with L, thedeflecting coils are placed as far down the
neck of the tube as possible, usually directly
after the accelerating anode.
A modern CRT TV tube has a screen diameter comparable with the
length of the tube neck. Hence the angle, is too large for the
approximation tan =to be valid. Under these circumstances it is
found that the deflection is no longer proportional to B . If the magnetic-deflection coil is driven by a saw-tooth current waveform, the deflection
of the beam on the face of the tube will notbe linear with time. For
such wide-angle deflection tubes, special linearity Correcting networks
must be added.
A TV tube has two sets of magnetic-.deflection coils mounted around
the neck at right angles to each other, corresponding to the two sets of
plates in the oscilloscope tube Sweep currents are applied to both coils,
with the horizontal signal much higher in frequency than that of the
vertical Sweep. The result is a rectangular raster of closely spaced lines
which cover the entire face of the tube and impart a uniform intensity to
the screen. When the video signal is applied to the electron gun, it
modulates the intensity of the beam and thus forms the TV picture.
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linearly increasing Magnetic field
Direction into the Paper
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MAGNETIC FOCUSING
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MAGNETIC FOCUSING
A magnetic field of the type here considered is
obtained through the use of a long solenoid,
the tube being placed within the coil. revealsthe motion. The Y axis represents the axis of
the cathode-ray tube.
Imagine that a cathode-ray tube is placed in a
constant longitudinal magnetic field, the axis
of the tube coinciding with the direction of the
magnetic field.
The origin 0 is the point at which the electrons
emerge from the anode. The velocity of the
origin is v0, the initial transverse velocity due tothe mutual repulsion of the electrons being
voxIt is now shown that the resulting motion is
a helix, as illustrated.
A force eBv normal to the path will exist, resulting from
the transverse velocity. This force gives rise to circularmotion, the radius of the circle being mv/eB,
with v, a constant, and equal to v0x. The resultant
path is a helix whose axis is parallel to the Y axis and
displaced from it by a distance R along theZ axis, as
illustrated.
The electronic motion can most easily be
analyzed by resolving the velocity into two
components, vyand v, along and transverse to
the magnetic field, respectively.
Since the force is perpendicular to B, there is
no acceleration in the Y direction. Hence vy
constant and equal to voy,.The pitch of the helix, defined as the distance
traveled along the direction of the magnetic
field in one revolution, is given by
p = voyTwhere T is the period, or the time for one
revolution.
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If the electron beam is defocused, a smudge is seen on theUnder these conditions an image of the anode hole will be observed on theBy continuing to increase the strength of the field beyond this critical value,
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If the electron beam is defocused, a smudge is seen on the
screen when the applied magnetic field is zero.
This means that the various electrons in the beam
pass through the anode hole with different transverse velocities v0x, and
so strike the screen at different points. This accounts for the
appearance of a broad, faintly illuminated area instead of a bright point
on the screen.
As the magnetic field is increased from zero the electrons will move in
helices of different radii, since the velocity v0xthat controls the radius of the
path will be different for different electrons.
However, the period, or the time to trace out the path, is independent of
v0x, and so the period will be the same for all electrons. If, then, the
distance from the anode to the screen is made equal to one pitch,
all the electrons will be brought back to the Y axis (the point 0 in Fig.),since they all will have made just one revolution.
screen. As the field is increased from zero, the smudge on the screen
resulting from the defocused beam will contract and will become a tiny sharp
spot (the image of the anode hole) when a critical value of the field is
reached. This critical field is that which makes the pitch of the helical path just
equal to the anode-screen distance, as discussed above.
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the pitch of the helix decreases, and the electrons travel through more than
one complete revolution. The electrons then strike the screen at various points,
so that a defocused spot is againvisible. A magnetic field strength will
ultimately
be reached at which the electrons make twocomplete revolutions in their pathfrom the anode to the screen, and once again the spot will be focused on the
screen.
This process may be continued, numerous foci being obtainable.
In fact, the current rating of the solenoid is the factor that generally furnishesa practical limitation to the order of the focus
The foregoing considerations may begeneralized in the following way:
If the screen is perpendicular to the Y axis at a distance L from the point ofemergence of the electron beam from the anode, then, for an anode-cathode
potential equal to Va,the electron beam will come to a focus at the center of the
screen provided that L is an integral multiple of p. Under these conditions,equationmay be rearranged to read .
where nis an integer representing the order of the focus.
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PARALLEL ELECTRIC AND MAGNETIC FIELDS
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If both electric and magnetic fields exist
simultaneously and parallel to each otherand the initial velocity of the electron either
is zero or is directed along the fields, the
magnetic field exerts no force on the
electron.
The resultant motion depends solely upon
the electric field intensity . In other words,
the electron will move in a direction parallel
to the fields with a constant acceleration. If
the fields are chosen as in Fig. the completemotion is specified by
vy=v0y - at ; y=v0yt(at2)
where a = e/m is the magnitude of the
acceleration. The negative sign results from
the fact that the direction of the acceleration
of an electron is opposite to the direction of
the electric field intensity .
