C19 Magnetic Field Student

8/13/2019 C19 Magnetic Field Student

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UNIT 19 : MAGNETIC FIELD

19.1 Magnetic field

19.2 Magnetic field produced by current-carrying conductor

19.3 Force on a moving charged particle in a

uniform magnetic field19.4 Force on a current-carrying conductor ina uniform magnetic field

19.5 Forces between two parallel current-carrying conductors

19.6 Torque on a coil19.7 Motion of charged particle in magnetic

field and electric field


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19.1 Magnetic Field (1 hour)

Learning Outcomes : At the end of this lesson, the students should be able to ;

(i) Define magnetic field.

(ii) Identify magnetic field sources.

(iii) Sketch the magnetic field lines.


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19.1 Magnetic field is defined as a region surrounding a magnet

or a conductor carrying current where amagnetic force is experienced . Magnets always have two poles :

a) North and south poles.b) Like poles repel and unlike poles attract.Magn etic f ie ld l ines A magnetic field can be represented by

magnetic field lines (straight lines or curves) . Arrows on the lines show the direction of the

field : the arrows point away from north polesand towards south poles.


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unit cross-sectional area

Magnet ic f ie ld l ines 19.1 Magnetic field

A uniform field is represented by parallel

lines . This means that the number of linespassing perpendicularly through unit areaat all cross-sections in a magnetic fieldare the same as shown below.


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A non-uniform field is represented by non-

parallel lines. The number of magnetic fieldlines varies at different unit cross-sectionsas shown below.

A1 A2


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19.1 Magnetic fieldMagnet ic f ie ld l ines

The tangent to a curved field line at apoint indicates the direction of themagnetic field at that point.

direction ofmagnetic field atpoint P.

P

Magnetic field lines do not intersect

one another.


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19.1 Magnetic fieldMagnet ic f ie ld l ines

The number of lines per unit cross sectionarea is an i ndication of the strength ofthe field. The number of lines per unitcross-sectional area is proportional tothe magnitude of the magnetic field .

stronger field in A1

A1 A2


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Magnetic field linesenter the pageperpendicularly

Magnetic field linesleave the pageperpendicularly

Magnetic field can also be represented by

crosses or by dotted circles as shownbelow.


B into the pageB out of page


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19.1 Magnetic field

The magnetic field lines pattern can beobtained by using iron filings or a plottingcompass .

the arrowhead of acompass needle isa no r th p ole .

Field Pattern s


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19.1 Magnetic field

The direction of the magnetic field at a point

is defined as the direction of a compassneedle points when placed at that point .

Field Patterns


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19.1 Magnetic fieldField Patterns

a. A bar magnet b. Horseshoe or U magnet

c. Two bar magnets (unlike pole) - attractive


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d. Two bar magnets (like poles) - repulsive

Neutral point(point where the resultantmagnetic force/fieldstrength is zero)

Field Patterns 19.1 Magnetic field


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19.1 Magnetic field

h. EarthMagnetic Field

Field Patterns


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19.1 Magnetic fieldMagn etic Flux,

is a measure of the number of field lines

that cross a surface area. is defined as the scalar product between

the magnetic flux de nsity, B and the vector

of the surface area, A . BA A B cos==

fluxmagnetic=

.andof directionthebetweenangle A B =

densityfluxmagnetic

strengthfieldmagnetic

=

= B

throughpasslinesfieldthatareathe= A

area, A

A

B


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19.1 Magnetic fieldMagnet ic Flux ,

scalar quantity.

unit : weber(Wb)/ tesla-meter squared(T.m2)

1 T.m 2 = 1 Wb Consider a uniform magnetic field B

passing through a surface area A as shownin figure below.

B

A

area, A

0cos= BA

BA B

In Figure below, = 0


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19.1 Magnetic fieldMagnet ic Flux ,

90cos= BA

0=

If = 90

A

B

Magn et ic Flux Dens i ty, B

is defined as the magnetic flux per unitarea at right angles to the magnetic field .

