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PhysicsPhysicsap erap er

1919

1

CHAPTER 19:CHAPTER 19:Magnetic fieldMagnetic field

(7 Hours)(7 Hours)

w w w .

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PhysicsPhysicsap erap er1919

2

Overview:

w w w .

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y

w w w .

k m p

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a t ri k

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y ElectrostaticsElectrostatics

Coulomb’sCoulomb’s

lawlaw

Charge in uniformCharge in uniform

Electric fieldElectric field

Electric fieldElectric field EquipotentialEquipotential

surfacesurface

Electric fieldElectric field

strengthstrength

ElectricElectric

potentialpotential

PotentialPotential

energyenergy

I N P R O G

R E S S

I N P R O G

R E S S




3

At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:

DefineDefine magnetic field.magnetic field.

IdentifyIdentify magnetic field sources.magnetic field sources.

SketchSketch the magnetic field lines.the magnetic field lines.

Learning Outcome:

19.1 Magnetic field (1 hour)

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4

is defined as a region around a magnet where a magnetica region around a magnet where a magnetic

force can be experiencedforce can be experienced.

A stationary electric chargestationary electric charge is surrounded by an electricsurrounded by an electric

field onlyfield only.

When an electric charge moveselectric charge moves, it is surrounded by ansurrounded by anelectric field and a magnetic fieldelectric field and a magnetic field. The motion of the electricmotion of the electric

charge produces the magnetic fieldcharge produces the magnetic field.

Magnetic field has two poles, called north (N)north (N) and south (S)south (S).This magnetic poles are always found in pairsfound in pairs whereas asingle magnetic pole has never been found.

Like poles (N-N or S-S) repelLike poles (N-N or S-S) repel each other.

Opposite poles (N-S) attractOpposite poles (N-S) attract each other.

19.1 Magnetic field




5

Magnetic field lines are used to represent a magnetic field.

By convention, magnetic field lines leave the north poleleave the north pole andenters the south poleenters the south pole of a magnet.

Magnetic field lines can be represented by straight lines or curves. The tangent to a curved field linetangent to a curved field line at a pointindicates the direction of the magnetic fielddirection of the magnetic field at that point

as shown in Figure 19.1.

Magnetic field can be represented by crossescrosses or by dotteddotted

circlescircles as shown in Figures 19.2a and 19.2b.

19.1.1 Magnetic field lines

Figure 19.1Figure 19.1

direction of magnetic field

at point P.PP

Figure 19.2a : magnetic field linesFigure 19.2a : magnetic field lines

enter enter the page perpendicularlythe page perpendicularly

XX XX XX XX

XX XX XX XX

Figure 19.2b : magnetic field linesFigure 19.2b : magnetic field lines

leaveleave the page perpendicularlythe page perpendicularly




6

A uniform fielduniform field is represented by parallel lines of forceparallel lines of force. Thismeans that the number of lines passing perpendicularlynumber of lines passing perpendicularly

through unit area at all cross-sections in a magnetic fieldthrough unit area at all cross-sections in a magnetic fieldare the sameare the same as shown in Figure 19.3.

A non-uniform field is represented by non-parallel lines. The

number of magnetic field lines varies at different unit cross-number of magnetic field lines varies at different unit cross-

sectionssections as shown in Figure 19.4.


unit cross-sectional areaunit cross-sectional area




7

The number of lines per unit cross-sectional area isnumber of lines per unit cross-sectional area isproportional to the magnitude of the magnetic fieldproportional to the magnitude of the magnetic field.

Magnetic field lines do not intersectdo not intersect one another.


stronger field instronger field in A A11

A A11

A A

22

weaker field inweaker field in A A22




8

The pattern of the magnetic field lines can be determined by

using two methods. compass needlescompass needles (shown in Figure 19.5)

sprinkling iron filings on paper sprinkling iron filings on paper (shown in Figure 19.6).

19.1.2 Magnetic field lines pattern

Figure 19.5: plotting a magnetic field line of a bar magnetic.Figure 19.5: plotting a magnetic field line of a bar magnetic.

Figure 19.6: thin iron filing indicate the magnetic field lines.Figure 19.6: thin iron filing indicate the magnetic field lines.




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Figures 19.7 shows the various pattern of magnetic field linesaround the magnets.

Figure 19.7aFigure 19.7a

a. Bar magnet

b. Horseshoe or U magnet

Figure 19.7bFigure 19.7b




10

d. Two bar magnets (like poleslike poles) - repulsiverepulsive

Neutral point (point wherewhere

the resultant magneticthe resultant magnetic

force is zeroforce is zero).

c. Two bar magnets (unlike poleunlike pole) - attractiveattractive

Figure 19.7cFigure 19.7c

Figure 19.7dFigure 19.7d




Southmagnetic pole

Southgeographical pole

Northmagnetic pole

11.5°

11

The Earth’s magnetic field is like that of a giant bar magnet as

illustrated in Figure 19.8 with a pole near each geographic poleof the Earth.

19.1.3 Earth’s magnetic field


The magnetic poles are tilted away

from the rotational axis by an angle of

11.5°.

Since the north pole of a compass

needle (Figure 19.8) points toward thesouth magnetic pole of the Earth, and

since opposite attract, it follows that

Figure 19.8 also shows that the field

lines are essentially horizontalhorizontal (parallel

to the Earth’s surface) near thenear the

equator equator but enter or leave the Earthenter or leave the Earth

vertically near the polesvertically near the poles.

the north geographical pole of thenorth geographical pole of the

EarthEarth is actually near the south polenear the south pole

of the Earth’s magnetic fieldof the Earth’s magnetic field.

h i




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Using the permanent magnetUsing the permanent magnet

One permanent magnet A permanent magnet is bring near to the soft iron and

touching the surface of the soft iron by following the path inthe Figure 19.9.

This method is called induced magnetizationinduced magnetization.

The arrowsarrows in the soft iron represent the magnetizationdirection with the arrowhead being the north pole and arrowtail being the south pole. It is also known as domainsdomains ( thethe

tiny magnetized region because of spin magnetictiny magnetized region because of spin magneticmoment of the electronmoment of the electron).

19.1.4 Magnetization of a Soft Iron

Figure 19.9Figure 19.9 NN SS

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In an unmagnetized piece of soft iron, these domains arearranged randomly but it is aligned in one direction when the

soft iron becomes magnetized. The soft iron becomes a temporary magnet with its south

pole facing the north pole of the permanent magnet and viceversa as shown in Figure 19.9.

