Unit 1

EDC

Unit 1

INTRODUCTION: Electronics means electron mechanics. Electronics deals with the movement of electrons under the influence of

externally applied electric & magnetic field. IEEE defines electronics as “that field of science and engineering which deals

with the electron devices and their utilization”. Electron device may be defined as “a device in which conduction takes place by

movement of electros through vacuum, gas or a semiconductor”. Applications

1. communication 2. entertainment3. defense4. mics5. instrumentation6. medical science7. industry

CHARGED PARTICLE:In physics, a charged particle is a particle with an electric charge. Particles either

have a positive, negative or no charge (being neutral). An electron is a subatomic

particle that carries a negative electric charge. In many physical phenomena, such as

electricity, magnetism, and thermal conductivity, electrons play an essential role. An

electron generates a magnetic field while moving, and it is deflected by external

magnetic fields. When an electron is accelerated, it can absorb or radiate energy in the

form of photons. Electrons, together with atomic nuclei made of protons and neutrons,

make up atoms.

Electron Charge------ 1.6x10-19 coulombs

Mass------- 9.11x10-19 Kg

Radius----- 10-15 meters

Motion of charged particle is due to either of electric or magnetic or gravitational fields.

When electron is subjected to these E or M or G fields it get acceleration and its

trajectory is determined by Newton’s laws provided that the force acting on the particle

is known. Their are two models of electron, for large scale phenomena classical model

is used and for small scale phenomena wave model is used.

MOTION OF CHARGED PARTICLE IN ELECTRIC FIELD:The force on a unit positive charge at any point in an electric field is the electric

field intensity at that point. Field intensity is represented by ‘E’. The force on a positive

charge q in an electric field of intensity E is given by ‘q E’.

f q = q E---------------------------------------------------------------------------------- 1

Where f q is in Newton’s, q is in coulombs, E is in volts/meter.

In order to calculate the path of a charged particle in on electric field the force, given by

equ.1, must related to the mass and the acceleration of the particle by Newton’s second

law of motion. Hence

f q = q E = m a = m dv /dt.------------------------------------------------------------- 2

Note: ‘f’,’E’,’a’,‘v’ represents scalar quantities, ‘f’,’E’,’a’,‘v’ represents vector

quantities. Where

m is mass in Kgs

a is acceleration in m/sec2

v is velocity in m/sec.

The solution of equ.2 subject to approximate initial conditions, gives the path of the

particle resulting from the action of the electric force.

Force on electron

f = - e E --------------------------------------------------------------------------------- 3

The minus sign denotes that the force is in the direction opposite to the field.

Constant electric field:

Electron is situated between the two plates of a parallel plate capacitor which are

contained in an evacuated envelop.

Figure: 1 the two dimensional electric field between the parallel plates of capacitor.

A difference of potential is applied between the plates, and E is in –X –axis direction. If

d is small compared to dimensions of plates, E.F is considered as uniform. In initial

conditions, characteristics of the motion are as follows.

vx = vox, x=xo when t=0------------------------------------------------------------ 4

This means that initial velocity vox is chosen along E. the lines of force, and that the

initial position x0 of the electron is along the X-axis.

d

Y

E o

Z

X

_ +

As in Y-axis — f = 0, a = 0 (Newton’s law)

Z-axis — f = 0, a = 0

a = 0 means velocity is constant, since initial velocity along these axes is zero the

particle will not move along these directions.

Newton‘s law applies to the X – axis direction yields

e E = max-------------------------------------------------------------------------------- 5

ax = e E / m constant velocity in constant E.F

In case of freely falling body in the uniform gravitational field of the earth.

vx = vox +ax t, x = xo + vox t + ½ ax t2

------------------------------------- 6

Here ax constant and independent of time.

Motion is determined by differentiating equ 6

d vx / dt = ax, d x /dt = vx

These are definitions of acceleration and velocity

Potential:

Ex need not be uniform and is function of distance, not function of time from Newton’s

2nd law

-e Ex / m = d vx / dt multiply with d x = vx dt and integrating we get,

- e/m ∫ xox Ex dx = ∫vxo

vx vx d vx ----------------------------------------- 8

The definite integral ∫xox Ex dx is an expression for the work done by the field in

carrying a unit +ve charge from the xo to the point x.

The potential V (in volts) of point x w.r.t point xo is the work done against the field in

taking a unit +ve charge from xo to x.