= A

Bfluxmagnetic=

fieldmagnetic

thetoanglesrightatarea= A


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vector quantity and its direction follows

the direction of the magnetic field. unit : weber per metre squared (Wb m -2)

or tesla (T).

gauss(G)10=mWb1=T1 4-2

Magnet ic Flux Densi ty, B19.1 Magnetic field


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19.2 Magnetic field produced by current-carrying conductor (1 hour)

Learning outcomes : At the end of this lesson, the students should be able to ;

Apply magnetic field formula ;

(i) for a long straight wire

(ii) for a circular coil

(iii) for a solenoid

r I

B o

2

r

I B o

2

nI B o


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19.2 Magnetic field ( B ) produced bycurrent - carrying conductor

B is a vector quantity.Magnitude :

r

I B o

2

B

B

I

Current out of the page

View from the top

A longstraightwire

(4

x 10-7

H m-1

)

spacefreeof

typermeabili0 :


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19.2 Magnetic field ( B ) produced by current - carrying conductor

B = magnetic field strength / flux

density (T) I = current in the wire (A)r = perpendicularly distance of P from

the wire (m)

o = constant of proportionalityknown as the permeability offree space (vacuum)

= 4 x 10 -7 Henry per metre (H m -1)

r I

B o P 2=

P rDirection : right-hand grip rule

out of the page


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Example 19.2.1

Determine the magnetic field strength atpoint X and Y from a long, straight wirecarrying a current of 5 A as shown below.

2 cm

6 cm

X

Y

I = 5 A

r

I B o X 2

= = 5.0 x 10 -5 T , into the page

BY = 1.67 x 10 -5 T , out of the page


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Exercise (DIY)

1. Two straight parallel wires are 30 cmapart and each carries a current of 20 A.Find the magnitude and direction of themagnetic field at a point in the plane of thewires that is 10 cm from one wire and 20cm from the other if the currents are

(i) in the same direction,(ii) in the opposite direction.


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I I X

S

N

R I

I I

SN

A circular coil



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A circular coil

Magnetic field strengthat the center given as

r

I B o

2

r

r = radius of the coil

(m)

r NI B o

2=

* For N loops / numberof turns on the coil


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Example 19.2.3

A circular coil having 400 turns of wire in airhas a radius of 6 cm and is in the plane of thepaper. What is the value of current must existin the coil to produce a flux density of 2 mT atits center ?


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A solenoid

Magnetic field strengthat the center

nI B o

L

N n = where

= number of turnsper length

L


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Example 19.2.4


An air-core solenoid with 2000 loops is 60 cmlong and has a diameter of 2.0 cm. If acurrent of 5.0 A is sent through it, what will bethe flux density within it ?

nI B o= L N

n = where


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Exercise


1. A solenoid is constructed by winding 400 turns ofwire on a 20 cm iron core. The relative permeability ofthe iron is 13000. What current is required toproduce a magnetic induction of 0.5 T in the center of

the solenoid ?

1-2-7-

r mH10x63.1)10x4(x)13000(

core,theof ty permeabiliThe

o


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2. A student is provided with a 3.0 m long wire witha current of 0.15 A flowing through it. What is thestrength of the magnetic field at the centre of thewire if the wire is bent into a circular coil of oneturn ? ( B = 1.97 x 10 -7 T )

3. A circular coil has 15 turns and a diameter of45.0 cm. If the magnetic field strength at thecentre of the coil is 8.0 x 10 -4 T, find the current

flowing in the coil. ( I = 19.1 A )( 0 = 4 x 10 -7 Hm -1 )


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19.3 Force on a moving charged particlein a uniform magnetic field.

A charge q moving with speed v at angle with the direction of a uniform magnetic field ofmagnitude B experiences a magnetic force

of magnitude,sin= Bqv F

sin

x

qvB F

Bvq F

* Where = angle between B and v* For electron , q = e.

19 3 F i h d i l i if i fi ld


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19.3 Force on a moving charged particle in a uniform magnetic field.

sin= Bqv F Direction of F :

Flemings right hand rule : - negative charge Flemings left hand rule : - positive charge

Thumb direction of Force ( F )

First finger direction of Magnetic field ( B )

Second finger direction of Velocity ( v )

negative charge

B

v

F

positive charge

B

v

F

19 3 F i h d ti l i if ti fi ld


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Determine the direction of the magneticforce , exerted on a charge in each problembelow.

B

v B v


Example 19.3.1

B v

X X X X

X X X X

X X X X

v

I

a. b.

c.d.


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19 3 Force on a moving charged particle in a uniform magnetic field


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Exercise1. Calculate the magnitude of the force on a

proton travelling 3.1 x 10 7 m s -1 in the uniformmagnetic flux density of 1.6 Wb m -2,if :(i) the velocity of the proton is perpendicular

to the magnetic field.