Two permanent magnets

Bring and touch the first magnet to one end of the soft ironand another end with the second magnet as shown in Figure19.10.

NN NN SSSS


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NN SS

Switch, S

I I I I

SSNN

Current -Current -

anticlockwiseanticlockwiseCurrent - clockwiseCurrent - clockwise

Using the electrical circuitUsing the electrical circuit

A soft iron is placed inside a solenoid (a long coil of wire

consisting of many loops of wire) that is connected to the power supply as shown in Figure 19.11.

When the switch S is closed, the current I flows in the solenoid

and produces magnetic field.

The directions of the fields associated with the solenoid can be

found by viewing the current flows in the solenoid from bothviewing the current flows in the solenoid from both

endend or applying the right hand grip ruleright hand grip rule as shown in Figure19.11.

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Other examples:

If you dropdrop a permanent magnet on the floor or strikestrike it with a

hammer, you may jar the domains into randomnessdomains into randomness. Themagnet can thus lose some or alllose some or all of its magnetism.

HeatingHeating a magnet too can cause a loss of magnetism.

The permanent magnet also can be demagnetized by placing itplacing it

inside a solenoid that connected to an alternating sourceinside a solenoid that connected to an alternating source.

NNSS

I I I I I I I I

SS NN

Thumb – north polenorth pole

Other fingers –

direction of currentdirection of current

in solenoidin solenoid.

Note:Note:

Figure 19.12aFigure 19.12a Figure 19.12bFigure 19.12b

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is defined as the magnetic flux per unit area across anthe magnetic flux per unit area across an

area at right angles to the magnetic fieldarea at right angles to the magnetic field.

Mathematically,

It also known as magnetic inductionmagnetic induction (magnetic field intensitymagnetic field intensity

OR strengthOR strength) It is a vector quantityvector quantity and its direction follows the direction of the direction of

the magnetic fieldthe magnetic field.

Its unit is tesla (T)tesla (T) OR weber per metre squared (Wb mweber per metre squared (Wb m −22)). Unit conversion :

19.1.5 Magnetic flux density, B

⊥

= A

BΦ

wherefluxmagnetic:Φ

fieldmagneticthetoanglesrightatarea:⊥ A

)G(gauss10mWb1T142

== −

(19.1)(19.1)

Ph iPh iap erap er




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The direction of any magnetic field is taken to be in the directionthat an Earth-calibrated compass points. Explain why this mean

that magnetic field lines must leave from the north pole of a

permanent bar magnet and enter its south pole.

Solution :Solution :

Example 19.1 :

Near the north pole of a permanent bar magnet, the north poleNear the north pole of a permanent bar magnet, the north poleof a compass will point away from the bar magnet so the fieldof a compass will point away from the bar magnet so the field

lines leave the north pole.lines leave the north pole.

Near the south pole of a permanent bar magnet, the north poleNear the south pole of a permanent bar magnet, the north pole

of a compass will point toward the bar magnet so the field linesof a compass will point toward the bar magnet so the field lines

enter the south pole.enter the south pole.

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Exercise 19.1 :

1. Sketch the magnetic field lines pattern around the bar

magnets for following cases.

a.

b.

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ApplyApply magnetic field :magnetic field :

for a long straight wire,for a long straight wire,

for a circular coil,for a circular coil,

for a solenoid.for a solenoid.

Learning Outcome:

19.2 Magnetic produced by current-carryingconductor (1 hour)

r

I μ B

π=2

0

R I μ B

2

0=

nI μ B 0=

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When a current flows in a conductor wirecurrent flows in a conductor wire or coilcoil, the

magnetic field will be producedmagnetic field will be produced.

The direction of magnetic fielddirection of magnetic field around the wire or coil can be

determined by using the right hand grip ruleright hand grip rule as shown inFigure 19.13.

19.2 Magnetic field produced by current –

carrying conductor


Thumb – direction of currentdirection of currentOther fingers – direction of magneticdirection of magnetic

fieldfield (clockwiseclockwise OR

anticlockwiseanticlockwise)

Note:Note:





21

The magnetic field lines pattern around a straight conductor carrying current is shown in Figures 19.14 and 19.15.

19.2.1 Magnetic field of a long straight conductor

(wire) carrying current

OR

B

I

Current out of the pageCurrent out of the pageFigure 19.14Figure 19.14

B I

I B

B





22

Consider a straight conductor (wire) carrying a current I is

placed in vacuum as shown in Figure 19.16.

OR


I

I

I XX

Current into the pageCurrent into the pageXX

B

B

B

B





23

The magnitude of magnetic flux density (magnetic field

intensity), B at point P at distance r from the wire carrying

current is given by

r PP

I Figure 19.16Figure 19.16

XX B

into the page (paper)into the page (paper)

r

I μ B

π=2

0

where spacefreeof ty permeabili:0 μ 17AmT104

−−×π=

(wire)conductor straightafrom pointaof distance:r

(19.2)(19.2)





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The magnetic field lines pattern around a circular coil carryingcurrent is shown in Figures 19.17.

19.2.2 Magnetic field of a circular coil


I I XX

SS

NN

OR

SS

NN

I I

I





25

Consider a circular shaped conductor with radius R that carries

a current I as shown in Figure 19.18.

R

NI μ B

2

0=

where

(19.3)(19.3)

The magnitude of magnetic fieldmagnetic field

intensityintensity B B at point O (centre of centre of

the circular coil or loopthe circular coil or loop) , is givenby

R

O

coilcircular theof radius: R

(loops)coilsof number : N

spacefreeof ty permeabili:0

μ

current: I Figure 19.18Figure 19.18





26

A solenoid is an electrical device in which a long wire hasan electrical device in which a long wire has

been wound into a succession of closely spaced loops withbeen wound into a succession of closely spaced loops withgeometry of a helixgeometry of a helix.

The magnetic field lines pattern around a solenoid carrying

current is shown in Figure 19.19.

19.2.3 Magnetic field of a solenoid

SSNN


I I





27

OR

I

I XX XX XX XX

I

I

I

I

I

I

SSNN





28

The magnitude of magnetic field intensity at the end of magnitude of magnetic field intensity at the end of N N

turn solenoidturn solenoid is given by

nI μ B 0

2

1= (19.5)(19.5)

where lengthunit per turnsof number :n

The magnitude of magnetic field intensity at the centre (mid-magnitude of magnetic field intensity at the centre (mid-

point/ inside) of point/ inside) of N N turn solenoidturn solenoid is given by

l

NI μ B 0=

nl

N =and

nI μ B 0= (19.4)(19.4)





29

Two long straight wires are placed parallel to each other and

carrying the same current I . Sketch the magnetic field lines patternaround both wires

a. when the currents are in the same direction.

b. when the currents are in opposite direction.


a.

