chapter19 student

8/10/2019 Chapter19 Student

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PHYSICS CHAPTER 19

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CHAPTER 19:Magnetic field


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PHYSICS CHAPTER 19

Sub topics

19.1 Magnetic filed19.2 Magnetic field produced by current-carryingconductor

19.3 Force on a moving charged particle in auniform magnetic field19.4 force on a current carrying conductor in auniform magnetic field

19.5 Force between two parallel current-carryingconductors19.6 Torque on a coil19.7 Motion of charged particle in magnetic filed

and electric field 2


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PHYSICS CHAPTER 19

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

Define magnetic field.

Identify magnetic field sources.Sketch the magnetic field lines.

Learning Outcome:

19.1 Magnetic field


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PHYSICS CHAPTER 19

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is defined as a region around a magnet where a magneticforce can be experienced .

A stationary electric charge is surrounded by an electricfield only .

When an electric charge moves , it is surrounded by anelectric field and a magnetic field . The motion of the electriccharge produces the magnetic field .Magnetic field has two poles, called north (N) and south (S) .This magnetic poles are always found in pairs whereas asingle magnetic pole has never been found.

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

19.1 Magnetic field


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Magnetic field lines are used to represent a magnetic field.

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

Magnetic field lines can be represented by straight lines orcurves. The tangent to a curved field line at a pointindicates the direction of the magnetic field at that point as

shown in Figure 6.1.

Magnetic field can be represented by crosses or by dotted

circles as shown in Figures 19.2a and 19.2b.

22.1.1 Magnetic field lines

Figure 19.1

direction of magnetic fieldat point P.

P

Figure 19.2a : magnetic field linesenter the page perpendicularly

X X X X

X X X X

Figure 19.2b : magnetic field linesleave the page perpendicularly


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

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

A non-uniform field is represented by non-parallel lines. Thenumber of magnetic field lines varies at different unit cross-sections as shown in Figure 19.4.

Figure 19.3

unit cross-sectional area


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The number of lines per unit cross-sectional area isproportional to the magnitude of the magnetic field .Magnetic field lines do not intersect one another.

Figure 19.4

stronger field in A1

A1 A2weaker field in A2


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The pattern of the magnetic field lines can be determined by

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

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 barmagnetic.

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 19.7a

a. Bar magnet

b. Horseshoe or U magnet

Figure 19.7b


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d. Two bar magnets ( like poles ) - repulsive

Neutral point (point wherethe resultant magneticforce is zero ).

c. Two bar magnets ( unlike pole ) - attractive

Figure 19.7c

Figure 19.7d


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PHYSICS CHAPTER 19

Southmagnetic pole

Southgeographical pole

Northmagnetic pole

11.5

11

The Earths magnetic field is like that of a giant bar magnet asillustrated in Figure 19.8 with a pole near each geographic poleof the Earth.

19.1.3 Earths magnetic field

Figure 19.8

The magnetic poles are tilted awayfrom the rotational axis by an angle of11.5 .Since the north pole of a compassneedle (Figure 19.8) points toward thesouth magnetic pole of the Earth, andsince opposite attract, it follows that

Figure 19.8 also shows that the fieldlines are essentially horizontal(parallel to the Earths surface) nearthe equator but enter or leave theEarth vertically near the poles .

the north geographical pole of theEarth is actually near the southpole of the Earths magnetic field .


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Using the permanent magnet

One permanent magnet A permanent magnet is bring near to the soft iron andtouching the surface of the soft iron by following the path inthe Figure 19.9.

This method is called induced magnetization .The arrows 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 domains ( thetiny magnetized region because of spin magneticmoment of the electron ).

19.1.4 Magnetization of a Soft Iron

Figure 19.9 N S


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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 southpole 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.

N N SSFigure 19.10


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Figure 19.11

N S

Switch, S

Using 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 powersupply as shown in Figure 19.11.