V= - ∫ xox Ex dx ------------------------------------------------------------------- 9

From equ 8 & 9

eV = ½ m (vx 2- vox 2

) ------------------------------------------------------ 10

V is independent of variation of filed distribution and dependent on magnitude of field

above equ indicates law of conservation of energy valid for multidimensional. Let two

points A&B, B at a higher potential than A in general form.

q VAB = ½ m (vA 2- vB 2

) --------------------------------------------------- 11

q – Coulombs, qVAB – joules, vA &, vB – initial & final velocities.

The potential energy between two points equals to the potential multiplied by the

charge.

The eV unit of energy:

Energy ---joule- mks,

When joule is small convert to watt, watt =103 (KW), 106 (MW)

When joule is large convert to erga= joule x 10-7

1 eV = 1.6x 10-19 j used for electrical, mechanical, thermal etc.

MeV= million eV, BeV = billion eV.

Relation between field intensity and potential.

V= - ∫ xox Ex dx only when E.F is uniform with distance.

Ex (x - xo) = V

Ex = V/ (x - xo)

= -V/d volts / meter

When Ex = f (d) correct result is obtained by

Ex = - dV/dx

Minus sign shows that the electric field is directed from the region of higher potential to

the region of lower potential.

Two- dimensional motion:

Figure 2: two dimensional electric motion in a uniform field.

Initial conditions when t = 0

vx = vox x =0

vy = 0 y =0

l

d Vd

vox

-

+

Y

X

vz = 0 z = 0

Since there is no force in Z- axis direction acceleration in that direction is zero. Motion

is only in X-Y plane. In X – axis direction velocity is, vox, constant from which it

follows that

x = vox t. -------------------------------------------------------------------------- 17

For a constant velocity in Y – direction

Velocity vy = ay t ( as voy = 0 ) --------------------------- 18

Displacement y = ½ ay t2 ( as voy = 0 )------------------------------ 19

ay = - e E / m

= e V / m d

vy varies from point to point,

vx constant throughout the path.

By combining equ 17 & 18

We get the path of the particle as

y = (½ ay /vox2) x 2 ------------------------------------------------------------- 20

This shows that the particle moves in a parabolic path in the region between the plates.

MOTION OF CHARGED PARTICLE IN MAGNETIC FIELD:

Motion of a charged particle in magnetic field is characterized by the change in

the direction of motion. It is expected also as magnetic field is capable of only changing

direction of motion. In order to keep the context of study simplified, we assume

magnetic field to be uniform. This assumption greatly simplifies the description and lets

us easily visualize the motion of a charged particle in magnetic field.

Lorentz magnetic force law is the basic consideration here. Hence, we shall first take a

look at the Lorentz magnetic force expression:

F = q (v x B) ----------------------------------------------------------------------1

We briefly describe following important points about this expression:

1: There is no magnetic force on a stationary charge (v=0). As such, our study here

refers to situations in which charge is moving with certain velocity in the magnetic field.

This condition is met when the charge is released with certain velocity in the magnetic

field.

2: The magnetic field (B) is a uniform stationary magnetic field for our consideration in

the module. It means that the magnitude and direction of magnetic field do not change

during motion. The charged particle, however, is subjected to magnetic force acting side

way. The direction of motion of charged particle, therefore, changes. In turn, the

direction of magnetic force being perpendicular to velocity also changes. Important

point to underline here is that this loop of changing directions of velocity and magnetic

force is continuous. In other words, the directions of both velocity and magnetic force

keep changing continuously with the progress of motion.

Force in a magnetic field:

Force on a moving charge in a magnetic field is given by motor law. It gives

fm = BIL--------------------------------------------------------------------------------- 1

fm – force (Newton), B - M.F intensity (weber/m2), I-current (A), L-length (m) directions

of B & I are perpendicular. F perpendicular (I &B) like screw .If I not perpendicular B

only component of I perpendicular to B contributes to the force. Direction of current

depends on selection of particle.

Figure 3: pertaining to the determination of the direction of the fore on a charged

particle in a magnetic field.

N-total e- in L length, takes T sec to travel a distance of L then the total number of

electrons passing through any cross section of wire in unit time is N/T. Then

Current I = N e / T.

The force in Newton’s on a length Lm is

fm = BIL = B eN/T.L or

fm = BeNv.