Example 19.2 :

I

I I

I





30

I I

OR


a.





31

OR


b.

I I XX

I

I

I

I





32

A long wire (X) carrying a current of 50 A is placed parallel to and5.0 cm away from a similar wire (Y) carrying a current of 10 A.

a. Determine the magnitude and direction of the magnetic flux

density at a point midway between the wires :

i. when the current are in the same direction.

ii. when they are in opposite direction.

b. When the currents are in the same direction there is a point

somewhere between X and Y at which the magnetic flux density

is zero. How far from X is this point ?

(Given µ 0 = 4π × 10−7 H m−1)

Example 19.3 :





33


a. i.

By using the equation of magnetic field at any point near the

straight wire, then at point A

Magnitude of BX :

A10m;100.5A;50 Y2

X =×== − I d I

T100.44

X−×= B

X B

Y B

OR

m105.22

2

YX

−×===d

r r

X I

AXr Yr

Y I

X

X0X

2πr

I μ B =

Direction : into the page OR upwards

( )( )2

7

X105.22

50104−

−

××

=π

π B

X B

Y B

X I Y I

d

Xr Yr A





34


a. i. Magnitude of BY :

Therefore the total magnetic flux density at point A is

A10m;100.5A;50 Y2

X =×== − I d I

T100.85

Y−×= B

Y

Y0Y

2πr

I μ B =

Direction : out of page OR downwards

( )( )2

7

Y105.22

10104−

−

××=

π

π B

YXA B B B

+=

YXA B B B +−=

Direction : into the pageDirection : into the page OR upwardsupwardsSign convention of Sign convention of B B:

Out of the page ⇒positive (+)

Into the page ⇒negative (−)

Note:Note:

54A 100.8100.4 −− ×+×−= B

T102.3 4A

−×−= B





35


a. ii.

By using the equation of magnetic field at any point near the

straight wire, then at point A

Magnitude of BX :

A10m;100.5A;50 Y2

X =×== − I d I

T100.44

X−×= B

X B

Y B

OR

Direction : into the page OR

upwards

( )( )2

7

X105.22

50104−

−

××

=π

π B

X I

AXr Yr

Y I XX

X B

Y B

X I Y I

d

Xr Yr A





36


a. ii. Magnitude of BY :

Therefore the resultant magnetic flux density at point A is

A10m;100.5A;50 Y2

X =×== − I d I

T100.85

Y−×= B

Direction : into the page OR upwards

( )( )2

7

Y105.22

10104−

−

××=

π

π B

YXA B B B

+=

YXA B B B −−=

Direction : into the pageDirection : into the page OR upwardsupwards

54A 100.8100.4 −− ×−×−= B

T108.4 4A

−×−= B





37


b.

Since the resultant magnetic flux density at point C is zero

thus

A10m;100.5A;50 Y2

X =×== − I d I

X B

OR

r r =X

X I

CXr Yr

Y I

Y

B

r d r −=Y

YXC B B B

+=YX0 B B +−=

YX B B = where

X

X0X

2πr

I μ B = and

Y

Y0Y

2πr

I μ B =

X I Y I

d

Xr Yr C

X B

Y B





38


b.

Y

Y0

X

X0

22 πr I μ

πr I μ =

( )r d

I

r

I

−= YX

( )r r −×= −2100.5

1050

A10m;100.5A;50 Y2

X =×== − I d I

m102.4 2−×=r




Physicsyap erap er1919

39

Two long straight wires are oriented perpendicular to the page as

shown in Figure 19.20.

The current in one wire is I 1 = 3.0 A pointing into the page and the

current in the other wire is I 2= 4.0 A pointing out of page. Determine

the magnitude and direction of the nett magnetic field intensity at

point P.

(Given µ 0 = 4π × 10−7 H m−1)

Example 19.4 :





Physicsyap erap er1919

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By applying the equation of magnetic field intensity for straight wire,thus

m100.5A;0.4A;0.3 2121

−×=== r I I

( ) ( )2

2

2

22 100.5100.5 −− ×+×=r

1r 2 B

2r

m101.72

2

−×=r θ

θ 2

2

2

1

101.7

100.5cos −

−

×

×==

r

r θ

704.0cos =θ

704.0101.7

100.5sin

2

2

=××

= −

−

θ

1

101

2πr

I μ B =

T1020.15

1−×= B

( )( )2

7

1100.52

0.3104−

−

××

=π

π B

1 B

1 I 2 I

P

XXm100.5

2−×




yyap erap er1919

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and

m100.5A;0.4A;0.3 2121

−×=== r I I

2

202

2πr

I μ B =

T1013.15

2−×= B

( )( )2

7

2101.72

0.4104−

−

××

= π

π B

Vector x-component (T) y-component (T)

Vector

sum

51 1020.1

−×= B1 B 0

θ B cos2−

2 B

( )( )704.01013.15−×−=6

1096.7−×−=

θ B sin2−( )( )704.01013.1

5−×−=61096.7 −×−=

651096.71020.1

−− ×−×= x B

6104.04

−×=

61096.70

−×−= y B

610.967

−×−=




yyap erap er1919

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Therefore the magnitude of the nett magnetic field intensity at point

P is given by

and its direction is

m100.5A;0.4A;0.3 2121

−×=== r I I

22

y x B B B +=

T1093.86−

×= B

( ) ( )2626 1096.71004.4 −− ×−+×=

= −

x

y

B

Bθ

1tan

1.63−=θ

××−= −

−−6

61

1004.4

1096.7tan

(297(297°° from +x-axis anticlockwise)from +x-axis anticlockwise) OR

1.63 B

1 B

2 B

P




yyap erap er1919

43

a. A closely wound circular coil of diameter 10 cm has 500 turns

and carries a current of 2.5 A. Determine the magnitude of the

magnetic field at the centre of the coil.

b. A solenoid of length 1.5 m and 2.6 cm in diameter carries a

current of 18 A. The magnetic field inside the solenoid is

2.3 mT. Calculate the length of the wire forming the solenoid.(Given µ 0 = 4π × 10−7 T m A−1)


a. Given

By applying the equation for magnitude of the magnetic field at

the centre of the circular coil, thus

Example 19.5 :

A5.2;500m;105.02

1010 22

==×=×

= −−

I N R

R

NI μ B

2

0=

T1057.1 2−×= B

( )( )

( )2

7

100.52

5.2500104−

−

××

=π

B




yyap eap e1919

44


b. Given

By applying the equation of magnetic flux density inside the

solenoid, thus

Since the shaped for each coil in the solenoid is circle, then the

circumference for one turn is

Therefore the length of the wire forming the solenoid is

l

NI μ Bi

0=

turns153= N

( ) ( )

5.1

18104103.2

73 N π −

− ×=×

T103.2m;103.1

2

106.2m;5.1 3

i2

2−−

−

×=×=×

== Br l

A18= I

πr 2ncecircumfere = ( )2103.12ncecircumfere −×= π

m1017.8ncecircumfere2−

×=( )ncecircumfere×= N L

( )2108.17153 −××= L

m5.12= L




yypp1919

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Exercise 19.2 :

Given µ 0 = 4π × 10−7 T m A−1

1.