When the switch S is closed, the current I flows in the solenoidand produces magnetic field.The directions of the fields associated with the solenoid can befound by viewing the current flows in the solenoid from both

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

I I

SN

Current -anticlockwise

Current - clockwise


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Other examples:

If you drop a permanent magnet on the floor or strike it with ahammer, you may jar the domains into randomness . Themagnet can thus lose some or all of its magnetism.Heating a magnet too can cause a loss of magnetism.The permanent magnet also can be demagnetized by placing itinside a solenoid that connected to an alternating source .

NSI I

I I S N

Thumb north pole

Other fingers direction of currentin solenoid .

Note:

Figure 19.12a Figure 19.12b


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is defined as the magnetic flux per unit area across anarea at right angles to the magnetic field .Mathematically,

It also known as magnetic induction (magnetic field intensity

OR strength )It is a vector quantity and its direction follows the direction ofthe magnetic field .Its unit is tesla (T) OR weber per metre squared (Wb m 2).Unit conversion :

19.1.5 Magnetic flux density, B

wherefluxmagnetic:

fieldmagneticthetoanglesrightatarea: A

)G(gauss10mWb1T1 42

(19.1)


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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 meansthat magnetic field lines must leave from the north pole of apermanent bar magnet and enter its south pole.Solution :

Example 19.1 :

Near the north pole of a permanent bar magnet, the northpole of a compass will point away from the bar magnet sothe field lines leave the north pole.Near the south pole of a permanent bar magnet, the northpole of a compass will point toward the bar magnet so the

field lines 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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Apply magnetic field formula

for a long straight wire,

for a circular coil,

for a solenoid.

Learning Outcome:

19.2 Magnetic produced by current-carryingconductor ( 1 hour)


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When a current flows in a conductor wire or coil , themagnetic field will be produced .The direction of magnetic field around the wire or coil can be

determined by using the right hand grip rule as shownin Figure 19.13.

19.2 Magnetic field produced by current carrying conductor

Figure 19.13

Thumb direction of currentOther fingers direction of magnetic

field (clockwise ORanticlockwise )

Note:


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The magnetic field lines pattern around a straight conductorcarrying 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 pageFigure 19.14

B I

I B

B


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Consider a straight conductor (wire) carrying a current I isplaced in vacuum as shown in Figure 19.16.

OR

Figure 19.15

I

I

I X

Current into the pageX

B

B

B

B


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The magnitude of magnetic flux density (magnetic fieldintensity), B at point P at distance r from the wire carrying

current is given by

r P

I Figure 19.16

X B

into the page (paper)

where spacefreeof ty permeabili:0 17 AmT104

(wire)conductorstraightafrom pointaof distance:r

(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

Figure 19.17

I I X

S

NORS

N

I I

I


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Consider a circular shaped conductor with radius R that carriesa current I as shown in Figure 19.18.

where

(6.3)

The magnitude of magnetic fieldintensity B at point O ( centre ofthe circular coil or loop ) , isgiven by

RO

coilcirculartheof radius: R(loops)coilsof number: N

spacefreeof ty permeabili:0

current: I Figure 19.18


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

been wound into a succession of closely spaced loops withgeometry of a helix .The magnetic field lines pattern around a solenoid carryingcurrent is shown in Figure 19.19.

19.2.3 Magnetic field of a solenoid

SN

Figure 19.19

I I


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OR

I

I X X X X

I

I

I

I

I

I

SN


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The magnitude of magnetic field intensity at the end of N turn solenoid is given by

(19.5)

where lengthunit perturnsof number:n

The magnitude of magnetic field intensity at the centre(mid-point/ inside) of N turn solenoid is given by

and

(19.4)


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Two long straight wires are placed parallel to each other and

carrying the same current I . Sketch the magnetic field lines patternaround both wiresa. when the currents are in the same direction.b. when the currents are in opposite direction.Solution :a.

Example 19.2 :

I

I I

I


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I I

OR

Solution :a.


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OR

Solution :b.

I I X

I

I

I

I


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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 pointsomewhere between X and Y at which the magnetic flux densityis zero. How far from X is this point ?

(Given 0 = 4 10 7 H m 1)

Example 19.3 :


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Solution :a. i.

