EDC first unit for jntuk

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1

Electronic Devices and Circuits

According to JNTUK syllabus

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Index of first unit

Fundamentals of atomic theory


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3

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
http://www.chemicalelements.com/groups/metalloids.htmlhttp://www.chemicalelements.com/groups/metalloids.html


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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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http://www.chemicalelements.com/groups/metalloids.htmlhttp://www.chemicalelements.com/groups/metalloids.html


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26

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.

28

Return to

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

+

++

+

+

+

+

+

-

-

-

-

-

-

-

-

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

+

++

+

+

+

+

+

-

-

-

-

-

-

-

-


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V

+

++

+

+

+

+

+

-

-

-

-

-

-

-

-


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t t l t ifi d fi ld


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V

+

++

+

+

+

+

+

-

-

-

-

-

-

-

-


32

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34

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Index Bhimavaram


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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.

48

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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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50

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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51

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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Index

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52

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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Index

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DEFLECTION IN A CATHODE-RAY TUBE MAGNETIC


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53

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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54

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

56

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Index

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MAGNETIC FOCUSING


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57

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.

58

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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Index

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PARALLEL ELECTRIC AND MAGNETIC FIELDS


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60

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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61

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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Index

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