(ii) the velocity of the proton makes an angle 60

with the magnetic field.(charge of the proton = +1.60 x 10 -19 C)

N F 10x9.7 12-

N F 12-10x9.6

sin= Bqv F



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sin= Bqv F Example 19.3.4

A charge q1 = 25.0 C moves with a speed of4.5 x 10 3 m/s perpendicularly to a uniformmagnetic field. The charge experiences a

magnetic force of 7.31 x 10-3

N. A second chargeq2 = 5.00 C travels at an angle of 40.0 o withrespect to the same magnetic field andexperiences a 1.90 x 10 -3 N force. Determine

(i) The magnitude of the magnetic field and(ii) The speed of q 2.



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sin= Bqv F Solution 19.3.4

q1 = 25.0 C , v1 = 4.5 x 10 3 m/s, 1= 90.0 o F 1 = 7.31 x 10 -3 N, q2 = 5.00 C, 2 = 40.0 o, F 2 = 1.90 x 10 -3 N force.

(i) 22-

11

11 T10x50.6 Bvq

F B B

(ii) v2 = 9.10 x 10 3 m/s



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Circular Motion of a Charged Particle in aUniform Magnetic Field

Consider a charged particle moving in a uniformmagnetic field with its velocity ( v ) perpendicularly tothe magnetic field ( B ).

As the particle enters the region, it will experience amagnetic force ( F ) which the force is perpendicular tothe velocity of the particle. Hence the direction of itsvelocity changes but the magnetic force remainsperpendicular to the velocity.

This magnetic force causes the particle to move ina circle .



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vv

B F

v B F

X X X X

X X X X

X X X X

X X X XB into thepage

v v

B F

v

B F

B out ofthe page

Circular Motion of a Charged Particle in a Uniform Magnetic Field


The magnetic force provides the centripetalforce for the particle to move in circular motion.

c B F F r

mv Bqv

2

sin

90 Bqmv

r r= ?m =?



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The time for one rotation (period),

vr 2

T

Bq

mvr

Bqm2

T

f 1

T

m2 Bq

f

and

and



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Exercise (DIY)1. A proton is moving with velocity 3 x 10 5 m/svertically across a magnetic field 0.02 T.(mp = 1.67 x 10 -27 kg)

Calculate ;

a) kinetic energy of the protonb) the magnetic force exerted on the proton

c) the radius of the circular path of theproton.

7.52 x 10 -17 J , 9.6 x 10 -16 N, 0.16 m



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Exercise

1. An electron is projected from left to right into amagnetic field directed into the page. The

velocity of the electron is 2 x 106

ms-1

and themagnetic flux density of the field is 3.0 T. Findthe magnitude and direction of the magneticforce on the electron.

(charge of electron = 1.6 x 10 -19 C)(9.6 x 10 -13 N, downwards)


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2. A proton with a mass of 1.67 x 10 -27 kg is

moving in a circular orbit perpendicular to amagnetic field. The angular velocity of the protonis 1.96 x 10 4 rad s -1 . Determine ;(i) the period of revolution,(ii) the magnetic field strength of the field.(charge of proton = 1.6 x 10 -19 C)

(T = 3.2 x 10 -4 s , B = 2.05 x 10 -4 T)


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19.4 Force on a current- carrying

conductor in a uniform magnetic

field (1 hour)


(i) Use force, B L I F

19 4 F t i g


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19.4 Force on a current- carryingconductor in a uniform magneticfield

When a current-carrying conductor isplaced in a magnetic field B, thus amagnetic force will act on that conductor.

The magnitude of the magnetic forceexerts on the current-carrying conductor isgiven by

In vector form , BIL F sin

B L I F

19.4 Force on a current-carrying conductor in a uniform magnetic field.


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19.4 Force on a current carrying conductor in a uniform magnetic field.

BIL F sin

forcemagnetic: F

densityfluxmagnetictheof magnitude: B

current: I

B I andof directionbetweenangle:conductor theof length: L

Direction of F : Flemings left hand rule.

B

I

F

Thumb direction of Force ( F )First finger direction of Magnetic field ( B )

Second finger direction of Current (I )



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y g g

BIL F sin

F = 0 when =0 B0 I

B

90 I

F is maximum when =90 o

90 BIL F sinmax BIL F max

0 BIL F sin0 F



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Determine the direction of the magnetic force ,

exerted on a conductor carrying current, I in eachproblem below.a. b.

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.

B I

X X X X

X X X X

X X X X

F (to the left)

B I

X X X X

X X X X

X X X X

F (to the right)

y g g

Example 19.4.1

a.