By using the equation of magnetic field at any point near thestraight wire, then at point A

Magnitude of BX :

A10m;100.5A;50 Y2

X I d I

T100.4 4X B

X B

Y B

OR

m105.22

2YX

d r r

X I A

Xr Yr Y I

X

X0X 2r

I B

Direction : into the page OR upwards

2

7

X105.22

50104

B

X B

Y BX I Y I

d

Xr Yr A


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Solution :

a. i. Magnitude of BY :

Therefore the total magnetic flux density at point A is

A10m;100.5A;50 Y2

X I d I

T100.8 5Y BY

Y0Y 2r

I B

Direction : out of page OR downwards

2

7

Y105.22

10104

B

YXA B B B

YXA B B B

Sign convention of B :Out of the page positive ( +)

Into the page negative ( )

Note:

54A 100.8100.4

B


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Solution :a. ii.

By using the equation of magnetic field at any point near thestraight wire, then at point A

Magnitude of BX :

A10m;100.5A;50 Y2

X I d I

T100.4 4X B

X BY B

OR

Direction : into the page OR

upwards

2

7

X105.22

50104

B

X I A

Xr Yr Y I

X

X BY B

X I Y I

d

Xr Yr A


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Solution :

a. ii. Magnitude of BY :

Therefore the resultant magnetic flux density at point A is

A10m;100.5A;50 Y2

X I d I

T100.8 5Y BDirection : into the page OR upwards

2

7

Y105.22

10104

B

YXA B B B

YXA B B B 54A 100.8100.4 B


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Solution :b.

Since the resultant magnetic flux density at point C is zerothus

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 BYX0 B B

YX B B whereX

X0X 2r

I B and

Y

Y0Y 2r

I B

X I Y I

d

Xr Yr C

X B

Y B


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Solution :

b.

Y

Y0

X

X0

22 r I

r I

r d I

r I YX

r r 2100.51050

A10m;100.5A;50 Y2

X I d I


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Two long straight wires are oriented perpendicular to the page asshown in Figure 6.20.

The current in one wire is I 1 = 3.0 A pointing into the page and thecurrent in the other wire is I 2= 4.0 A pointing out of page. Determinethe magnitude and direction of the nett magnetic field intensity atpoint P.

(Given 0 = 4 10 7 H m 1)

Example 19.4 :

Figure 6.20


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Solution :

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

m100.5A;0.4A;0.3 2121 r I I

2

22

22 100.5100.5 r

1r 2 B 2

r

m101.7 22r

2

2

2

1

101.7

100.5cos

r

r

704.0cos 704.0

101.7100.5

sin 22

1

101 2r

I B

T1020.1 51 B

2

7

1100.52

0.3104

B

1 B

1 I 2 I

P

Xm100.5 2


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Solution :and

m100.5A;0.4A;0.3 2121 r I I

2202 2r

I B

T1013.1 52 B

2

7

2101.72

0.4104

B

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

Vectorsum

51 1020.1

B1 B 0

B cos22 B 704.01013.1 5

61096.7

B sin2 704.01013.1 5

61096.7 65 1096.71020.1 x B

6104.04

61096.70 y B610.967


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Solution :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

2626 1096.71004.4

x

y

B

B 1tan

6

61

1004.41096.7tan

OR

1.63 B

1 B

2 B

P

PHYSICS CHAPTER 19


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a. A closely wound circular coil of diameter 10 cm has 500 turnsand carries a current of 2.5 A. Determine the magnitude of themagnetic field at the centre of the coil.

b. A solenoid of length 1.5 m and 2.6 cm in diameter carries acurrent of 18 A. The magnetic field inside the solenoid is

2.3 mT. Calculate the length of the wire forming the solenoid.(Given 0 = 4 10 7 T m A 1)Solution :a. Given

By applying the equation for magnitude of the magnetic field atthe centre of the circular coil, thus

Example 19.5 :

A5.2;500m;105.021010 2

2 I N R

R

NI B

20

27

100.525.2500104

B

PHYSICS CHAPTER 19


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Solution :b. Given

By applying the equation of magnetic flux density inside thesolenoid, thus

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

Therefore the length of the wire forming the solenoid is

l

NI Bi

0

turns153 N

5.1

18104103.2

73 N

T103.2m;103.12

106.2m;5.1 3i

22

Br l A18 I

r 2ncecircumfere 2103.12ncecircumfere m1017.8ncecircumfere

2

ncecircumfere N L 2108.17153 L

PHYSICS CHAPTER 19


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Exercise 19.2 :Given 0 = 4 10 7 T m A 1

1.