Where v. average or drift speed m/sec. force per electron

fm = eBv

To summarize:

The force on a negative charge e (coulombs) moving with a component of

velocity v (meters per sec’), normal to a field B (Weber’s per square meter) is given by

eBv (Newton’s) and in a direction perpendicular to the plane of B and v.

v

B

fm

I 90o

o

CURRENT DENSITY:

Figure 4: Pertaining to the determination of the magnitude of the force fm on a charged

particle in a M. field.

Current density J = current per unit area of the conducting medium assuming a uniform

current distribution.

J ≡ I/A

J-A/m2, A (m) from equ.

J = Ne/TA = Nev/LA

Electron concentration n= N/LA (per cubic meter) and we get

J = nev = pv

Where p-charge density coulombs/cubic meter v-m/sec.

Independent of conducting medium. ρ & v need not be constant vary with time &

distance.

MOTION IN A MAGNETIC FIELD:

Path of a charged particle.

If the particle is at rest, fm = 0 and the particle remains at rest.

If the initial velocity of the particle is along the lines of the magnetic flex, there is no

force acting on the particle. (As fm = eBv)

Hence a particle whose initial velocity has no component normal to a uniform magnetic

field will continue to move with constant speed along the lines of flex.

L

A

N electrons

Figure 5: Circular motion of an e in a transverse M-field.

Initial speed v0, constant M.F. Since the force fm is perpendicular to v and so to the

motion at every instant, no work is done on the electron. So no increase in K.E. So no

change in speed.

Since v & B are constant in magnitude, then fm is constant in magnitude and

perpendicular to the direction of motion of the particle. So this type of force results in

motion in a circular path with constant speed. To find the radius of circle, it is recalled

that a particle moving in a circle path with a constant speed v, has an acceleration

toward the center of the circle of magnitude v2 / R. where R is the radius of the path in

meters then

m v2 / R = eBv

R = mv/eB

The corresponding angular velocity in radians/sec is given by

w = v / R = e B / m.

The time in sec for one complete revolution, called the period, is

T = 2∏ / w= 2∏ m/eB

For an electron T = 2∏ m/B

It is noted that radius of the path is directly perpendicular to the speed of the particle.

The period and the angular velocity are independent of speed or radius. This mean of

course, that faster moving particles will traverse larger circles in the same time that a

slower particle mores in its smaller circles.

Ex: Cyclotron and magnetic focusing apparatus.

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

R

Magnetic field into paper

P

vo

Field free region

O

P

CATHOD RAY OSCILLOSCOPE (CRO):

The block diagram of CRO is as shown in:-

Figure 6: CRO block diagram

Attenuator: It is a potential divider ckt that attenuates the amplitude of the input signal

to the required amplitude.

Vertical Amplifier: To deflect the beam of the CRO, voltages, in the range of 100V are

required. Generally, the input signal will be in the range of mV. Hence, the vertical

amplifier amplifies the weak input signal to a level required by CRT. Gain of the

amplifier should be large.

Delay line: To be able to observe the complete waveform of the input signal. The

horizontal of vertical deflection of the e- beam should start simultaneously.

Signal processing in horizontal amplifier system takes a finite amount of

time. Such a delay doesn’t occur in vertical deflection system. Hence, the signal given

to the vertical input should be delayed to observe the waveform property.

CRT: It is the heart of the CRO. It provides a high velocity e- beam that passes through

a set of deflection plates. One set of plates is oriented to deflect the e - beam vertically

when appropriate voltage is applied between them. The 2nd set of plates deflects the

beam horizontally.

Time base Generator:

Need for time base: When we apply voltage across the vertical plates, the spot moves up

and down and we can observe a vertical line. This is due to the fact that we have not

provided a time axis.

To provide a time axis, we have to move the spot horizontally from one

end of the screen to other end at a ray proportional to the time

attenuatorVertical amplifier

Delay line

Triggering circuit

Time base generator Horizontal

amplifier

In put

Signal

Vertical deflection plates screen

Electron gun

Horizontal deflectionplates

Time Base Ckt:

The simplest ckt to generate time base is a

Capacitor that is charged by a constant current source.

Measurement of frequency:

The signal for which the frequency ‘f’’ is to be measured is given to the vertical

input. The no. of divisions occupied by 1 complete cycle of the wave form is measured.

