Matriculation Physics ( Magnetic Field )

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PHYSICS CHAPTER 6

CHAPTER 6: CHAPTER 6: Magnetic fieldMagnetic field

(7 Hours)(7 Hours)

PHYSICS CHAPTER 6

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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:

6.1 Magnetic field (1 hour)

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PHYSICS CHAPTER 6

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is defined as a region around a magnet where a magnetic a region around a magnet where a magnetic force can be experiencedforce can be experienced.

A stationary electric chargestationary electric charge is surrounded by an electric surrounded by an electric field onlyfield only.

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

6.1 Magnetic field

PHYSICS CHAPTER 6

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Magnetic field lines are used to represent a magnetic field. By convention, magnetic field lines leave the north poleleave the north pole and

enters 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 point indicates the direction of the magnetic fielddirection of the magnetic field at that point as shown in Figure 6.1.

Magnetic field can be represented by crossescrosses or by dotted dotted circlescircles as shown in Figures 6.2a and 6.2b.

6.1.1 Magnetic field lines

Figure 6.1Figure 6.1

direction of magnetic field at point P.

PP

Figure 6.2a : magnetic field lines Figure 6.2a : magnetic field lines enterenter the page perpendicularly the page perpendicularly

XX XX XX XX

XX XX XX XX

Figure 6.2b : magnetic field lines Figure 6.2b : magnetic field lines leave leave the page perpendicularlythe page perpendicularly

PHYSICS CHAPTER 6

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A uniform fielduniform field is represented by parallel lines of forceparallel lines of force. This means that the number of lines passing perpendicularly number of lines passing perpendicularly through unit area at all cross-sections in a magnetic field through unit area at all cross-sections in a magnetic field are the sameare the same as shown in Figure 6.3.

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 6.4.


unit cross-sectional areaunit cross-sectional area

PHYSICS CHAPTER 6

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The number of lines per unit cross-sectional area is number of lines per unit cross-sectional area is proportional 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 in stronger field in AA11

AA11 AA22

weaker field in weaker field in AA22

PHYSICS CHAPTER 6

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The pattern of the magnetic field lines can be determined by using two methods. compass needlescompass needles (shown in Figure 6.5)

sprinkling iron filings on papersprinkling iron filings on paper (shown in Figure 6.6).

6.1.2 Magnetic field lines pattern

Figure 6.5: plotting a magnetic field line of a bar Figure 6.5: plotting a magnetic field line of a bar magnetic.magnetic.

Figure 6.6: thin iron filing indicate the magnetic field lines.Figure 6.6: thin iron filing indicate the magnetic field lines.

PHYSICS CHAPTER 6

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Figures 6.7 shows the various pattern of magnetic field lines around the magnets.

Figure 6.7aFigure 6.7a

a. Bar magnet

b. Horseshoe or U magnet

Figure 6.7bFigure 6.7b

PHYSICS CHAPTER 6

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

Neutral point (point where where the resultant magnetic the resultant magnetic force is zeroforce is zero).

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

Figure 6.7cFigure 6.7c

Figure 6.7dFigure 6.7d

PHYSICS CHAPTER 6

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The Earth’s magnetic field is like that of a giant bar magnet as illustrated in Figure 6.8 with a pole near each geographic pole of the Earth.

6.1.3 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 6.8) points toward the north magnetic pole of the Earth, and since opposite attract, it follows that

Figure 6.8 also shows that the field lines are essentially horizontalhorizontal (parallel to the Earth’s surface) near near the equatorthe equator but enter or leave the enter or leave the Earth vertically near the polesEarth vertically near the poles.

the north geographical pole of the north geographical pole of the EarthEarth is actually near the south near the south pole of the Earth’s magnetic fieldpole of the Earth’s magnetic field.

PHYSICS CHAPTER 6

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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 in the Figure 6.9.

This method is called induced magnetizationinduced magnetization. The arrowsarrows in the soft iron represent the magnetization

direction with the arrowhead being the north pole and arrow tail being the south pole. It is also known as domains domains ( the the tiny magnetized region because of spin magnetic tiny magnetized region because of spin magnetic moment of the electronmoment of the electron).

6.1.4 Magnetization of a Soft Iron

Figure 6.9Figure 6.9 NN SS

PHYSICS CHAPTER 6

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In an unmagnetized piece of soft iron, these domains are arranged 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 vice versa as shown in Figure 6.9.

Two permanent magnets Bring and touch the first magnet to one end of the soft iron

and another end with the second magnet as shown in Figure 6.10.

NN NN SSSSFigure 6.10Figure 6.10

PHYSICS CHAPTER 6

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NN SS

Switch, S

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 6.11.

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 both viewing the current flows in the solenoid from both endend or applying the right hand grip rule right hand grip rule as shown in Figure 6.11.

II II

SSNN

Current - Current - anticlockwiseanticlockwise

Current - clockwiseCurrent - clockwise

PHYSICS CHAPTER 6

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Other examples:

If you drop drop a permanent magnet on the floor or strikestrike it with a hammer, you may jar the domains into randomnessdomains into randomness. The magnet 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 it placing it

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

NNSS

II IIII II

SS NN

Thumb – north polenorth pole

Other fingers – direction of current direction of current in solenoidin solenoid.