The two wires shown in Figure 6.21 carry currents of 5.00 A inopposite 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 P 1 and

P 2.

ANS. : 1.33 10 5 T, out of page; 2.67 10 6 T, out of page

Figure 19.21A00.5 A00.5

cm0.10

cm0.15cm0.52P

1P

PHYSICS CHAPTER 19


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Exercise 19.2 :2.

Four long, parallel power wires each carry 100 A current. Across sectional diagram for this wires is a square, 20.0 cm oneach 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 thesquare.

ANS. : 4.0 10 4 T , to the left (180 )

Figure 19.22

X X

PHYSICS CHAPTER 19


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Use formulae:

Describe circular motion of a charge in a uniformmagnetic field.

Use relationship F B=

F C.

Learning Outcome:19.3 Force on a moving charged particle in a

uniform magnetic field

PHYSICS CHAPTER 19


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19.3.1 Magnetic force A stationary electric charge in a magnetic field will notexperience a magnetic force . But if the charge is moving witha velocity, v in a magnetic field, B then it will experience amagnetic force .The magnitude of the magnetic force can be calculated byusing the following equation:

19.3 Force on a moving charged particle

in a uniform magnetic field

where forcemagnetic: F densityfluxmagnetic: B

chargeaof velocity:vchargetheof magnitude:q

Bv and betweenangle:

(19.6)

PHYSICS CHAPTER 19


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PHYSICS CHAPTER 19

B

v

F B

v

F

49

Figure 6.23

In vector form ,

The direction of the magnetic force can be determined by using theFlemings hand rule.

Flemings right hand rule for negative charge

Flemings left hand rule for positive charge

(19.7)

shown inFigures 6.23

and 6.24

Thumb direction of Force

First finger direction of Field

Second finger direction of Velocity

Figure 6.24Note:

PHYSICS CHAPTER 19


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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

X X X X

X X X X

X X X X

v

I

I

v

B v

PHYSICS CHAPTER 19


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Solution :a. By using Flemings left hand rule, thus

b. By using Flemings right hand rule, thus

c. By using Flemings right hand rule, thus

B

v(into the page) F

B v F (to the left)

B

v

X X X X

X X X X

X X X X

F (to the left)

PHYSICS CHAPTER 19


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Solution :

d.

e. I

v

BX X X X

X X X F (to the left)

Using right hand grip rule to determine the direction of magneticfield produces by the current I on the charge position. Thenapply the Flemings right hand rule, thus

Using right hand grip rule to determine the direction of magneticfield forms by the current I on the charge position. Then applythe Flemings left hand rule, thus

v

I

(upwards)

BXX

X

X

X

X

X X

F

PHYSICS CHAPTER 19


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Calculate the magnitude of the force on a proton travelling

5.0 107

m s1

in the uniform magnetic flux density of 1.5 Wb m2

, 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)

Solution :a. Given

Therefore

b. GivenHence

Example 19.7 :

90 217 mWb5.1;sm105.0 Bv

qvB F sin 90sin5.1100.51060.1 719

50 50sin5.1100.51060.1 719 F

PHYSICS CHAPTER 19


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Consider a charged particle moving in a uniform magnetic fieldwith its velocity perpendicular to the magnetic field . As the particle enters the region, it will experiences a magneticforce which the force is perpendicular 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 acircular 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

Figure 19.25a

v v B F

v

B F

X X X X

X X X X

X X X X

X X X X

v v

B F

v

B F

Figure 19.25b

PHYSICS CHAPTER 19


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Since the path is a circle therefore the magnetic force F Bcontributes the centripetal force F c (nett force) in this motion.