The two wires shown in Figure 19.21 carry currents of 5.00 Ain opposite directions and are separated by 10.0 cm.

a. Sketch the magnetic field lines pattern around both wires.

b. Determine the nett magnetic flux density at points P1 and

P2.

ANS. :ANS. : 1.331.33×× 1010−−55 T, out of page; 2.67T, out of page; 2.67×× 1010−−66 T, out of pageT, out of page

Figure 19.21Figure 19.21A00.5 A00.5

cm0.10

cm0.15

cm0.52P

1P




yypp1919

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2.

Four long, parallel power wires each carry 100 A current. A

cross sectional diagram for this wires is a square, 20.0 cm on

each side as shown in Figure 19.22.

a. Sketch the magnetic field lines pattern on the diagram.

b. Determine the magnetic flux density at the centre of the

square.

ANS. :ANS. : 4.04.0 ×× 1010−−44 T , to the left (180T , to the left (180°°))


XX XX




yypp1919

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UseUse force:force:

DescribeDescribe circular motion of a charge in a uniformcircular motion of a charge in a uniform

magnetic field.magnetic field.

UseUse relationshiprelationship F F BB== F F

CC..

Learning Outcome:

19.3 Force on a moving charged particle in a

uniform magnetic field (1 hour)

( ) Bvq F

×=

w w w

.k m p

h . m

a t ri k

. e d

u . m

y

w w w

.k m p

h . m

a t ri k

. e d

u . m

y




pp1919

48

19.3.1 Magnetic force A stationarystationary electric charge in a magnetic field will notnot

experience a magnetic forceexperience a magnetic force. But if the charge is movingcharge is moving with

a velocity, v in a magnetic field, B then it will experience ait will experience a

magnetic forcemagnetic force. The magnitudemagnitude of the magnetic force can be calculated by

using the following equation:

19.3 Force on a moving charged particle

in a uniform magnetic field

θ qvB F sin=

where forcemagnetic: F densityfluxmagnetic: B

chargeaof velocity:vchargetheof magnitude:q

Bvθ

and betweenangle:

(19.6)(19.6)




pp1919

49

B

v

F


In vector formvector form,

The direction of the magnetic force can be determined by using

the Fleming’s hand rule.

Fleming’s right handFleming’s right hand rule for negativenegative charge

Fleming’s left handFleming’s left hand rule for positivepositive charge

(19.7)(19.7)( ) Bvq F

×=

B

v

F

shown in

Figures

19.23 and

19.24

ThumbThumb – direction of – direction of ForceForce

First finger First finger – direction of – direction of FieldField

Second finger Second finger – direction of – direction of VelocityVelocity

Figure 19.24Figure 19.24Note:Note:




pp1919

50

Determine the direction of the magnetic force, F exerted on a

charge in the following problems:a. b.

c. d.

e.

Example 19.6 :

+

B

v

−

B

v

XX XX XX XX

XX XX XX XX

XX XX XX XX

v

+

I

I

−v

− B

v




pp1919

51


a. By using Fleming’s left hand rule, thus

b. By using Fleming’s right hand rule, thus

c. By using Fleming’s right hand rule, thus

+

B

v

(into the page)(into the page) F

− B

v

F

(to the left)(to the left)

−

B

v

XX XX XX XX

XX XX XX XX

XX XX XX XX

F





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52


d.

e. I

−v

B

XX XX XX XX

XX XX XX F


Using right hand grip rule to determine the direction of magnetic

field produces by the current I on the charge

position. Then apply the Fleming’s right hand rule, thus

Using right hand grip rule to determine the direction of magnetic

field forms by the current I on the charge position. Then apply

the Fleming’s left hand rule, thus

v

+

I

(upwards)(upwards)

B

XXXX

XX

XX

XX

XX

XX XX

F




pp1919

53

Calculate the magnitude of the force on a proton travelling

5.0× 107

m s−1

in the uniform magnetic flux density of 1.5 Wb m−2

, if :a. the velocity of the proton is perpendicular to the magnetic field.

b. the velocity of the proton makes an angle 50°with the magnetic

field.

(Given the charge of the proton is +1.60× 10−19 C)


a. Given

Therefore

b. Given

Hence

Example 19.7 :

90=θ

217 mWb5.1;sm105.0 −− =×= Bv

θ qvB F sin=( )( )( ) 90sin5.1100.51060.1 719 ××= −

N1020.1 11−×= F 50=θ

N1019.9

12−

×= F

( )( )( ) 50sin5.1100.51060.1 719 ××= − F




pp1 91 9

54

Consider a charged particle moving in a uniform magnetic fieldwith its velocity perpendicular to the magnetic fieldvelocity perpendicular to the magnetic field.

As the particle enters the region, it will experiences a magneticmagnetic

forceforce which the force is perpendicular to the velocityperpendicular to the velocity of theparticle. Hence the direction of its velocity changes but themagnetic force remains perpendicular to the velocity.

This magnetic force, F B

makes the path of the particle is apath of the particle is a

circular circular as shown in Figures 19.25a, 19.25b, 19.25c and 19.25d.

19.3.2 Motion of a charged particle in a uniformmagnetic field


+ v

v

+ B F

+

v

B F

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

+ v

v

+ B F

+

v

B F





1919

55

Since the path is a circle therefore the magnetic force F B

contributes the centripetal force F c (nett force) in this motion.