Note:Note:

Figure 6.12aFigure 6.12a Figure 6.12bFigure 6.12b

PHYSICS CHAPTER 6

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

Mathematically,

It also known as magnetic inductionmagnetic induction (magnetic field intensity magnetic 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 m22)). Unit conversion :

6.1.5 Magnetic flux density, B

A

BΦ

whereflux magnetic :Φ

field magnetic the toanglesright at area :A

)G(gauss 10 m Wb1T 1 42

(6.1)(6.1)

PHYSICS CHAPTER 6

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The direction of any magnetic field is taken to be in the direction that 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 1 :

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

Near the south pole of a permanent bar magnet, the north Near the south pole of a permanent bar magnet, the north pole of a compass will point toward the bar magnet so the pole of a compass will point toward the bar magnet so the field lines enter the south pole.field lines enter the south pole.

PHYSICS CHAPTER 6

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Exercise 6.1 :

1. Sketch the magnetic field lines pattern around the bar magnets for following cases.

a.

b.

PHYSICS CHAPTER 6

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ApplyApply magnetic field formula magnetic field formula

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:

6.2 Magnetic produced by current-carrying conductor (1 hour)

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r

IμB

20

R

IμB

20

nIμB 0

PHYSICS CHAPTER 6

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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 in Figure 6.13.

6.2 Magnetic field produced by current – carrying conductor


Thumb – direction of currentdirection of current

Other fingers – direction of magnetic direction of magnetic field field (clockwise clockwise OR anticlockwiseanticlockwise)

Note:Note:

PHYSICS CHAPTER 6

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The magnetic field lines pattern around a straight conductor carrying current is shown in Figures 6.14 and 6.15.

6.2.1 Magnetic field of a long straight conductor (wire) carrying current

OR

B

I

Current out of the pageCurrent out of the pageFigure 6.14Figure 6.14

BI

I B

B

PHYSICS CHAPTER 6

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

placed in vacuum as shown in Figure 6.16.

OR


I

I

IXX

Current into the pageCurrent into the pageXX

B

B

B

B

PHYSICS CHAPTER 6

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The magnitude of magnetic flux density (magnetic field

intensity), B at point P at distance r from the wire carrying current is given by

rPP

IFigure 6.16Figure 6.16

XX B

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

r

IμB

20

where space free ofty permeabili :0μ 17 A m T 10 4 (wire)conductor straight a frompoint a of distance :r

(6.2)(6.2)

PHYSICS CHAPTER 6

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The magnetic field lines pattern around a circular coil carrying current is shown in Figures 6.17.

6.2.2 Magnetic field of a circular coil


I IXX

SS

NNOR

SS

NN

I I

I

PHYSICS CHAPTER 6

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Consider a circular shaped conductor with radius R that carries

a current I as shown in Figure 6.18.

R

NIμB

20

where

(6.3)(6.3)

The magnitude of magnetic fieldmagnetic field intensityintensity BB at point O (centre of centre of the circular coil or loopthe circular coil or loop) , is given by

R

O

coilcircular theof radius :R(loops) coils ofnumber :N

space free ofty permeabili :0μ

current :IFigure 6.18Figure 6.18

PHYSICS CHAPTER 6

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A solenoid is an electrical device in which a long wire has an electrical device in which a long wire has been wound into a succession of closely spaced loops with been wound into a succession of closely spaced loops with geometry of a helixgeometry of a helix.

The magnetic field lines pattern around a solenoid carrying current is shown in Figure 6.19.

6.2.3 Magnetic field of a solenoid

SSNN

Figure 6.19Figure 6.19I I

PHYSICS CHAPTER 6

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OR

I

IXX XX XX XX

I

I

I

I

I

I

SSNN

PHYSICS CHAPTER 6

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

nIμB 02

1 (6.5)(6.5)

where lengthunit per turnsofnumber :n

The magnitude of magnetic field intensity at the centre magnitude of magnetic field intensity at the centre

(mid-point/ inside) of (mid-point/ inside) of NN turn solenoid turn solenoid is given by

l

NIμB 0

nl

Nand

nIμB 0 (6.4)(6.4)

PHYSICS CHAPTER 6

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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 pattern around both wires

a. when the currents are in the same direction.

b. when the currents are in opposite direction.


a.

Example 2 :

I

I I

I

PHYSICS CHAPTER 6

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I I

OR


a.

PHYSICS CHAPTER 6

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OR


b.

I IXX

I

I

I

I

PHYSICS CHAPTER 6

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A long wire (X) carrying a current of 50 A is placed parallel to and 5.0 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 0 = 4 107 H m1)

Example 3 :

PHYSICS CHAPTER 6

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a. i.