Thus

Figure 19.25c Figure 19.25d

v v B F

v

B F

v v

B F

v

B F

X X X X

X X X X

X X X X

X X X X

c B F F

r mv

qvB2

sin 90 and

PHYSICS CHAPTER 19


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The period of the circular motion,T

makes by the particle isgiven by

And the frequency of the circular motion is

where particlechargedtheof mass:m velocitytheof magnitude:v

pathcirculartheof radius:r particlechargedtheof magnitude:q

(19.8)

r v

T

r v

2T

2and

and

v

r T

2

(19.9)

T f

1

PHYSICS CHAPTER 19


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An electron at point A in Figure 6.26 has a speed v of 2.50 10 6m s -1. Determinea. 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

Figure 19.26

PHYSICS CHAPTER 19


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Solution :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 :OR

m100.20;sm1050.2 216 d v

Be

mvr

2d

r

Be

mvd

2

19

6312

1060.1

1050.21011.9

2

100.20

B

and

v

BA

B

F

PHYSICS CHAPTER 19


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Solution :b. The period of the electron is

Since the path is the semicircular then the time required for theelectron moves from A to B is given by

m100.20;sm1050.2 216 d v

r v T 2

T

d v

22

T

26 100.201050.2

and

T t 21

71051.221 t

PHYSICS CHAPTER 19


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

a. b.

ANS. : positive; positive2. Determine the direction of the magnetic force exerted on a

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

ANS. : into the page; out of page

Bv

F B

v F

Switch, S

v

Switch, S

v

PHYSICS CHAPTER 19


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Exercise 19.3 :3. An electron experiences the greatest force as it travels

2.9 10 6 m s 1 in a magnetic field when it is moving north. Theforce is upward and of magnitude 7.2 10 13 N. Determine themagnitude and direction of the magnetic field.(Given the charge of the electron is 1.60 10 19 C)(Physics for scientists & engineers ,3 rd edition, Giancoli, Q22,p.705)

ANS. : 1.6 T to the east4. An electron moving with a speed of 9.1 10 5 m s 1 in the

positive x direction experiences zero magnetic force. When itmoves in the positive y direction, it experiences a force of2.0 10 13 N that points in the negative z direction. What is thedirection and magnitude of the magnetic field?

(Given e=1.60 10 19 C and me= 9.11 10 31 kg)(Physics, 3 rd edition, James S. Walker, Q8, p.762)

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

PHYSICS CHAPTER 19


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Exercise 19.3 :5. Two charged particles with different speeds move one at a

time through a region of uniform magnetic field. The particlesmove in the same direction and experience equal magneticforces.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 rd edition, James S. Walker, Q9, p.762)

ANS. : 1/46. A 12.5 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 rd edition, James S. Walker, Q18, p.763)

ANS. : 12.1 m s 1; 13.9 s

PHYSICS CHAPTER 19


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Use force:

Learning Outcome:19.4 Force on a current-carrying conductor in a

uniform magnetic field

PHYSICS CHAPTER 19


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When a current-carrying conductor is placed in a magneticfield B, thus a magnetic force will acts on that conductor .The magnitude of the magnetic force exerts on the current-carrying conductor is given by

In vector form ,

19.4 Force on a current-carrying

conductor in a uniform magnetic field

(19.10)

(19.11)

where forcemagnetic: F densityfluxmagnetictheof magnitude: B

current: I conductor theof length:l

B I andof direction betweenangle:

PHYSICS CHAPTER 19


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PHYSICS CHAPTER 19

B

I

F

65

The direction of the magnetic force can be determined by using

the Flemings left hand rule as shown in Figure 19.27.

From the equation (19.10),

the magnetic force on the conductor has its maximumvalue when the conductor (and therefore the current) andthe magnetic field are perpendicular (at rightangles) to each other then = 90 (shown in Figure 19.28a).

Thumb direction of Force

First finger direction of Magnetic Field

Second finger direction of Current

Figure 19.27

Note:

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the magnetic force on the conductor is zero when theconductor (and therefore the current) is parallel to themagnetic field then =0 (shown in Figure 19.28b).

Figure 19.28a

IlB F max B

90 I

90sinmax IlB F

B0

I

Figure 19.28b

0 F

0sin IlB F

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1 Tesla is defined as the magnetic flux density of a field in whicha force of 1 Newton acts on a 1 metre length of a conductorwhich carrying a current of 1 Ampere and is perpendicular tothe field.