Thus

Figure 19.25cFigure 19.25c Figure 19.25dFigure 19.25d

− v

v

− B F

−

v

B F

− v

v

−

B F

−

v

B F

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

c B F F =

r

mvθ qvB

2

sin = 90=θ and




1919

56

The period of the circular motion, T makes by the particle isgiven by

And the frequency of the circular motion is

Bq

mvr =

where particlechargedtheof mass:mvelocitytheof magnitude:v

pathcircular theof radius:r particlechargedtheof magnitude:q

(19.8)(19.8)

ω r v =

T

r v

π 2=

T

π ω

2=and

Bq

mT

π 2=

Bq

mvr =and

v

r T

π 2=

(19.9)(19.9)

T

f 1

=




1919

57

An electron at point A in Figure 19.26 has a speed v of 2.50

× 106 m s-1. Determine

a. the magnitude and direction of the magnetic field that will cause

the electron to follow the semicircular path from A to B.

b. the time required for the electron to move from A to B.

(Given e=1.60× 10−19 C and me= 9.11× 10−31 kg)

Example 19.8 :

−

v

BA cm0.20





1919

58


a. Since the path makes by the electron is a semicircular thus the

the magnitude of the magnetic field is given by

Direction of magnetic field : into the pageinto the page

OR

m100.20;sm1050.2 216 −− ×=×= d v

Be

mvr =

T1042.14−×= B

2

d r =

Be

mvd =

2

( )( )

( )19

6312

1060.1

1050.21011.9

2

100.20

−

−−

×

××=

×

B

and

−

v

BA

B

F




1919

59


b. The period of the electron is

Since the path is the semicircular then the time required for the

electron moves from A to B is given by

m100.20;sm1050.2 216 −− ×=×= d v

rωv =

s1051.27−

×=T

T π ω 2=

=

T

π d v

2

2

( )T

π 26 100.20

1050.2−×

=×

and

T t 2

1=

( )71051.22

1 −×=t

s1026.1 7−×=t




1919

60

Exercise 19.3 :1. Determine the sign of a charge in the following problems.

a. b.

ANS. :ANS. : positive; positivepositive; positive

2. Determine the direction of the magnetic force exerted on a

positive charge in each problem below when a switch S isclosed.

a. b.

ANS. :ANS. : into the page; out of pageinto the page; out of page

B

v

F

B

v F

Switch, S

+ v

Switch, S

+v




1919

61

3. An electron experiences the greatest force as it travels

2.9× 106 m s−1 in a magnetic field when it is moving north. The

force is upward and of magnitude 7.2× 10−13 N. Determine the

magnitude and direction of the magnetic field.

(Given the charge of the electron is 1.60× 10−19 C)

(Physics for scientists & engineers ,3(Physics for scientists & engineers ,3rdrd edition, Giancoli, Q22, p.705)edition, Giancoli, Q22, p.705)

ANS. :ANS. : 1.6 T to the east1.6 T to the east4. An electron moving with a speed of 9.1× 105 m s−1 in the

positive x direction experiences zero magnetic force. When it

moves in the positive y direction, it experiences a force of

2.0× 10−13 N that points in the negative z direction. What is the

direction and magnitude of the magnetic field?

(Given e=1.60× 10−19 C and me= 9.11× 10−31 kg)

(Physics, 3(Physics, 3rdrd edition, James S. Walker, Q8, p.762)edition, James S. Walker, Q8, p.762)

ANS. :ANS. : 1.37 T to the left (in the negative1.37 T to the left (in the negative y y direction)direction)




1919

62

5. Two charged particles with different speeds move one at a

time through a region of uniform magnetic field. The particles

move in the same direction and experience equal magnetic

forces.

a. If particle 1 has four times the charge of particle 2, which

particle has the greater speed? Explain.

b. Calculate the ratio of the speeds, v1/v2.(Physics, 3(Physics, 3rdrd edition, James S. Walker, Q9, p.762)edition, James S. Walker, Q9, p.762)

ANS. :ANS. : 1/41/4

6. A 12.5 µ C particle of mass 2.80× 10−5 kg moves

perpendicular to a 1.01 T magnetic field in a circular path of

radius 26.8 m.

a. How fast is the particle moving?

b. How long will it take the particle to complete one orbit?

(Physics, 3(Physics, 3rdrd edition, James S. Walker, Q18, p.763)edition, James S. Walker, Q18, p.763)

ANS. :ANS. : 12.1 m s12.1 m s−−11

; 13.9 s; 13.9 s




1919

63


UseUse force:force:

Learning Outcome:

19.4 Force on a current-carrying conductor in a

uniform magnetic field (1 hour)

( ) Bl I F

×=

w w w

.k m p

h . m

a t ri k

. e d

u . m

y

w w w

.k m p

h . m

a t ri k

. e d

u . m

y




1919

64

When a current-carrying conductor current-carrying conductor is placed in a magneticin a magnetic

fieldfield B, thus a magnetic force will acts on that conductor magnetic force will acts on that conductor .

The magnitudemagnitude of the magnetic force exerts on the current-carrying conductor is given by

In vector formvector form,

19.4 Force on a current-carrying

conductor in a uniform magnetic field

θ IlB F sin= (19.10)(19.10)

( ) Bl I F

×= (19.11)(19.11)

where forcemagnetic: F densityfluxmagnetictheof magnitude: B

current: I conductor theof length:l

B I θ

andof direction betweenangle:




1919

65

The direction of the magnetic force can be determined by using

the Fleming’s left hand ruleFleming’s left hand rule as shown in Figure 19.27.

From the equation (19.10),

the magnetic forcemagnetic force on the conductor has its maximummaximum value

when the conductor (and therefore the current) and theconductor (and therefore the current) and themagnetic field are perpendicular magnetic field are perpendicular (at right angles) to eachother then θ θ ==9090°° (shown in Figure 19.28a).

ThumbThumb – direction of – direction of ForceForce

First finger First finger – direction of – direction of FieldField

Second finger Second finger – direction of – direction of CurrentCurrent


Note:Note: B

I

F




1919

66

the magnetic forcemagnetic force on the conductor is zerozero when theconductor (and therefore the current) is parallel to theconductor (and therefore the current) is parallel to the

magnetic fieldmagnetic field then θ θ =0=0°° (shown in Figure 19.28b).


IlB F =max

B

90=θ

I

90sinmax IlB F =

B

0=θ

I


0= F

0sin IlB F =

One teslaOne tesla is defined as the magnetic flux density of a field in which aas the magnetic flux density of a field in which a

force of 1 newton acts on a 1 metre length of a conductor which carryingforce of 1 newton acts on a 1 metre length of a conductor which carrying

a current of 1 ampere and is perpendicular to the field.a current of 1 ampere and is perpendicular to the field.