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

Magnitude of BX :

A 10m; 100.5A; 50 Y2

X IdI

T 100.4 4X

B

XB

YB

XB

YB

XIYI

d

Xr YrA OR

m 105.22

2YX

d

rr

XI

AXr Yr

YI

X

X0X 2πr

IμB

Direction : into the page OR upwards

2

7

X105.22

50104

π

πB

PHYSICS CHAPTER 6

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a. i. Magnitude of BY :

Therefore the total magnetic flux density at point A is

A 10m; 100.5A; 50 Y2

X IdI

T 100.8 5Y

BY

Y0Y 2πr

IμB

Direction : out of page OR downwards

2

7

Y105.22

10104

π

πB

YXA BBB

YXA BBB

Direction : into the page Direction : into the page OR upwardsupwardsSign convention of Sign convention of BB:

Out of page positive (+)

Into the page negative ()

Note:Note:

54A 100.8100.4 B

T 102.3 4A

B

PHYSICS CHAPTER 6

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a. ii.

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

Magnitude of BX :

A 10m; 100.5A; 50 Y2

X IdI

T 100.4 4X

B

XB

YB XB

YB

XIYI

d

Xr YrA OR

Direction : into the page OR upwards

2

7

X105.22

50104

π

πB

XI

AXr Yr

YIXX

PHYSICS CHAPTER 6

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a. ii. Magnitude of BY :

Therefore the resultant magnetic flux density at point A is

A 10m; 100.5A; 50 Y2

X IdI

T 100.8 5Y

BDirection : into the page OR upwards

2

7

Y105.22

10104

π

πB

YXA BBB

YXA BBB

Direction : into the page Direction : into the page OR upwardsupwards

54A 100.8100.4 B

T 108.4 4A

B

PHYSICS CHAPTER 6

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b.

Since the resultant magnetic flux density at point C is zero thus

A 10m; 100.5A; 50 Y2

X IdI

XB

XIYI

d

Xr YrC OR

rr X

XI

CXr Yr

YI

XB

YB

YB

rdr Y

YXC BBB

YX0 BB

YX BB whereX

X0X 2πr

IμB and

Y

Y0Y 2πr

IμB

PHYSICS CHAPTER 6

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b.

Y

Y0

X

X0

22 πr

Iμ

πr

Iμ

rd

I

r

I

YX

rr 2100.5

1050

A 10m; 100.5A; 50 Y2

X IdI

m 102.4 2r

PHYSICS CHAPTER 6

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

The current in one wire is I1 = 3.0 A pointing into the page and the

current in the other wire is I2= 4.0 A pointing out of page. Determine

the magnitude and direction of the nett magnetic field intensity at point P.

(Given 0 = 4 107 H m1)

Example 4 :


PHYSICS CHAPTER 6

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

m 100.5A; 0.4A; 0.3 2121

rII

22222 100.5100.5 r

1r2B

2r

m 101.7 22

rθ

θ2

2

2

1

101.7

100.5cos

r

rθ

704.0cos θ

704.0101.7

100.5sin

2

2

θ

1

101 2πr

IμB

T 1020.1 51

B

2

7

1100.52

0.3104

π

πB

1B

1I 2I

P

XXm 100.5 2

PHYSICS CHAPTER 6

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and

m 100.5A; 0.4A; 0.3 2121

rII

2

202 2πr

IμB

T 1013.1 52

B

2

7

2101.72

0.4104

π

πB

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

Vector sum

51 1020.1 B1B

0

θB cos2

2B 704.01013.1 5

61096.7

θB sin2 704.01013.1 5

61096.7 65 1096.71020.1 xB

6104.04

61096.70 yB610.967

PHYSICS CHAPTER 6

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Therefore the magnitude of the nett magnetic field intensity at point P is given by

and its direction is

m 100.5A; 0.4A; 0.3 2121

rII

22yx BBB

T 1093.8 6B

2626 1096.71004.4

x

y

B

Bθ 1tan

1.63θ

6

61

1004.4

1096.7tan

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

1.63

B

1B

2B

P

PHYSICS CHAPTER 6

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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 0 = 4 107 T m A1)


a. Given

By applying the equation for magnitude of the magnetic field at

the centre of the circular coil, thus

Example 5 :

A 5.2;500m; 105.02

1010 22

INR

R

NIμB

20

T 1057.1 2B

2

7

100.52

5.2500104

π

B

PHYSICS CHAPTER 6

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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

NIμBi

0

turns153N

5.1

18104103.2

73 Nπ

T; 103.2m; 103.1 2

106.2m; 5.1 3

i2

2

BrlA 18I

πr2ncecircumfere 2103.12ncecircumfere πm1017.8ncecircumfere 2

ncecircumfereNL

2108.17153 Lm 5.12L

PHYSICS CHAPTER 6

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Exercise 6.2 :

Given 0 = 4 107 T m A1

1.

The two wires shown in Figure 6.21 carry currents of 5.00 A in 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. : 1.33 1.33101055 T, out of page; 2.67 T, out of page; 2.67101066 T, out of page T, out of page

Figure 6.21Figure 6.21A 00.5 A 00.5

cm 0.10

cm 0.15cm 0.5

2P

1P

PHYSICS CHAPTER 6

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Exercise 6.2 :

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 6.22.

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

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

square.

ANS. :ANS. : 4.0 4.0 10 1044 T , to the left (180 T , to the left (180))


XX XX

PHYSICS CHAPTER 6

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

DescribeDescribe circular motion of a charge in a uniform circular motion of a charge in a uniform magnetic field.magnetic field.