Note:

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

carrying conductor in the following cases.a. b.

Solution :For both cases, use Flemings left hand rule :a.

Example 19.9 :

B I

X X X X

X X X X

X X X X B I

X X X X

X X X X

X X X X

B I

X X X X

X X X X

X X X X

F (to the left)

b.

B I

X X X X

X X X X

X X X X

F (to the right)

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

of 1.20 Wb m2

.a. Calculate the magnitude of the force on the wire when a currentof 15 A is flowing.

b. For the same current in (a), determine the magnitude of the forceon 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 currentflows is 12 A, calculate the magnitude of magnetic field supplied.

Solution :a. Given

Example 19.10 :

90;mWb20.1;m00.1 2 Bl

A15 I IlB F sin

90sin20.100.115

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S CS C 9

70

Solution :b. Given

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

c. GivenThe magnitude of the magnetic field is given by

m501A;15 .l I

IlB F sin 90sin20.150.115

N1060m;501A;12 2 F .l I

IlB F sin

90sin50.11210602

B

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71

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 rods 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)Solution :a.

b. Since the magnetic force acting on the rod just balances the rodsweight, therefore

Example 19.11 :

90;T2.0;m5.0g;1050 3 Bl m

IlB F sin IlBmg sin 90sin2.05.081.91050 3 I

I F

g m

B

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72


Derive force per unit length of two parallel current-carrying conductors.

Use formulae:

Define one ampere.

Learning Outcome:22.5 Forces between two parallel current-

carrying conductors ( 1 hour)

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19.5.1 Force per unit lengthConsider two identical straight conductors 1 and 2 carryingcurrents I 1 and I 2 with length l are placed parallel to each otheras shown in Figure 19.29.

19.5 Forces between two parallel current-carrying conductors

Figure 19.29

d

1 21 I

1 I

2 I

2 I

P1 B

12 F

21 F Q

2 B

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The conductors are in vacuum and their separation is d .The magnitude of the magnetic flux density, B1 at point P on theconductor 2 due to the current in the conductor 1 is given by

Conductor 2 carries a current I 2 and in the magnetic field B1thus the conductor 2 will experiences a magnetic force,

F 12.

The magnitude of F 12 is given by

d I

B

2

101 Direction : into the page

sin1212 lB I F 90 and

90sin210

2

d I

l I

d l I I

F

2

21012

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The magnitude of F 21 is given by

Conclusion :

and the type of the force is attractive .From the equation (6.12), thus the force per unit length is givenby

sin2121 lB I F 90 and

90sin2

201

d I l I

d

l I I F

2

21021

d l I I

F F F

2

2102112

(6.12)

(19.13)

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76

If the direction of current in the conductor 2 is change to upsidedown as shown in Figure 6.30.

The magnitude of F 12 and F 21 can be determined by using theeq. (6.12) and their direction can be determined by applyingFlemings left hand rule.Conclusion : Type of the force is repulsive .

Figure 19.30

2 I

2 I

1 I

1 I d

12

Note:

The currents are in thesame direction 2conductors attracteach other.The currents are inopposite direction 2conductors repel eachother.

21 F Q 2

B

12 F 1 BP

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77

Two long straight parallel wires are placed 0.25 m apart in avacuum. 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 forceper unit length between the wires as in (b).

(Given 0 = 4 10 7 T m A 1)Solution :

a. The diagram is

Example 19.12 :

12 F 21 F

d 1

1 I

2

2 I

m250A;4.221 .d I I

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

c. GivenTherefore the current needed in the second wire is

m250A;4.221 .d I I

d I I

l

F

2

210 25.024.24.2104 7

l F

A64.01 I

d

I I l

F

2

210

25.02

64.0104106.4 2

76

I

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79

From the eq. (6.13), if two long straight parallel conductors areplaced 1.0 m apart in a vacuum and carry equal currents of1.0 A thus the force per unit length that each conductor exertson each other is given by