If, initially, a component of velocity v0xperpendicular to themagnetic field exists, this component, together with the magnetic
field, will give rise to circular motion, the radius of the circular path
being independent of . However, because of the electric field ,
the velocity along the field changes with time. Consequently, the
resulting path is helical with a pitch that changes with the time.That is, the distance traveled along the Y axis per revolution
increases with each revolution.
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PERPENDICULAR ELECTRIC AND MAGNETIC FIELDS
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Thus there is no component of force along the
Y direction, and the Y component of
acceleration is zero. Hence the motion along Y
assuming that the electron starts at the origin.
is given by
fy =0; vy=v0y; y=v0yt
The directions of the fields are shown in Fig.
The magnetic field is directed along
the Y axis,
and the Electric field in directedalong thex axis.
The force on an electron due to the electric
field is directed along the +X axis.
Any force due to the magnetic field is always
normal to B, and hence lies in a plane parallel
to theXZ plane.
If the initial velocity component parallel to B is
zero, the path lies entirely in a plane
perpendicular to B.
It is desired to investigate the path of an electron
starting at rest at the origin.
The initial magnetic force is zero, since the velocity
is zero
The electric force is directed along the +X axis, and the electron will
be accelerated in this direction.
As soon as the electron is in motion, the magnetic force will no longer be zero.
There will then be a component of this force which will be proportional to the X
component of velocity and will be directed along the +Z axis.
The path will thus bend away from the +X direction toward the +Z direction.
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The arguments given above do indicate the manner inThe force due to the electric field is
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which the electron starts on its path.
This path will now he shown to be a cycloid.
The force due to the electric field is
e along the +X direction.
The force due to the magnetic field is found as
follows:
At any instant, the velocity is determined
by the three components vx,vyand vzalong the
three coordinate axes.
Since B is in the Y direction, no force will be
exerted on the electron due to vy.
Because of vx the force is eBvzin the +Zdirection,
Similarly, the force due tovzis eBvz in the X direction.
62Bhimavaram
A straight forward procedure is involved in the
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Hence Newtons law, when expressed in terms
of the three components, yields
By writing for convenience
the foregoing equations may be written in the form
63
g p
solution of these equations. If the first
equation of (3) is differentiated and combined
with the second, we obtain
In order to find the coordinates xand z from these expressions, each equation
must be integrated. Thus, subject to the initial conditions x = z = 0,
This linear differential equation with constant coefficients is readily solved for vx .
Substituting this expression for vx in Eq. (3) this equation can be solved for vz
Subject to the initial conditions
vx=vz = 0, we obtain
vx = u sin t vz = u u cos t
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If, for convenience,
Bhimavaram
Equations (5) are the parametric equations of acommon cycloid, defined as the path generated by aThe physical interpretation of the symbols introducedabove merely as abbreviations is as follows:From these interpretations and from Fig. it isclear that the maximum displacement of theStraight Line Path As a special case ofimportance, consider that the electron is
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The angle gives the number of radians through
which the circle has rotated. From the diagram, itreadily follows that
x = Q Q cos ; z = QQ sin ---(6)
which are identical with Eqs. (5), thus proving that
the path is cycloidal as predicted.
y , p g y
point on the circumference of a circle of radius Q,
which rolls along a straight line, theZ axis. This is
illustrated in Fig.
The point P, whose coordinates are x and z (y = 0),represents the position of the electron at any time.
The dark curve is the locus of the point P. The
reference line CC is drawn through the center
of the generating circle parallel to the X axis.
Since the circle rolls on theZ axis, then OC representsthe length of the circumference that has already come
in contact with theZ axis. This length is evidently
equal to the arc PC (and equals Q).
y
represents the angular velocity of rotation of the
rolling circle.
represents the number of radians through which
the circle has rotated.
Q repre8ents the radius of the rolling circle.
Since u = Q, then u represents the velocity of
translation of the center of the rolling circle.
p
electron along the X axis is equal to the
diameter of the rolling circle, or 2Q. Also, the
distance along theZ axis between cusps is
equal to the circumference of the rolling circle,
or 2Q. At each cusp the speed of the electron
is zero, since at this point the velocity is
reversing its direction Fig.
This is also seen from the fact that each cusp is
along theZ axis, and hence at the same
potential. Therefore the electron has gained no
energy from the electric field, and its speed
must again be zero.
If an initial velocity exists that is directed
parallel to the magnetic field, the projection of
the path on theXZ plane will still be a cycloidbut the particle will now have a constant
velocity normal to the plane. This path might
be called a cycloidal helical motion.
p ,
released perpendicular to both the electric and
magnetic fields so that
v0x,v0y=0and v0z 0.
Note that this velocity u is independent of the charge
or mass of the ions. and net force is zero . In such a system of perpendicular
fields will act as a velocity filter and allow only those particles whose velocity
is given by the ratio /Bto be selected.
The electric force is e along the + X direction(Fig.), and the magnetic force is eBv0z along the X
direction. If the net force on the electron is zero, it
will continue to move along theZ axis with the
constant speed This conditions is realized when
e= eBv0z
v0z =/B=u
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