UseUse relationship relationship FFBB = = FFCC..

Learning Outcome:

6.3 Force on a moving charged particle in a uniform magnetic field (1 hour)

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BvqF

PHYSICS CHAPTER 6

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6.3.1 Magnetic force A stationarystationary electric charge in a magnetic field will not not

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 a it will experience a magnetic forcemagnetic force.

The magnitudemagnitude of the magnetic force can be calculated by using the following equation:

6.3 Force on a moving charged particle in a uniform magnetic field

θqvBF sin

where force magnetic :Fdensityflux magnetic :B

charge a of velocity :vcharge theof magnitude:q

Bvθ

and between angle :

(6.6)(6.6)

PHYSICS CHAPTER 6

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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

(6.7)(6.7) BvqF

B

v

F

shown in Figures 6.23 and 6.24

Thumb Thumb – direction of – direction of ForceForce

First finger First finger – direction of – direction of FieldField

Second fingerSecond finger – direction of – direction of VelocityVelocity

Figure 6.24Figure 6.24Note:Note:

PHYSICS CHAPTER 6

49

Determine the direction of the magnetic force, F exerted on a charge in the following problems:

a. b.

c. d.

e.

Example 6 :

B

v

B v

XX XX XX XX

XX XX XX XX

XX XX XX XX

v

I

I

v

B

v

PHYSICS CHAPTER 6

50


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


PHYSICS CHAPTER 6

51


d.

e.

I

v

B

XX XX XX XX

XX XX XXF


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

PHYSICS CHAPTER 6

52

Calculate the magnitude of the force on a proton travelling 5.0107 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 +1.601019 C)


a. Given

Therefore

b. Given Hence

Example 7 :

90217 m Wb5.1;s m 105.0 Bv

θqvBF sin 90sin5.1100.51060.1 719

N 1020.1 11F50

N 1019.9 12F 50sin5.1100.51060.1 719 F

PHYSICS CHAPTER 6

53

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

As the particle enters the region, it will experiences a magnetic magnetic forceforce which the force is perpendicular to the velocityperpendicular to the velocity of the particle. Hence the direction of its velocity changes but the magnetic force remains perpendicular to the velocity.

This magnetic force, FB makes the path of the particle is a path of the particle is a circularcircular as shown in Figures 6.25a, 6.25b, 6.25c and 6.25d.

6.3.2 Motion of a charged particle in a uniform magnetic field


v v

BF

v

BF

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

v

v

BF

v

BF


PHYSICS CHAPTER 6

54

Since the path is a circle therefore the magnetic force FB

contributes the centripetal force Fc (nett force) in this motion.

Thus

Figure 6.25cFigure 6.25c Figure 6.25dFigure 6.25d

v v

BF

v

BF

v v

BF

v

BF

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

cB FF

r

mvθqvB

2

sin 90θand

PHYSICS CHAPTER 6

55

The period of the circular motion, T makes by the particle is given by

And the frequency of the circular motion is

Bq

mvr

where particle charged theof mass :m velocity theof magnitude :v

pathcircular theof radius :rparticle charged theof magnitude :q

(6.8)(6.8)

rv

T

rv

2

T

2and

Bq

mT

2

Bq

mvr and

v

rT

2

(6.9)(6.9)

Tf

1

PHYSICS CHAPTER 6

56

An electron at point A in Figure 6.26 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=1.601019 C and me= 9.111031 kg)

Example 8 :

v

BA cm 0.20


PHYSICS CHAPTER 6

57


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 pageOR

m 100.20; s m 1050.2 216 dv

Be

mvr

T 1042.1 4B

2

dr

Be

mvd

2

19

6312

1060.1

1050.21011.9

2

100.20

B

and

v

BA

B

F

PHYSICS CHAPTER 6

58


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

m 100.20; s m 1050.2 216 dv

rωv

s 1051.2 7T

T

πω

2

T

πdv

2

2

T

π26 100.20

1050.2

and

Tt2

1

71051.22

1 t

s 1026.1 7t

PHYSICS CHAPTER 6

59

Exercise 6.3 :1. Determine the sign of a charge in the following problems.

a. b.

ANS. :ANS. : positive; positive positive; positive

2. Determine the direction of the magnetic force exerted on a positive charge in each problem below when a switch S is closed.

a. b.

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

B

v

F B

vF

Switch, S

v

Switch, S

v

PHYSICS CHAPTER 6

60

Exercise 6.3 :3. An electron experiences the greatest force as it travels

2.9106 m s1 in a magnetic field when it is moving north. The force is upward and of magnitude 7.21013 N. Determine the magnitude and direction of the magnetic field.