1 ampere is defined as the constant current, whichflowing in each of two infinitely long parallel straightconductors of negligible of cross sectional area separatedby a distance of 1.0 metre in vacuum, would produce aforce per unit length between the conductors of2.0 10 7 N m 1. ( Ampre's force law)

19.5.2 The ampere / One Ampere

d

I I l

F

2

210

12

11104 7

17 m N100.2

l

F

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SF02780

m

Fig. 6.9a

D

Q

A

E

F

B

C

G

H

P

I A

d

l

Ampere (current) Balance

F

g mW

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Exercise 19.4 :Given 0 = 4 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 th edition, Nelkon&Parker, Q6, p.336)

ANS. : 0.2 N; 60

2. A current-carrying conductor experiences no magnetic forcewhen it is placed in a uniform magnetic field. Explain thestatement.

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82


Use formulae:

where N = number of turns

Explain the working principles of a moving coil

galvanometer.

Learning Outcome:

19.6 Torque on a coil ( 1 hour)

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83

19.6.1 Formula of torqueConsider a rectangular coil (loop) of wire with side lengths a andb 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 withthe direction of the magnetic field. A current I is flowing roundthe coil as shown in Figure 6.31.

19.6 Torque on a coil

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84

Q

P

b

a

Figure 19.31a

B B

B B

B F

F

1 F I I

I

I 1 F

A

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85

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

sin2b

sin2b

B

B

B

1 F

1 F A

Q

2b

rotationrotation

Figure 19.31b: side view

90sin1 IlB F al and IaB F 1

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86

From the Figure 19.31a, the forces F lie along the axis PQ.The resultant force on the coil is zero but the nett torque is

not zero because the forces F 1 are perpendicular to the axisPQ as shown in Figure 19.31a.

The forces F 1 cause the coil to rotate in the clockwisedirection about 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

sin2

sin2 11

b F

b F

IaB F 1

sin2

21

b F and

sin2

2 b

IaB

sin IabB coil)of area( Aaband

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since thus

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

sin IAB 90

IAB

90sin IABcos

OR

(19.14)

(19.15)

where coilon thetorque: densityfluxmagnetic: Bcoilin theflowscurrent: I

B A andareactor between veangle: B andcoiltheof planee between thangle:

(coils)turnsof number: N

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From the eq. (19.14), thus the formula of torque in the vectorform is given by

The torque is zero when = 90 or = 0 and is maximumwhen = 0 or = 90 as shown in Figures 19.32a and 19.32b.

(19.16)

0

90 B A

0sin NIAB

90cos NIAB OR

0

Figure 19.32a

B A

0

90

Figure 19.32b90sin NIAB

0cos NIAB OR

NIAB max

plane of the coil

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In a radial field , the plane of the coil is always parallel to themagnetic field for any orientation of the coil about the verticalaxis as shown in Figure 19.33.

Hence the torque on the coil in a radial field is alwaysconstant and maximum given by

Radial field is used in moving coil galvanometer.

0

90 ORSN

coilfixed softiron cylinder

radial field

Figure 19.33: Plan view of moving coil meter

90sin NIAB 0cos NIAB ORmaximum

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90

A 50 turns rectangular coil with sides 10 cm 20 cm is placedvertically in a uniform horizontal magnetic field of magnitude 2.5 T. Ifthe current flows in the coil is 7.3 A, determine the torque acting onthe coil when the plane of the coil isa. perpendicular to the field,b. parallel to the field,

c. at 75 to the field.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, = 90 and = 0 , thus thetorque on the coil is sin NIAB NIAB cos OR

B A90

90cos NIAB 0sin NIAB

0 PHYSICS CHAPTER 19


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91

Solution :b.

c.

B A

90

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

NIAB cos 0cos5.2100.23.750 2

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

B

A

15 75

From the figure, = 75 and = 15 ,thusthe torque on the coil is

NIAB cos

75cos5.2100.23.7502

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92

A galvanometer consists of a coil of wire suspended in themagnetic field of a permanent magnet. The coil is rectangularshape and consists of many turns of fine wire as shown inFigure 6.34.