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A wire of length 0.655 m carries a currentof 21.0 A. In the presence of a 0.470 Tmagnetic field, the wire experiences aforce of 5.46 N . What is the angle (less

than 90o

) between the wire and themagnetic field?

BIL F 1-sin=

y g g

Example 19.4.2

BIL F sin



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1. A square coil of wire containing a single turn isplaced in a uniform 0.25 T magnetic field. Eachside has a length of 0.32 m, and the current inthe coil is 12 A. Determine the magnitude of the

magnetic force on each of the four sides.

y g g

Exercise BIL F sin

90 oB

I0.96 N (top and bottom sides)0 N (left and right sides)


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


(i) Derive force per unit length of two parallel

current-carrying conductors.

(ii) Use force per unit length,

(iii) Define one ampere. d

I I

L

F

2

210

19 5 Forces between two parallel


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19.5 Forces between two parallelcurrent- carrying conductors.

Consider two identical straight conductors Xand Y carrying currents I 1 and I 2 with length L are placed parallel to each other as shownbelow.

The conductorsare in vacuumand theirseparation is d .

d

2 I

2 I

1 I

1 I

X Y

1 B

2 B

P 12 F

21 F Q

19.5 Forces between two parallel current- carrying conductors.


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The magnitude of the magnetic flux density, B1 atpoint P on conductor Y due to the current in

conductor X is given by

d

2 I

2 I

1 I

1 I

X Y

1 B

2 B

P 12 F

21 F Q

Conductor Y

carries a current I 2 and in the magneticfield B1 thenconductor Yexperiences amagnetic force, F 12 .

d 2 I

B 101

into the page



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The magnitude of F 12 :

d

2 I

2 I

1 I

1 I

X Y

1 B

2 B

P 12 F

21 F Q

d I

B2

101 =sin= L I B F 2112

902 210

12 sin= L I d I

F

90

L I d I

F 21012 2= to the left(towards X)

The magnitude of F 21 : sin= L I B F 1221d

I B

220

2 =

902 120

21 sin= L I d I

F

L I d

I F 1

2021 2

=

to the right (towards Y)

d L I I

F F F 2

2102112 ===

Attractive force


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d L I I

F F F 2

2102112 === Rearrange, d

I I L F

2

210

If I 1 = I 2 = 1 A and d = 1 m , then

N/m10x2

121110x4

7-

-7

=

)())((=

L

F

L F

Definition of 1 Ampere



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One ampere is defined as the constantcurrent that, when it is flowing in each oftwo infinitely long, straight, parallel

conductors which have negligible of crosssectional areas and are 1.0 metre apart invacuum, would produce a force per

unit length between the conductors of2.0 x 10 -7 N m -1.

Definition of 1 Ampere



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Example 19.5.1

Two very long parallel wires are placed 2.0 cmapart in air. Both wires carry a current of 8.0 Aand 10 A respectively. Find the magnitude of

the magnetic force in newton, on each metrelength of wire.

d 2

L I I F 210


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l ( h )


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19.6 Torque on a coil (1 hour)


(i) Use torque , where N = number ofturns.

(ii) Explain the working principles of a moving coil

galvanometer.

B A NI

19 6 Torque on a coil


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19.6 Torque on a coil. Consider arectangular loop

with length a andwidth b is pivotedso that it can

rotate about avertical axis(shown in figure)which is at rightangle to a uniformmagnetic field offlux density B.

Axis of rotation

Top view


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Th t g ti19.6 Torque on a coil.


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The two magneticforces on sides 2

and 4 each oflength a are equaland opposite andhave the value F where,F 2 = F 4 = BI L = BI a

The forcesexerted a torquethat tends to rotate

the coil clockwise.

Side view

19.6 Torque on a coil.


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The magnitude ofthis torque for each

side is

turns N

forsin

sinsin

)sin2

()sin2

(

sin2

sin2

42

NBIA

BIA BIab

b BIa

b BIa

b F

b F

Fd

ab=A (area of the coil) = angle between B andthe normal to plane of the coil

B A NI



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A 20 turns rectangular coil with sides 6.0 cmx 4.0 cm is placed vertically in a uniformhorizontal magnetic field of magnitude 1.0 T.If the current flows in the coil is 5.0 A,determine the torque acting on the coil whenthe plane of the coil is(a) perpendicular to the field,(b) parallel to the field,(c) at 60 to the field.