(Given the charge of the electron is 1.601019 C)(Physics for scientists & engineers ,3(Physics for scientists & engineers ,3rdrd edition, Giancoli, Q22, edition, Giancoli, Q22, p.705)p.705)

ANS. :ANS. : 1.6 T to the east 1.6 T to the east

4. An electron moving with a speed of 9.1105 m s1 in the

positive x direction experiences zero magnetic force. When it

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

2.01013 N that points in the negative z direction. What is the direction and magnitude of the magnetic field?

(Given e=1.601019 C and me= 9.111031 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 negative 1.37 T to the left (in the negative yy direction) direction)

PHYSICS CHAPTER 6

61

Exercise 6.3 :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. : 1/4 1/4

6. A 12.5 C particle of mass 2.80105 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 s 12.1 m s11; 13.9 s; 13.9 s

PHYSICS CHAPTER 6

62


Learning Outcome:

6.4 Force on a current-carrying conductor in a uniform magnetic field (1 hour)

ww

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BlIF

PHYSICS CHAPTER 6

63

When a current-carrying conductorcurrent-carrying conductor is placed in a magnetic in a magnetic fieldfield B, thus a magnetic force will acts on that conductormagnetic 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,

6.4 Force on a current-carrying conductor in a uniform magnetic field

θIlBF sin (6.10)(6.10)

BlIF

(6.11)(6.11)

where force magnetic :Fdensityflux magnetic theof magnitude :B

current :Iconductor theoflength :l

BIθ

and ofdirection between angle :

PHYSICS CHAPTER 6

64

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 6.27.

From the equation (6.10), the magnetic forcemagnetic force on the conductor has its maximummaximum

value when the conductor (and therefore the current) and conductor (and therefore the current) and the magnetic field are perpendicularthe magnetic field are perpendicular (at right angles) to each other then ==9090 (shown in Figure 6.28a).

Thumb Thumb – direction of – direction of ForceForce

First finger First finger – direction of – direction of FieldField

Second fingerSecond finger – direction of – direction of CurrentCurrent


Note:Note:B

I

F

PHYSICS CHAPTER 6

65

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

magnetic fieldmagnetic field then =0=0 (shown in Figure 6.28b).


IlBF max

B

90θI

90sinmax IlBF

B

0θI


0F

0sinIlBF

One teslaOne tesla is defined as the magnetic flux density of a field in which as the magnetic flux density of a field in which a force of 1 newton acts on a 1 metre length of a conductor which a force 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.carrying a current of 1 ampere and is perpendicular to the field.

Note:Note:

PHYSICS CHAPTER 6

66

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 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)

PHYSICS CHAPTER 6

67

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 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 60102 N and the current

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

supplied.


a. Given

Example 10 :

90;m Wb20.1 ;m 00.1 2 θBlA 15I

θIlBF sin 90sin20.100.115

N 18F

PHYSICS CHAPTER 6

68


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

m 501A; 15 .lI

θIlBF sin 90sin20.150.115

N 27F

N 1060m; 501A; 12 2 F.lI

θIlBF sin

90sin50.1121060 2 B

T 1033.3 2B

PHYSICS CHAPTER 6

69

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 force acting 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 s2)


a.

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

weight, therefore

Example 11 :

90;T 2.0 ;m 5.0 g;1050 3 θBlm

θIlBF sinθIlBmg sin

A 91.4I 90sin2.05.081.91050 3 I

IF

gm

B

PHYSICS CHAPTER 6

70

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DeriveDerive force per unit length of two parallel current- force per unit length of two parallel current-

carrying conductors.carrying conductors. UseUse formulae: formulae:

DefineDefine one ampere. one ampere.

Learning Outcome:

6.5 Forces between two parallel current-carrying conductors (1 hour)

ww

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πd

IIμ

l

F

2210

PHYSICS CHAPTER 6

71

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

currents I1 and I2 with length l are placed parallel to each other

as shown in Figure 6.29.

6.5 Forces between two parallel current- carrying conductors


d

1 2

1I

1I

2I

2I

P1B

12F

21F

Q

2B

PHYSICS CHAPTER 6

72

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 I2 and in the magnetic field B1

thus the conductor 2 will experiences a magnetic force, F12.

The magnitude of F12 is given by

d

IB

2

101 Direction : into the page

sin1212 lBIF 90and

90sin2

102

d

IlI

d

lIIF

2210

12

PHYSICS CHAPTER 6

73

The magnitude of F21 is given by

Conclusion :

and the typetype of the force is attractiveof the force is attractive. From the equation (6.12), thus the force per unit length is given

by

sin2121 lBIF 90and

90sin2

201

d

IlI

d

lIIF

2210

21

d

lIIFFF

2210

2112

(6.12)(6.12)

πd

IIμ

l

F

2210 (6.13)(6.13)

PHYSICS CHAPTER 6

74

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

The magnitude of F12 and F21 can be determined by using the eq. (6.12) and their direction can be determined by applying Fleming’s left hand rule.Conclusion : Type of the force is repulsiveType of the force is repulsive.


2I

2I

1I

1I d

1 2Note:Note: The currents are in the

same directionsame direction – 2 conductors attractattract each other.

The currents are in opposite directionopposite direction – 2 conductors repelrepel each other.

