C19 Magnetic Field Student
Transcript of 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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Magnet ic f ie ld l ines 19.1 Magnetic field
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.
Magnet ic f ie ld l ines 19.1 Magnetic field
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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19.2 Magnetic field ( B ) produced by current - carrying conductor
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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19.2 Magnetic field ( B ) produced by current - carrying conductor
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
19.2 Magnetic field ( B ) produced by current - carrying conductor
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19.2 Magnetic field ( B ) produced by current - carrying conductor
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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19.2 Magnetic field ( B ) produced by current - carrying conductor
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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19.2 Magnetic field ( B ) produced by current - carrying conductor
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
19.2 Magnetic field ( B ) produced by current - carrying conductor
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
19.2 Magnetic field ( B ) produced by current - carrying conductor
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
19.3 Force on a moving charged particle in a uniform magnetic field.
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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19.3 Force on a moving charged particle in a uniform magnetic field.
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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19.3 Force on a moving charged particle in a uniform magnetic field.
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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19.3 Force on a moving charged particle in a uniform magnetic field.
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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19.3 Force on a moving charged particle in a uniform magnetic field.
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 .
19 3 Force on a moving charged particle in a uniform magnetic field
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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
19.3 Force on a moving 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 =?
19 3 Force on a moving charged particle in a uniform magnetic field
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19.3 Force on a moving charged particle in a uniform magnetic field.
Circular Motion of a Charged Particle in a Uniform Magnetic Field
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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19.3 Force on a moving charged particle in a uniform magnetic field.
Circular Motion of a Charged Particle in a Uniform Magnetic Field
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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19.3 Force on a moving charged particle in a uniform magnetic field.
Circular Motion of a Charged Particle in a Uniform Magnetic Field
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)
Learning Outcomes : At the end of this lesson, the students should be able to ;
(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 )
19.4 Force on a current-carrying conductor in a uniform magnetic field.
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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
19.4 Force on a current-carrying conductor in a uniform magnetic field.
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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.
19.4 Force on a current-carrying conductor in a uniform magnetic field.
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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
19.4 Force on a current-carrying conductor in a uniform magnetic field.
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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)
Learning Outcomes : At the end of this lesson, the students should be able to ;
(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
19.5 Forces between two parallel current- carrying conductors.
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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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19.5 Forces between two parallel current- carrying conductors.
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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
19.5 Forces between two parallel current- carrying conductors.
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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
19.5 Forces between two parallel current- carrying conductors.
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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)
Learning Outcomes : At the end of this lesson, the students should be able to ;
(i) Use torque , where N = number ofturns.
(ii) Explain the working principles of a moving coil
galvanometer.
B A NI
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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
19.6 Torque on a coil.
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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
19.6 Torque on a coil.
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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.=
19.6 Torque on a coil.
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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
19.6 Torque on a coil.
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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
19.6 Torque on a coil.
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A moving coil galvanometer
Structure of a moving-coil galvanometer
Structure of a moving-coil galvanometer
19.6 Torque on a coil.
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A moving coil galvanometer
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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19.6 Torque on a coil.
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A moving coil galvanometer
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.
19.6 Torque on a coil.
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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 )
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19.7 Motion of charged particle inmagnetic field and electric field (1hour)
Learning Outcomes : At the end of this lesson, the students should be able to ;
(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
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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
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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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19.7 Motion of charged particle in magnetic field and electric field
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