The no. of divisions multiplied by the time base setting in sec. is equal to the time

period (T) of one cycle. The frequency of wave form is inverse of the time period. ‘T’.

Measurement of phase Difference:

The phase difference between 2 sinusoidal signals of same frequency can be

calculated from the amplitude of the 2 signals as when A&B

are the amplitudes of 2 signals.

To measure the phase difference of two signals, the 2 sine waves are applied

simultaneously to the vertical & horizontal deflecting plate of CRO.

Horizontal amplifier: This amplifies the output of the time base generator. Some times,

it amplifies the horizontal input signal.

Trigger Ckt: It is a ckt that synchronies horizontal & vertical inputs. It makes sure that

the horizontal sweep and vertical input always start at the same point on the input signal.

Application of CRO:

It is used to measure the voltage, current, frequency & phase difference

of the given signals.

Measurement of voltage:

If the signal is applied to the vertical deflection plate only a vertical line

appears on the screen. The height of the line is proportional to peak voltage of the

Ø = 0o

Ø = 30o

Ø = 90o

Ø = 150o

Ø = 180o

applied signal. The vertical scale on the CRT screen is marked in Cms. Each centimeter

is further sub-divided into 5 parts, so that each part represents 0.2cm.

Measurement of current:

When a current is to be measured it is passed through a known resistance

and voltage across it is measured.

CRO: OSCILLOSCOPES DEFINITION:

Waveforms having a frequency as low as approximately 1 hertz (Hz) or as high

as several megahertz (MHz). High-end oscilloscopes can display signals having

frequencies up to several hundred gigahertzes (GHz). The display is broken up into so-

called horizontal divisions (hor div) and vertical divisions (vert div). Time is displayed

from left to right on the horizontal scale. Instantaneous voltage appears on the vertical

scale, with positive values going upward and negative values going downward.

CATHODE-RAY TUBE

Power and Scale Illumination:  Turns instrument on and controls illumination of the

graticule.

Focus:  Focus the spot or trace on the screen.

Intensity:  Regulates the brightness of the spot or trace

ELECTROSTATIC focusing:

Figure: electrostatic focusing system of a CRT

The accelerating beam would be scattered now because of variation in energy and

would produce a brad ill-defined spot on the screen. This electron beam is focused on

the screen by an electrostatic lens consisting of two more accelerating anode.

Figure 7: electrostatic focusing

Preaccelerating anode

Focusing anode

Equipotential surfaces

focuse

Accelerating anode

High voltage supply

Control grid

Va

Anode voltage

Screen

C

Electron

gun

At equipotential surfaces the electron changes its direction. These equipotential surfaces act as electrostatic lens.

MAGNETIC FOCUSING:

Axis of tube along M. Field lines

Figure 8: The helical path of an electron introduced at an angle (not 90o) with a constant

M. Field.

Velocity of origin is v0.

Initial transverse velocity clue to repulsion v0x

Velocity = vy + vθ along & transverse to the M.F

Since F perpendicular B no ay = 0 as vy is constant and equal to v0y

F = eBvθ perpendicular to path exist, resulting from transverse velocity. This

force gives rise to circular motion. The radius of the circle is

R = mvθ /eB where vθ = v0x.

The pitch of the helix, defined as the distance traveled along the direction of the

magnetic field in one revolution is given by

P = v0yT T-time for one revolution = period

P = 2Πm/eB v0y

When applied M.F = 0 smudge is seen on screen different transverse velocities v0x as

different points on screen at M.F. increase e- follow helix path with diff R at v0x is diff

period is independent of v0x so the period will be 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 o| ) here only one spot.

Critical field smallest points here distance between A & S is P at critical field increase P

decrease and e- travels more than one revolution.

Electronic path

f = e B vox

ZR= m vox/ e B

X

Y O’

B

vo

voy vox

O

The current rating of the solenoid is the factor that generally furnishes a practical

limitation to the order of the focus. In General

If the screen is perpendicular to the y axis at a distance L from the point of emergence of

the electron beam from the anode, then, for an anode cathode potential equal to v a, 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, we can write.

e/m = 8Π2 Van2/L2 B2 where n is an integer representing the order of the focus.

Here eVa = ½ mv0y2 the only effect of the anode potential is to accelerate the electron

along the tube axis this implies that the transverse velocity v0x, which is variable and

unknown, is negligible in comparison with v0y.