21F

Q 2B

12F1B

P

PHYSICS CHAPTER 6

75

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 0 = 4 107 T m A1)


a. The diagram is

Example 12 :

12F

21F

d1

1I

2

2I

m 250A; 4.221 .dII

PHYSICS CHAPTER 6

76


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

c. Given

Therefore the current needed in the second wire is

m 250A; 4.221 .dII

πd

IIμ

l

F

2210

25.02

4.24.2104 7

π

π

l

F

16 m N106.4 l

F

A 64.01 I

πd

IIμ

l

F

2210

25.02

64.0104106.4 2

76

π

Iπ

A 98.82 I

PHYSICS CHAPTER 6

77

From the eq. (6.13), 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, which the constant current, which flowing in each of two infinitely long parallel straight flowing in each of two infinitely long parallel straight conductors of negligible of cross sectional area separated conductors of negligible of cross sectional area separated by a distance of 1.0 metre in vacuum, would produce a by 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 2.02.0101077 N m N m 1 1..

6.5.2 The ampere

πd

IIμ

l

F

2210

12

11104 7

π

π

17 m N100.2 l

F

PHYSICS CHAPTER 6

78

Exercise 6.4 :Given 0 = 4 107 T m A1

1. A vertical straight conductor Y of length 0.5 m is situated in a uniform 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; 60 0.2 N; 602. A current-carrying conductor experiences no magnetic force

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

PHYSICS CHAPTER 6

79


where where NN = number of turns = number of turns ExplainExplain the working principles of a moving coil the working principles of a moving coil

galvanometer.galvanometer. ExplainExplain the DC electrical measuring instruments. the DC electrical measuring instruments.

Learning Outcome:

6.6 Torque on a coil (1 hour)

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BANIτ

PHYSICS CHAPTER 6

80

6.6.1 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 with

the direction of the magnetic field. A current I is flowing round the coil as shown in Figure 6.31.

6.6 Torque on a coil

PHYSICS CHAPTER 6

81

Q

P

b

a


θ

BB

B

B

BF

F

1F

I I

I

I1F

A

PHYSICS CHAPTER 6

82

From the Figure 6.31b, the magnitude of the force F1 is given by

From the Figure 6.31a, the forces F lie along the axis PQ.

sin2

b

sin2

b

θB

B

B

1F

1F

A

Q

2

brotationrotationrotationrotation

Figure 6.31b: side viewFigure 6.31b: side view

90sin1 IlBF al and

IaBF 1

PHYSICS CHAPTER 6

83

From the Figure 6.31a, the forces F lie along the axis PQ. The resultant forceresultant force on the coil is zerozero but the nett torquenett torque is

not zeronot zero because the forces F1 are perpendicular to the axis

PQ as shown in Figure 6.31a.

The forces F1 cause the coil to rotate coil to rotate in the clockwise clockwise

direction about the axis PQ direction about the axis PQ as shown in Figure 6.31b. The magnitude of the nett torque about the axis PQ (refer to

Figure 6.31b) is given by

sin

2sin

2 11

bF

bFτ

IaBF 1

sin

22 1

bF and

sin

22

bIaB

sinIabB coil) of area(Aab and

PHYSICS CHAPTER 6

84

since thus

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

sinIABτ θ 90

θIAB 90sin

θIABcos

sinNIABτ

OR

θNIABτ cos

(6.14)(6.14)

(6.15)(6.15)

where coil on the torque:τdensityflux magnetic :B

coil in the flowscurrent :IBA

and areactor between ve angle :Bθ

and coil theof plane ebetween th angle :(coils) turnsofnumber :N

PHYSICS CHAPTER 6

85

From the eq. (6.14), thus the formula of torque in the vector form is given by

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

when = 0= 0 or or = 90= 90 as shown in Figures 6.32a and 6.32b.

BANIτ

(6.16)(6.16)

0

90θB A

0sinNIABτ

90cosNIABτ OR

0τ


B

A

0θ

90

Figure 6.32bFigure 6.32b90sinNIABτ

0cosNIABτ OR

NIABτ max

plane of the coil

PHYSICS CHAPTER 6

86

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 6.33.

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

Radial field is used in moving coil galvanometer.

0θ

90ORSSNN

coilfixed soft iron cylinder

radial field

Figure 6.33: Plan view of moving coil meterFigure 6.33: Plan view of moving coil meter

90sinNIABτ 0cosNIABτ OR

NIABτ maximummaximum

PHYSICS CHAPTER 6

87

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 acting on 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.


The area of the coil is given by

a.

Example 13 :

A 7.3 T; 5.2 turns;50 IBN

2222 m 100.210201010 A

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

sinNIABτ θNIABτ cos OR

B

A

90θ

90cosNIAB 0sinNIAB0τ

PHYSICS CHAPTER 6

88


b.

c.

BA

90

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

θNIABτ cos 0cos5.2100.23.750 2

m N 3.18τ

A 7.3 T; 5.2 turns;50 IBN

B

A

1575θ

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

θNIABτ cos 75cos5.2100.23.750 2

m N 72.4τ

PHYSICS CHAPTER 6

89

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 6.34.