Note:Note:




1919

67

Determine the direction of the magnetic force, exerted on a current-

carrying conductor in the following cases.a. b.


For both cases, use Fleming’s left hand rule :

a.

Example 19.9 :

B I

XX XX XX XX

XX XX XX XX

XX XX XX XX B

I

XX XX XX XX

XX XX XX XX

XX XX XX XX

B I

XX XX XX XX

XX XX XX XX

XX XX XX XX

F


b.

B I

XX XX XX XX

XX XX XX XX

XX XX XX XX

F

(to the right)(to the right)




1919

68

A wire of 100 cm long is placed perpendicular to the magnetic field

of 1.20 Wb m−2

.a. Calculate the magnitude of the force on the wire when a current

of 15 A is flowing.

b. For the same current in (a), determine the magnitude of the force

on the wire when its length is extended to 150 cm.

c. If the force on the wire in part (b) is 60× 10−2 N and the current

flows is 12 A, calculate the magnitude of magnetic field was

supplied.


a. Given

Example 19.10 :

90;mWb20.1;m00.1 2 === − θ Bl

A15= I θ IlB F sin=

( )( )( ) 90sin20.100.115= N18= F




1919

69


b. Given

The magnitude of the magnetic force on the wire is given by

c. Given

The magnitude of the magnetic field is given by

m501A;15 .l I ==

θ IlB F sin=( )( )( ) 90sin20.150.115=

N27= F

N1060m;501A;12 2−×=== F .l I

θ IlB F sin=

( )( )

90sin50.1121060

2 B

=×

−

T1033.3 2−×= B




1919

70

A straight horizontal rod of mass 50 g and length 0.5 m is placed in

a uniform magnetic field of 0.2 T perpendicular to the rod. The forceacting on the rod just balances the rod’s weight.

a. Sketch a labelled diagram shows the directions of the current,

magnetic field, weight and force.

b. Calculate the current in the rod.

(Given g = 9.81 m s−2)


a.

b. Since the magnetic force acting on the rod just balances the

rod’s

weight, therefore

Example 19.11 :

90;T2.0;m5.0g;1050 3 ===×= − θ Bl m

θ IlB F sin=θ IlBmg sin=

A91.4= I

( ) ( )( ) 90sin2.05.081.91050

3 I =× −

I

F

g m

B




1919

71


DeriveDerive force per unit length of two parallel current-force per unit length of two parallel current-

carrying conductors.carrying conductors.

UseUse force per unit length:force per unit length:

DefineDefine one ampere.one ampere.

Learning Outcome:

19.5 Forces between two parallel current-

carrying conductors (1 hour)

πd

I I μ

l

F

2

210=

w w w

.k m p

h . m

a t ri k

. e d

u . m

y

w w w

.k m p

h . m

a t ri k

. e d

u . m

y

PhysicsPhysicsap erap er191919 5 Forces between two parallel current



1919

72

19.5.1 Force per unit length Consider two identical straight conductors 1 and 2 carrying

currents I 1and I

2with length l are placed parallel to each other

as shown in Figure 19.29.

19.5 Forces between two parallel current-

carrying conductors


d

1 21 I

1 I

2 I

2 I

P1 B

12 F

21 F

Q

2 B




1919

73

The conductors are in vacuum and their separation is d .

The magnitude of the magnetic flux density, B1 at point P on the

conductor 2 due to the current in the conductor 1 is given by

Conductor 2 carries a current I 2 and in the magnetic field B1

thus the conductor 2 will experiences a magnetic force, F 12 . The magnitude of F 12 is given by

d

I B

π

µ

2

101 = Direction : into the page

θ sin1212 lB I F = 90=θ and

90sin2

10

2

= d

I

l I π

µ

d

l I I F

π

µ

2

21012 =




1919

74

The magnitude of F 21 is given by

Conclusion :

and the type of the force is attractivetype of the force is attractive.

From the equation (19.12), thus the force per unit length is given

by

θ sin2121 lB I F = 90=θ and

90sin2

201

=

d

I l I

π

µ

d

l I I F

π

µ

2

21021 =

d

l I I F F F

π

µ

2

2102112 ===

(19.12)(19.12)

πd

I I μ

l

F

2

210= (19.13)(19.13)




1919

75

If the direction of current in the conductor 2 is change to upside

down as shown in Figure 19.30.

The magnitude of F 12 and F 21 can be determined by using the

eq. (19.12) and their direction can be determined by applyingFleming’s left hand rule.

Conclusion : Type of the force is repulsiveType of the force is repulsive.


2 I

2 I

1 I

1 I d

1 2Note:Note:

The currents are in the

same directionsame direction – 2

conductors attractattract

each other.

The currents are inopposite directionopposite direction – 2

conductors repelrepel each

other.21 F Q 2

B

12 F 1 B

P




1919

76

Two long straight parallel wires are placed 0.25 m apart in a

vacuum. Each wire carries a current of 2.4 A in the same direction.

a. Sketch a labelled diagram to show clearly the direction of the

force on each wire.

b. Calculate the force per unit length between the wires.

c. If the current in one of the wires is reduced to 0.64 A, calculate

the current needed in the second wire to maintain the same force

per unit length between the wires as in (b).

(Given µ 0 = 4π × 10−7 T m A−1)


a. The diagram is

Example 19.12 :

12 F

21 F

d

1

1 I

2

2 I

m250A;4.221 .d I I ===




199

77


b. The force per unit length between the wires is given by

c. Given

Therefore the current needed in the second wire is

m250A;4.221 .d I I ===

πd I I μ

l F

2

210= ( )( )( )

( )25.024.24.2104 7

π π

l F

−

×=

16 m N106.4 −−×=

l

F

A64.01 = I

πd

I I μ

l

F

2

210=

( )( )( )25.02

64.0104106.4 2

76

π

I π −

− ×=×

A98.82 = I




19

78

From the eq. (19.13), if two long straight parallel conductors are

placed 1.0 m apart in a vacuum and carry equal currents of

1.0 A thus the force per unit length that each conductor exerts

on each other is given by

The ampereThe ampere is defined as the constant current, whichthe constant current, whichflowing in each of two infinitely long parallel straightflowing in each of two infinitely long parallel straight

conductors of negligible of cross sectional area separatedconductors of negligible of cross sectional area separated

by a distance of 1.0 metre in vacuum, would produce aby a distance of 1.0 metre in vacuum, would produce a

force per unit length between the conductors of force per unit length between the conductors of

2.02.0× 1010−77

N mN m− 11..