A short Focusing Coil:

Longitudinal M.Field over the entire length of a commercial tube is not too practical.

A short coil is wound around the neck of the tube.

Because of the fringing of the magnetic lines of flux a radial component of B exists in

addition to the component along the tube axis.

Two components of force on electron.

f = f (axial com of v + radial c of field) + f (axial com of field + radial c of v).

The motion will be a rotation about axis of the tube and if conditions are correct, the

electron on leaving the region of the coil may turn sufficiently so as to move in a line

toward the center of the screen.

A rough adjustment of focus is done by positioning of coil along neck. A fine

adjustment of focus is done by controlling the coil current.

ELECTROSTATIC DEFLECTION IN A CATHODE RAY

TUBE:

Those e- which are not collected by anode pass through the tiny anode hole and

strike the end of the glass envelope.

Figure 9: Electrostatic deflection in a cathode-ray tube.

D depends on Vd (defection plate’s potential)

V0x = (2e Va/m) ½ where initial velocity is negligible.

Between plate e- move in parabolic path given by

Y = ½ (ay/v0x2)x2

From point M at the edge of the plate path is straight line towards screen as it is field

free. This path is tangent to parabola at the point M slope of the line is

Tan θ = dy/dx where x=l

= ayl/v0x2

Straight line, is (from figure)

Y = ayl / v0x2 (x-l/2) --------- 1

Since x=l & y= ½ ay l2/v0x2 at the point M

AT POINT O | e- move towards p| in straight line path regardless of va & vd.

AT POINT P | Y=D, x=1+ ½ l equ 1 reduces to

D = = aylL/v0x2

Substituting equ ay & v0x

We get D = lLvd/2dva ------------ 2

D α Vd

Mean CRT can be used at a linear – voltage indicating device.

The electrostatic deflection sensitivity of a CRT is defined at the deflection (in meters)

on the screen per volt of deflecting voltage.

S ≡ D/vd = lL/2dVa. --------------- 3

S is independent of Vd & e/m & S α 1/vd.

Correction is needed in measured value of CRT.

Magnetic Deflection in a cathode – ray tube:

P’Vertical deflection plates

+ Vd + DCathode M θ d X O O’

P_ _ L

Va l/2 l/2AnodeVoltage

L flurorescent screen

Figure 10: Magnetic deflection in a cathode ray tube.

B is uniform in l region & zero outside.

Cathode to O: Straight line, in magnetic field: force of magnitude eBv, v.velocity. The

OM is arc of circle whose center is at Q the speed of e- is constant and equal to

v = v0x = (2eVa/m)½

The angle φ is by definition of radian measure, equal to the length of the arc OM/R, R

radius of circle. We assume a small angle of deflection then

φ ≈ l/R

We know R = mv/eB

If MP| projected backward will pass through the center O| of the region of the M.F. then

D ≈ L tan φ ≈ L φ = lL/R = lLeB/mv = lLB/(va)½ (e/2m)½

The deflection per unit M.F. intensity, D/B given by

D/B = lL/(va)½ (e/2m)½ is called magnetic deflection sensitivity of the tube.

MDS is independent of B.

MDS α 1/(va) ½ , MDS α (e/m) ½

MDS α L, magnetic coils are placed as far as down the neck of the tube as possible.

Usually directly affect the accelerating anode.

Deflection in a TV tube:

If magnetic deflection coil is driven by a saw tooth current waveform, the deflection of

the beam on the face of the tube will not be linear with time for such wide angle

deflection tubes, special linearity correcting n/w’s must be added. When the video signal

is applied to the electron gun, it modulates the intensity of the beam and thus forms the

TV picture.

Parallel Electric and Magnetic Fields:

Q P’ ** * * * * Magnetic field out of paper

θ * * * * * * * * * * * * DCathode * * * * * * * M θ * * * * * * * * * * * * * * X

O * * O’* * * P * * * * * * * * * * * * * * L

Va * * * * * * AnodeVoltage l/2 l/2

l flurorescent screen

If the initial velocity of the electron either is zero or is directed along the fields

the magnetic field exerts no force on the electron and the resulting motion depends

solely upon the electric field intensity E.

When e- parallel E & M.F with constant a.

If fields are selected as follow the motion of e- will be.

vy = v0y –at

y = v0y t -1/2 at2

‘-‘ indicate e- motion in opposite to E. Field.