6.6.2 Moving-coil galvanometer


PHYSICS CHAPTER 6

90

When the current II flows through the coil flows through the coil, the magnetic field magnetic 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, s

given by

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

NIABτ

kθτ s where constant torsional:k

radianin coil theof anglerotation :θ

sττ kθNIAB

NAB

kθI

(6.17)(6.17)

(6.18)(6.18)

PHYSICS CHAPTER 6

91

A rectangular coil of 10 cm 4.0 cm in a galvanometer has 50 turns and a magnetic flux density of 5.0 102 T. The resistance of the coil is 40 and a potential difference of 12 V is applied across the 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 14 :

2322 m 100.4100.41010 A

4012 IIRV A 3.0I

NIABmax

m N 100.3 3max

τ

; 04 T; 100.5 turns;50 2 RBNV 12V

23 100.5100.43.050

PHYSICS CHAPTER 6

92

OhmmeterOhmmeter It is used to measure the unknown resistance of the resistormeasure the unknown resistance of the resistor. Figure 6.35 shows the internal connection of an Ohmmeter.

6.6.3 Electrical instruments

QP

0

MR SR

ε

XR

where

resistance (coil)meter :MRresistance variable:SR

resistanceunknown :XR


PHYSICS CHAPTER 6

93

When nothing is connected to terminals P and Qnothing is connected to terminals P and Q, so that the circuit is open (that is, when RR ), there is no current and no current and no deflectionno deflection.

When terminals P and Q are short circuitedP and Q are short circuited (that is when RR = 0 = 0), the ohmmeter deflects full-scaledeflects full-scale.

For any value of RX the meter deflection depends on the value

of RX.

AmmeterAmmeter It is used to measure a current flows in the circuitmeasure a current flows in the circuit. Ammeter is connected in seriesconnected in series with other elements in the

circuit because the currentcurrent to be measured must pass directly must pass directly through the ammeterthrough the ammeter.

An ammeter should have low internal resistancelow internal resistance (RM) so that the currentcurrent in the circuit would not affectedwould not affected.

PHYSICS CHAPTER 6

94

The maximum readingmaximum reading from the ammeter is known as full full scale deflectionscale deflection (fs).

If the full scale current passing through the ammeter then the potential difference (p.d.) across that ammeter is given by

If the meter is used to measure currents that are larger than its

full scale deflection (I >II >Ifsfs), some modification has to be done. A resistor has to be connected in parallel with the meter resistor has to be connected in parallel with the meter

(coil) resistance (coil) resistance RRMM so that some of the current will current will bypasses the meter (coil) resistancebypasses the meter (coil) resistance.

This parallel resistor is called a shuntshunt denoted as RS. Figure 6.36 shows the internal connection of an ammeter

with a shunt in parallel.

Mfsfs RIV where resistance )meter(coil :MR

current scale full :fsI(p.d.) difference potential scale full :fsV

PHYSICS CHAPTER 6

95

Since shunt is connected in parallel with the meter (coil) resistance then

0 max

A

MR

SR

I fsI

SI

I


SM RR VV and fsS III SSMfs RIRI

SfsMfs RIIRI Mfs

fsS R

II

IR

(6.19)(6.19)

PHYSICS CHAPTER 6

96

VoltmeterVoltmeter It is used to measure a potential difference (p.d.) across measure a potential difference (p.d.) across

electrical elements in the circuitelectrical elements in the circuit. Voltmeter is connected in parallelconnected in parallel with other elements in the

circuit therefore its resistance must be largerresistance must be larger than the than the resistance of the element so that a very small amount of resistance of the element so that a very small amount of current only can flows through itcurrent only can flows through it. An ideal voltmeterideal voltmeter has infinity resistanceinfinity resistance so that no current exist in it.

To measure a potential difference that are larger than its full

scale deflection (V > VV > Vfsfs), the voltmeter has to be modified. A resistor has to be connected in seriesA resistor has to be connected in series with the meter with the meter

(coil) resistance (coil) resistance RRMM so that only a fraction of the total p.d.

appears across the RM and the remainder appears across the serial resistor.

This serial resistorserial resistor is called a multipliermultiplier OR bobbinbobbin

denoted as RB.

PHYSICS CHAPTER 6

97

Figure 6.37 shows the internal connection of a voltmeter with a multiplier in series.

Since the multiplier is connected in series with the meter (coil)

resistance then the current through them are the same, current through them are the same, IIfsfs.

0 max

V

MRBR

Electrical Electrical elementelement

V1I

fsI

I


PHYSICS CHAPTER 6

98

The p.d. across the electrical element is given by

Hence the multiplier resistance is MB RR VVV

MfsBfs RIRIV

fs

MfsB I

RIVR (6.20)(6.20)

Note:Note: To convert a galvanometer to ammetergalvanometer to ammeter, a shunt shunt

(parallel resistor)(parallel resistor) is used. To convert a galvanometer to voltmetergalvanometer to voltmeter, a multiplier multiplier

(serial resistor)(serial resistor) is used.

PHYSICS CHAPTER 6

99

A milliammeter with a full scale deflection of 20 mA and an internal resistance of 40 is to be used as an ammeter with a full scale deflection of 500 mA. Calculate the resistance of the shunt required.


By applying the formula of shunt resistor, thus

Example 15 :

A 10 500 ; 40A; 1020 3M

3fs

IRI

Mfs

fsS R

II

IR

67.1SR

40102010500

102033

3

PHYSICS CHAPTER 6

100

A galvanometer has an internal resistance of 30 and deflects full scale for a 50 A current. Describe how to use this galvanometer to make

a. an ammeter to read currents up to 30 A.

b. a voltmeter to give a full scale deflection of 1000 V.


a. We make an ammeter by putting a resistor in parallel (RS) with

the internal resistance, RM of the galvanometer as shown in

figure below.