19.5.2 The ampere

πd

I I μ

l

F

2

210=

( )( ) ( )( )12

11104 7

π

π −×

=

17 m N100.2 −−×=

l

F




19

79

Exercise 19.4 :Given µ 0 = 4π × 10−7 T m A−1

1. A vertical straight conductor Y of length 0.5 m is situated in auniform horizontal magnetic field of 0.1 T.

a. Sketch a labelled diagram to show the directions of the

current, field and force.

b. Calculate the force on Y when a current of 4 A is passed

into it.c. Through what angle must Y be turned in a vertical plane

so that the force on Y is halved?

(Advanced level physics, 7(Advanced level physics, 7thth edition, Nelkon&Parker, Q6, p.336)edition, Nelkon&Parker, Q6, p.336)

ANS. :ANS. : 0.2 N; 600.2 N; 60°°

2. A current-carrying conductor experiences no magnetic force

when it is placed in a uniform magnetic field. Explain the

statement.




80


UseUse torque:torque:

wherewhere N N = number of turns= number of turns

ExplainExplain the working principles of a moving coilthe working principles of a moving coil

galvanometer.galvanometer.

Learning Outcome:

19.6 Torque on a coil (1 hour)

( ) B A NI τ

×=

w w w

.k m p

h . m

a t ri k

. e d

u . m

y

w w w

.k m p

h . m

a t ri k

. e d

u . m

y




81

19.6.1 Formula of torque

Consider a rectangular coil (loop) of wire with side lengths a and

b that it can turn about axis PQ. The coil is in a magnetic field of

flux density B and the plane of the coil makes an angle θ withthe direction of the magnetic field. A current I is flowing round

the coil as shown in Figure 19.31a.

19.6 Torque on a coil




82

Q

P

b

a


θ

B B

B

B

B F

F

1 F

I I

I

I 1 F

A

φ




83

From the Figure 19.31b, the magnitude of the force F 1 is givenby

From the Figure 19.31a, the forces F lie along the axis PQ.

φsin2

b

φsin2

b

θ B

B

B

1 F

1 F

A

φ

Q

φ

φ

2b

rotationrotationrotationrotation

Figure 19.31b: side viewFigure 19.31b: side view

90sin1 IlB F = al =and

IaB F =1




84

From the Figure 19.31a, the forces F lie along the axis PQ.

The resultant forceresultant force on the coil is zerozero but the nett torquenett torque is notnot

zerozero because the forces F 1 are perpendicular to the axis PQ asshown in Figure 19.31a.

The forces F 1 cause the coil to rotatecoil to rotate in the clockwise directionclockwise direction

about the axis PQabout the axis PQ as shown in Figure 19.31b.

The magnitude of the nett torque about the axis PQ (refer toFigure 19.31b) is given by

−

−= φ φ sin

2sin

211

b F

b F τ

IaB F =1

−=φ sin

22

1

b F and

( )

−= φ sin

22

b IaB

φ sin IabB−= coil)of area( Aab =and




85

since thus

For a coil of N turns, the magnitude of the torque is given by

φ sin IABτ =θ −=φ 90

( )θ IAB −=τ

90sin

θ IABcos=τ

φ sin NIABτ =OR

θ NIABτ cos=

(19.14)(19.14)

(19.15)(19.15)

where coilon thetorque:τ densityfluxmagnetic: Bcoilin theflowscurrent: I

B A

andareactor between veangle:φ Bθ

andcoiltheof planee between thangle:

(coils)turnsof number : N




86

From the eq. (19.14), thus the formula of torque in the vector

form is given by

The torque is zerotorque is zero when θ θ = 90= 90°° or or φφ = 0= 0°°and is maximummaximum

when θ θ = 0= 0°° or or φφ = 90= 90°°as shown in Figures 19.32a and

19.32b.

( ) B A NI τ

×= (19.16)(19.16)

0=φ

90=θ B

A

0sin NIABτ =

90cos NIABτ =OR

0=τ


B

A

0=θ

90=φ

Figure 19.32bFigure 19.32b90sin NIABτ =

0cos NIABτ =OR

NIABτ =max

plane of the coil




87

In a radial fieldradial field, the plane of the coilplane of the coil is always parallelalways parallel to the

magnetic fieldmagnetic field for any orientation of the coil about the vertical

axis as shown in Figure 19.33.

Hence the torquetorque on the coil in a radial fieldradial field is always constantconstantand maximummaximum given by

Radial field is used in moving coil galvanometer.

0=θ

90=φ ORSSNN

coilfixed soft

iron cylinder

radial field

Figure 19.33: Plan view of moving coil meter Figure 19.33: Plan view of moving coil meter

90sin NIABτ = 0cos NIABτ =OR

NIABτ = maximummaximum




88

A 50 turns rectangular coil with sides 10 cm × 20 cm is placed

vertically in a uniform horizontal magnetic field of magnitude 2.5 T.

If the current flows in the coil is 7.3 A, determine the torque actingon the coil when the plane of the coil is

a. perpendicular to the field,

b. parallel to the field,

c. at 75° to the field.Solution :Solution :

The area of the coil is given by

a.

Example 19.13 :

A7.3T;5.2turns;50 === I B N

( )( ) 2222 m100.210201010 −−− ×=××= A

From the figure, θ = 90°and φ = 0° , thusthe torque on the coil is

φ sin NIABτ =θ NIABτ cos= OR

B

A

90=θ

90cos NIAB=

0sin NIAB=

0=τ




89


b.

c.

B

A

90=φ

From the figure, θ = 0°and φ = 90° ,

thus the torque on the coil isθ NIABτ cos=

( ) ( ) ( )( ) 0cos5.2100.23.750 2−×=m N3.18=τ

A7.3T;5.2turns;50 === I B N

B

A

15=φ

75=θ

From the figure, θ = 75°and φ =15°,thus the torque on the coil is

θ NIABτ cos=

( ) ( ) ( )( )

75cos5.2100.23.750

2−

×= m N72.4=τ




90

A galvanometer consists of a coil of wire suspended in the

magnetic field of a permanent magnet. The coil is rectangular

shape and consists of many turns of fine wire as shown in

Figure 19.34.