If, initially, a component of velocity v0x perpendicular to the M.F. exists this component,

together with the M.F., will give rise to circular motion, the radius of the circular path

being independent of E.

E v various along the field with time.

The resulting path is helical with a pitch that changes with the time. i.e., the distance

traveled along the y axis per revolution increase with each revolution.

Perpendicular Electric and Magnetic field:

Assume E is in –x, B is in –y, force (due to B) perpendicular B is in xz plane.

ay = 0. Hence fy = 0, vy = v0y y = v0yt.

If the initial velocity component parallel to B is zero, the path lies entirely in a plane

perpendicular to B.

When e- starting at rest at the origin fMF = 0 since vy = 0. and since E is in –x fEM is in +x

e- moves, as e- move force due to Magnetic field is not 0 and that force will exist in +z

direction on e-.Therefore e- turns from +x to +z this will result to cycloid.

Force due to E.F. is eE along +x direction.

Force due to M.F. is as follows.

Due to vx the force is eBvx in +z direction

Due to vz the force is eBvz in –x direction

From Newton’s law

fx = m dvx/dt = eE -eBvz.

Figure 11:Parallel E.F & M.F

Y

E B

X

Z

fz = m dvz/dt = eBvx

we can write w ≡ eB/m and u≡E/B

we get dvx/dt = wu - wvz ------------- 1

dvz/dt = + wvx ---------------- 2

Differentiating equation 1 and combining with equation 2 we get

dvx2 / dt2 = -w dv2 / dt = -w2 vx -------------- 3

vx, vz = 0 initial conditions

vx = u sin wt, vz = u – u cos wt ------------ 4

In order to find coordinates x & z from these expressions each equation must be

integrated subject to the initial condition x=z=0.

X= u (1- cos wt)/w, z = ut- u/w sin wt -------------- 5

Θ = wt and Q = u/w

Then x = Q (1- cos Θ), z = Q (Θ- sin Θ)

Cycloidal path:

x = Q (1- cos Θ) ,, z = Q (Θ- sin Θ) are the parametric equ of a common cycloid ,

defined as the path generated by a point on the circumference of a circle of radius Q

which rolls along a straight line the Z-axis.

Figure 12: the cycloid path of an electron in perpendicular electric and magnetic fields

when the initial velocity is zero

oc’ is circumference that has already come in contact with Z-axis

oc’= pc’, arc = Q Θ

Θ gives number of radians through which the circle has rotated. We get from diagram

x = Q (1- cos Θ),

z = Q (Θ- sin Θ)

This path is cycloid.

W – Angular velocity of rotation of the rolling circle.

Θ – Number of radians through which the circle has rotated

Q – Radius of the rolling circle.

U = wQ is velocity of translation of the circle of center of the rolling circle.

Max displacement of e- along x is zQ (diameter of r circle).

Distance between cusps along z is 2Π Q (circumference).

At each cusp v=0 since v change its direction & same V. Therefore the electron has

gained no energy from the E.F. 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 the xz plane will still be cycloid but the particle will now have a constant

Q x Qcosθ

O X

p Qθ rolling Qsinθ circle

C’ C

2ΠQ cycloidal path 2Q

velocity normal to the plane. This path may be called a “cycloidal helical motion” is

given by x = Q (1- cos Θ), z = Q (Θ- sin Θ) and vy = v0y, y = v0yt.

Straight line path:

When e- is perpendicular to both E.F. and M.F.

And V0x = v0y = 0 and v0x ≠ 0

The E.F. is eE along +x, M.F. is eBv0x along –x so net force is zero; it will continue to

move along the z axis with the constant speed v0z. This conditions is realized when

eE = eBv0z

v0z = E/B = u is independent of the charge or mass of the ions.

This system of perpendicular field will act as a ‘velocity filter’ and allow only those

particles whose velocity is given by the ratio E/B to be selected.

Trochoidal Paths:

If initial velocity component in the direction perpendicular to the M.F. is not zero, it can

be shown that the path is a trochoid. This curve is the locus of a point a “spoke” of a

wheel rolling on a straight line.

a. Q| (spoke length) > Q (radius) of rolling circle – prolate cycloid

b. Q| = Q common cycloid

c. Q| < Q curtate cycloid

Figure 13: The trochoidal paths of e- in perpendicular E & M.F.

X

b B magnetic a Field c ZInto paper

E