Example 16 :

30A; 1050 M6

fs RI

SI

I MRfsI

GG

SR

PHYSICS CHAPTER 6

101


a. Given

Since RS in parallel with RM therefore

b. We make a voltmeter by putting a resistor in series (RB) with the

internal resistance, RM of the galvanometer as shown in figure

below.

30A; 1050 M6

fs RIA 30I

SM RR VV and fsS III

SfsMfs RIIRI

in parallel.in parallel.

SSMfs RIRI

S66 105030301050 R

100.5 5SR

VfsI

MR

GG

BR

fsI

PHYSICS CHAPTER 6

102


b. Given

Since RB in series with RM therefore

30A; 1050 M6

fs RIV 1000V

MB RR VVV

MfsBfs RIRIV

30105010501000 6B

6 R

100.2 7BR in series.in series.

PHYSICS CHAPTER 6

103

Exercise 6.5 :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 3.0106 N m rad1. Determine the current is required to give a deflection of 0.5 rad.

ANS. :ANS. : 1.0 1.0101033 A A

2. A milliammeter of negligible resistance produces a full scale deflection when the current is 1 mA. How would you convert the milliammeter to a voltmeter with full scale deflection of 10 V?

ANS. :ANS. : 1.0 1.0101044 in series in series3. A moving-coil meter has a resistance of 5.0 and full scale

deflection is produced by a current of 1.0 mA. How can this meter be adapted for use as :

a. a voltmeter reading up to 10 V,

b. a ammeter reading up to 2?

ANS. :ANS. : 9995 9995 in series; 2.5 in series; 2.5101033 in parallel in parallel

PHYSICS CHAPTER 6

104

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ExplainExplain the motion of a charged particle in both the motion of a charged particle in both

magnetic field and electric field.magnetic field and electric field. Derive and useDerive and use formulae formulae

in a velocity selector.in a velocity selector.

Learning Outcome:

6.7 Motion of charged particle in magnetic field and electric field (1 hour)

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B

Ev

PHYSICS CHAPTER 6

105

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 to each other as shown in Figure 6.38.

6.7 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

BF

EF

PHYSICS CHAPTER 6

106

The charged particle will experiences the electric force FE is

downwards with magnitude qE and the magnetic force FB is

upwards with magnitude qvB as shown in Figure 6.38. If the particle travels in a straight line with a constant velocity

hence the electric and magnetic forces are equal in electric and magnetic forces are equal in magnitudemagnitude. Therefore

Only the particles with velocity equal to velocity equal to E/BE/B can pass can pass through without being deflected by the fieldsthrough without being deflected by the fields.

Eq. (6.21) also works for electron or other negatively charged particles.

EB FF qEqvB 90sin

B

Ev (6.21)(6.21)

PHYSICS CHAPTER 6

107

Figure 6.38 known as velocity selectorvelocity selector. Normally, after the charged particle passing through the velocity

selector it will enter the next region consist of a uniform uniform magnetic field onlymagnetic field only. This apparatus known as mass mass spectrometerspectrometer as shown in Figure 6.39.


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

EF

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

BF

v

BFr

v

PHYSICS CHAPTER 6

108

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 particle will make a semicircular

pathpath of radius r as shown in Figure 6.39.Therefore

From the eq. (6.22), the mass spectrometer can be used to

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

CB FF

rB

v

m

q and

r

mvqvB

2

B

Ev

2rB

E

m

q (6.22)(6.22)

PHYSICS CHAPTER 6

109

An electron with kinetic energy of 8.01016 J passes perpendicular through a uniform magnetic field of 0.40103 T. It is found to follow a circular path. Calculate

a. the radius of the circular path.

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

(Given e/m = 1.761011 C kg-1, me = 9.111031 kg)


a. The speed of the electron is given by

Example 17 :

T 1040.0J; 100.8 316 BK

2

2

1mvK

23116 1011.92

1100.8 v

17 s m 1019.4 v

PHYSICS CHAPTER 6

110


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

T 1040.0J; 100.8 316 BK

CB FF

r

mvevB

2

90sin

r

vB

m

e

r

7311 1019.4

1040.01076.1

m 595.0r

T

πrv

2

T

π 595.021019.4 7

s 1092.8 8T

PHYSICS CHAPTER 6

111

Exercise 6.6 :1. An electron moving at a steady speed of 0.50106 m s1

passes between two flat, parallel metal plates 2.0 cm apart with a potential difference of 100 V between them. The electron 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. : 0.50 0.50101044 V m V m11; 0.010 T; 0.010 T

2. 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(Physics for scientist and engineers, 6thth edition, Serway&Jewet, edition, Serway&Jewet, Q32, p.921)Q32, p.921)

(mp=1.671027 kg and charge of the proton, q=1.601019 C)

ANS. :ANS. : 6.56 6.56101022 T T

112

PHYSICS CHAPTER 6

Next Chapter…CHAPTER 7 :

Electromagnetic induction

Matriculation Physics ( Magnetic Field )

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