19.6.2 Moving-coil galvanometer

Figure 19.34

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When the current I flows through the coil , the magnetic fieldexerts a torque on the coil as given by

This torque is opposed by a spring which exerts a torque, sgiven by

The coil and pointer will rotate only to the point where thespring torque balances the torque due to magnetic field ,thus

NIAB

where constanttorsional:k radianincoiltheof anglerotation:

s k NIAB

(19.17)

(19.18)

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94

A rectangular coil of 10 cm 4.0 cm in a galvanometer has 50 turnsand a magnetic flux density of 5.0 10 2 T. The resistance of thecoil is 40 and a potential difference of 12 V is applied across thegalvanometer, calculate the maximum torque on the coil.Solution :

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

NIABmax

;04T;100.5turns;50 2 R B N V12V

23 100.5100.43.050

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95

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 density0.15 T and its suspension has a torsional constant of 3.0 10 6N m rad 1. Determine the current is required to give adeflection of 0.5 rad.

ANS. : 1.0 10 3 A

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At the end of this chapter, students should be able to:Explain the motion of a charged particle in both magneticfield and electric field.

Derive and use velocity

in a velocity selector.

Learning Outcome:19.7 Motion of charged particle in magnetic

field and electric field

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Consider a positively charged particle with mass m, charge qand velocity v enters a region of space where the electric andmagnetic fields are perpendicular to the particles velocity and toeach other as shown in Figure 19.38.

19.7 Motion of charged particle in

magnetic field and electric field

Figure 19.38

E

X X X X X X

X X X X X X

X X X X X X

X X X X X X

Bv v v

B F

E F

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The charged particle will experiences the electric force F E isdownwards with magnitude qE and the magnetic force F B is

upwards with magnitude qvB as shown in Figure19.38.If the particle travels in a straight line with a constant velocityhence the electric and magnetic forces are equal inmagnitude . Therefore

Only the particles with velocity equal to E/B can passthrough without being deflected by the fields .Eq. (19.21) also works for electron or other negatively chargedparticles.

E B F F qE qvB 90sin

(19.21)

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Figure 19.38 known as velocity selector .Normally, after the charged particle passing through the velocityselector it will enter the next region consist of a uniformmagnetic field only . This apparatus known as massspectrometer as shown in Figure 19.39.

Figure 19.39

E

X X X X X X

X X X X X X

X X X X X X

X X X X X X

Bv

E F

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

v

X X X X X

B F

v

B F

r

v

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100

When the charged particle entering the region consist ofmagnetic field only , the particle will make a semicircularpath of radius

r as shown in Figure 19.39.Therefore

From the eq. (19.22), the mass spectrometer can be used todetermine the value of q/ m for any charged particle.

C B F F

rBv

mq

and

r mv

qvB2

B E

v

(19.22)

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101

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

through a uniform magnetic field of 0.40 103

T. It is found to followa circular path. Calculatea. the radius of the circular path.b. the time required for the electron to complete one revolution.

(Given e/m = 1.76 10 11 C kg -1, me = 9.11 10 31 kg)Solution :a. The speed of the electron is given by

Example 19.15 :

T1040.0J;100.8 316 B K

2

21

mv K

23116 1011.921100.8 v17 sm1019.4 v

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102

Solution :

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

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

T1040.0J;100.8 316 B K

C B F F r

mvevB

2

90sin

r v

Bme

r 7

311 1019.41040.01076.1

T r

v2

T

595.021019.4 7

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Exercise 19.6 :1. An electron moving at a steady speed of 0.50 10 6 m s 1

passes between two flat, parallel metal plates 2.0 cm apartwith a potential difference of 100 V between them. Theelectron is kept travelling in a straight line perpendicular to theelectric field between the plates by applying a magnetic fieldperpendicular to the electrons path and to the electric field.Calculate :

a. the intensity of the electric field.b. the magnetic flux density needed.

ANS. : 0.50 10 4 V m 1; 0.010 T2. A proton moving in a circular path perpendicular to a constant

magnetic field takes 1.00 s to complete one revolution.Determine the magnitude of the magnetic field.(Physics for scientist and engineers, 6 th edition, Serway&Jewet,Q32, p.921)

chapter19 student

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