19.6.2 Moving-coil galvanometer





91

When the current I I flows through the coilflows through the coil, the magnetic fieldmagnetic field

exerts a torque on the coilexerts a torque on the coil as given by

This torque is opposed by a spring which exerts a torque, τ s given by

The coil and pointer will rotatecoil and pointer will rotate only to the point where the springspring

torque balances the torque due to magnetic fieldtorque balances the torque due to magnetic field, thus

NIABτ =

kθ τ s =

where constanttorsional:k radianincoiltheof anglerotation:θ

sτ τ =kθ NIAB =

NAB

kθ I =

(19.17)(19.17)

(19.18)(19.18)




92

A rectangular coil of 10 cm × 4.0 cm in a galvanometer has 50

turns and a magnetic flux density of 5.0 × 10−2 T. The resistance of

the coil is 40 Ω and a potential difference of 12 V is applied acrossthe galvanometer, calculate the maximum torque on the coil.


The area of the coil is given by

The current through the galvanometer is

Therefore the maximum torque on the coil is

Example 19.14 :

( )( ) 2322 m100.4100.41010 −−− ×=××= A

( )4012 I = IRV =

A3.0= I

NIAB=maxτ

m N100.33

max

−

×=τ

;04T;100.5turns;502 Ω=×== −

R B N V12=V

( ) ( ) ( )( )23100.5100.43.050

−− ××=




93

Exercise 19.5 :1. A moving coil meter has a 50 turns coil measuring 1.0 cm by

2.0 cm. It is held in a radial magnetic field of flux density

0.15 T and its suspension has a torsional constant of 3.0× 10−6 N m rad−1. Determine the current is required to give a

deflection of 0.5 rad.

ANS. :ANS. : 1.01.0×× 1010−−33 AA




94


ExplainExplain the motion of a charged particle in boththe motion of a charged particle in both

magnetic field and electric field.magnetic field and electric field.

Derive and useDerive and use velocity,velocity,

in a velocity selector.in a velocity selector.

Learning Outcome:

19.7 Motion of charged particle in magnetic field

and electric field (1 hour)

B

E v =

w w w

.k m p

h . m

a t r

i k . e

d u

. m y

w w w

.k m p

h . m

a t r

i k . e

d u

. m y




95

Consider a positively charged particle with mass m, charge q

and velocity v enters a region of space where the electric and

magnetic fields are perpendicular to the particle’s velocity and toeach other as shown in Figure 19.38.

19.7 Motion of charged particle in

magnetic field and electric field


E

++++++++++++++++++

−−−−−−−−−−−−−−−−−−

XX XX XX XX XX XX

XX XX XX XX XX XX

XX XX XX XX XX XX

XX XX XX XX XX XX

B

v

+ v

+ v

+

B F

E F




96

The charged particle will experiences the electric force F E is

downwards with magnitude qE and the magnetic force F B is

upwards with magnitude qvB as shown in Figure 19.38.

If the particle travels in a straight line with a constant velocity

hence the electric and magnetic forces are equal in magnitudeelectric and magnetic forces are equal in magnitude.

Therefore

Only the particles with velocity equal tovelocity equal to E/B E/B can pass throughcan pass through

without being deflected by the fieldswithout being deflected by the fields.

Eq. (19.21) also works for electron or other negatively charged

particles.

E B F F =qE qvB =90sin

B

E v = (19.21)(19.21)




97

Figure 19.38 known as velocity selector velocity selector .

Normally, after the charged particle passing through the velocityselector it will enter the next region consist of a uniformuniform

magnetic field onlymagnetic field only. This apparatus known as massmass

spectrometer spectrometer as shown in Figure 19.39.


E

++++++++++++++++++

−−−−−−−−−−−−−−−−−−

XX XX XX XX XX XX

XX XX XX XX XX XX

XX XX XX XX XX XX

XX XX XX XX XX XX

B

v

+

E F

XX XX XX XX XX

XX XX XX XX XX

XX XX XX XX XX

XX XX XX XX XX

XX XX XX XX XX

XX XX XX XX XX

XX XX XX XX XX

XX XX XX XX XX

XX XX XX XX XX

v

+

XX XX XX XX XX

B F

v

B F

r

+

v

+




98

When the charged particle entering the region consist of charged particle entering the region consist of

magnetic field onlymagnetic field only, the particle will make a semicircular pathparticle will make a semicircular path of

radius r as shown in Figure 19.39.Therefore

From the eq. (19.22), the mass spectrometer can be used to

determine the value of determine the value of q/ q/ mm for any charged particle.

C B F F =

rB

v

m

q

= and

r

mvqvB

2

=

B

E v =

2rB

E

m

q= (19.22)(19.22)




99

An electron with kinetic energy of 8.0× 10−16 J passes perpendicular

through a uniform magnetic field of 0.40× 10−3

T. It is found to followa circular path. Calculate

a. the radius of the circular path.

b. the time required for the electron to complete one revolution.

(Given e/m = 1.76× 1011 C kg-1, me = 9.11× 10−31 kg)


a. The speed of the electron is given by

Example 19.15 :

T1040.0J;100.8 316 −− ×=×= B K

2

2

1mv K =

( ) 23116 1011.921100.8 v−− ×=×

17 sm1019.4 −×=v


316



100


a. Since the path made by the electron is circular, thus

b. The time required for the electron to complete one revolution is

given by

T1040.0J;100.8316 −− ×=×= B K

C BF F

=r

mvevB

2

90sin =

r

v B

m

e=

( )( ) r

7

311 1019.41040.01076.1 ×=×× −

m595.0=r

T

πr

v

2

=( )T

π 595.021019.4 7 =×

s1092.8 8−×=T


E i 19 6



101

Exercise 19.6 :1. An electron moving at a steady speed of 0.50× 106 m s−1

passes between two flat, parallel metal plates 2.0 cm apart

with a potential difference of 100 V between them. Theelectron is kept travelling in a straight line perpendicular to the

electric field between the plates by applying a magnetic field

perpendicular to the electron’s path and to the electric field.

Calculate :

a. the intensity of the electric field.

b. the magnetic flux density needed.

ANS. :ANS. : 0.500.50×× 101044 V mV m−−11; 0.010 T; 0.010 T

2. A proton moving in a circular path perpendicular to a constant

magnetic field takes 1.00 µ s to complete one revolution.

Determine the magnitude of the magnetic field.

(Physics for scientist and engineers, 6(Physics for scientist and engineers, 6thth edition, Serway&Jewet, Q32,edition, Serway&Jewet, Q32,

p.921)p.921)

(m p=1.67× 10−27 kg and charge of the proton, q=1.60× 10−19 C)

ANS. :ANS. : 6.566.56×× 1010−−22 TT




Next Chapter…CHAPTER 20 :

Electromagnetic induction

w w w

.k m p

h . m

a t r

i k . e

d u

. m y

w w w

.k m p

h . m

a t ri

k . e

d u

. m y

4 magnetic field

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