Example 19.6.1



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Solution 19.6.1

N = 20 turns , A = 24 x 10-4

m2 ,

B = 1.0 T , I = 5.0 A

B

A90= B

A

90= B

A

30=60=

sin= NABI

(a) (b) (c)

0= mN240.= mN120.=



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Exercise

1. A rectangular loop of wire has an area of0.30 m 2 . The plane of the loop makes anangle of 30 o with a 0.75 T magnetic field.What is the torque on the loop if the currentis 7.0 A ?Solution

sin NABI

B

A

60=30=

m N 36.1



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Exercise

2. Calculate the magnetic flux densityrequired to give a coil of 100 turns a torqueof 0.5 Nm when its plane is parallel to thefield. The dimension of each turn is 84 cm 2 ,and the current is 9.0 A.

sin= NABI

Solution

o90=

mT 1.66 B



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A moving coil galvanometer

Structure of a moving-coil galvanometer

Structure of a moving-coil galvanometer



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The galvanometer is the main component inanalog meters for measuring current andvoltage.

It consists of a magnet, a coil of wire, aspring, a pointer and a calibrated scale.


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When there is a current in the coil, the coilrotates in response to the torque ( = NABI )applied by the magnet.

This causes the pointer ( attached to thecoil) to move in relation to the scale.

The basic operation of the galvanometeruses the fact that a torque acts on a currentloop in the presence of a magnetic field.



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A moving coil galvanometer The torque experienced by the coil isproportional to the current in it; the larger thecurrent, the greater the torque and the morethe coil rotates before the spring tightensenough to stop the rotation.

Hence, the deflection of the pointer attached

to the coil is proportional to the current .

The coil stops rotating when this torque is

balanced by the restoring torque of the spring.


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Exercise


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Exercise

1. The moving coil of a galvanometer has 100turns and an area of 1.5 x 10 -4 m 2 . It issuspended by a wire with a torsional constantof 2.6 x 10 -8 Nm rad -1 . The coil is placed in a

radial magnetic field of 0.1 T. Calculate thecurrent flowing in the coil if a deflection of 1.2rad is observed.

( I = 2.08 x 10 -5 A )

19 7 Motion of charged particle in


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19.7 Motion of charged particle inmagnetic field and electric field (1hour)


(i) Explain the motion of a charged particle in both

magnetic field and electric field.

(ii) Derive and use velocity, in a velocity

selector. B E v

19.7 Motion of charged particle in


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g pmagnetic field and electric field

Consider a charged particle +q moves witha velocity v in combined electric andmagnetic fields (the electric and magneticfields are perpendicular), the particleexperiences no resultant force ( a = 0).

The particle will continue to move in the

same direction with the same velocity. For this to happen, the electric forcedownward must balance the magnetic force

upwards (refer diagram), 19.7 Motion of charged particle in magnetic field and electric field


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B E

v

BqvqE

F F B E o90,

xx

xx

xx

x

x

x x

x

x

x x x x

x

x

x

x

x

x

x

x

x x xxxx

+ + + + + + +

- - - - - - -

+ +v vF E

F B

sin= Bqv F B

19.7 Motion of charged particle in magnetic field and electric field


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Velocity Selector

A velocity selector uses this property ofcrossed electric and magnetic fields toselect a single velocity of particle; only

particles traveling at this velocity will beundeflected.



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Velocity selector



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Example 19.7.1

What is the velocity of protons (+1 e)injected through a velocity selector ifE = 3 x 10 5 V/m and B = 0.25 T ?

Solution= 1.20 x 10 6 m/s



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Exercise

1. A velocity selector is to be constructedto select ions (positive) moving to the rightat 6.0 km/s. The electric field is 300 Vm -1

upwards. What should be the magnitudeand direction of the magnetic field?

Solution

pageof outT,05.0v

E B



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When the magnetic field only is applied,the particle moves in an arc of a circle ofradius r under the action of the centrally-directed magnetic,

r B

E

m

q B

E v

r

mv Bqv

F F c B

2

2

but

=

==

=



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Both E andB are

applied

Only B isapplied



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A


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A mass spectrometer is a device used forseparating atoms or molecules accordingto their mass.

The atoms or molecules are ionized and

then accelerated through an electric field,giving them a speed which depends ontheir mass (their kinetic energies are allthe same).

Then they enter a region of uniformmagnetic field, which bends them in acircular path.

A mass spectrometer


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A mass spectrometer

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