Download Chapter 21: Magnetic Forces and Fields Magnetic poles, north and

Magnetic poles, north and south
Chapter 21: Magnetic Forces and Fields
• Magnetic fields
Magnetic poles occur only as
N-S pairs, unlike electrical
charges that exist separately as
+ and – entities.
• The force on a moving charge and on a current in a magnetic field
• The torque on a coil in a magnetic field. Electric motors
• Magnetic field produced by a current. Ampere’s law
Existence of a “magnetic
monopole”?
• Omit 21.9, magnetic materials
Only a theoretical possibility at
present.
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Friday, February 2, 2007
The north pole of a compass needle points along the magnetic field lines toward
the south pole of a magnet
Compass needle
points along
magnetic field line
toward the magnetic
south pole
2
Magnetic field lines of the earth
Magnetic field lines form a pattern
similar to an electric dipole.
! the north pole of the earth is a magnetic south pole...
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Force on a charge moving in a magnetic field
F = qvB sin !
!
q
Force is at right angles to both !v and !B
Right hand rule
Magnetic field
21.C5: Determine whether each particle is positively or negatively
charged, or neutral.
!B
#1: positive
!F
!v
#2: neutral
F in Newtons
Palm (Push, i.e. force)
Fingers (Field)
#3: negative
q in Coulombs
v in m/s
thuMb (Motion of charge)
!F
B in Tesla (T)
Earth’s magnetic field ! 10-4 T
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21.5/2: B = 0.3 T, v = 45 m/s, q = 8.4 µC. Find the force on the charge in
the three cases below.
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Motion of a charge in an electric field
There is in general a component of the
force in the direction of the displacement,
so the force does work and the speed of
the charge changes.
Motion of a charge in a magnetic field
The force is always perpendicular to
the velocity (and displacement), so the
force does no work and the charge
moves at constant speed. The charge
follows a circular path if the magnetic
field is uniform.
F = qvB sin !, qvB = 1.13 × 10−4 N
a) ! = 30◦, F = 0.57 × 10−4 N, into screen
b) ! = 90◦, F = 1.13 × 10−4 N, into screen
c) ! = 150◦, F = 0.57 × 10−4 N, into screen
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Motion of a charge in a uniform magnetic field
Prob. 21.11/12:
a) Is q positive or negative?
As B is at right angles to v:
Right hand rule: q is negative
F=qvB
b) If the magnitude of q is 8.2"10-4
C, what is the mass of the particle?
F always at right angles to velocity
! it does no work, v is constant
The circular path has radius
determined by the centripetal force
provided by the magnetic field –
mv2
Fc =
= qvB
r
r=
So, m =
v = 140 m/s
B = 0.48 T
r = 960 m
mv
qB
qBr 8.2 × 10−4 × 0.48 × 960
=
v
140
= 2.7 × 10−3 kg
mv
So, r =
qB
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Motion of a charge in a uniform magnetic field
Force on a charge moving in a magnetic field
!
Force is at right angles to both !v and B
As B is at right angles to v:
F!
q
F = qvB sin !
!
Right hand rule
Palm (Push, i.e. force)
Fingers (Field)
F=qvB
Magnetic field
!v
!B
F always at right angles to velocity
! it does no work, v is constant
F in Newtons
The circular path has radius
determined by the centripetal force
provided by the magnetic field –
mv2
Fc =
= qvB
r
q in Coulombs
v in m/s
thuMb (Motion of charge)
B in Tesla (T)
So, r =
Earth’s magnetic field ! 10-4 T
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mv
qBr
and, m =
qB
v
12
Mass spectrometer, to measure mass of ions
Mass Spectrum
In the magnetic field:
Isotopes of neon
qBr eBr
=
v
v
1 2
and mv = eV
2
!
so, v = 2eV /m
m=
!F
KE = eV
m=
m=
eB2r2
2V
eB2r2
2V
–V
m, q = e
v!0
" atomic mass of isotopes of neon
Removes an electron from neutral atoms,
forming singly-charged positive ions
21.18
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Velocity Selector
Prob. 21.18: The ion source in a mass spectrometer produces both singly
and doubly ionized species, X+ and X2+. The difference in mass is too
small to detect.
Magnetic Field
The charge +q feels an upward
force due to the magnetic field.
Both species are accelerated through the same electric potential difference
and experience the same magnetic field, which causes them to move on
circular paths.
F = qvB
Electric Field
The charge +q feels a downward
force due to the electric field.
Find the ratio of the radii of the paths.
With both the B and E fields
present, the net force on the charge
is zero if:
qvB = qE, that is,
v = E/B " measurement of velocity,
#
independent of q, m
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F = qE
16
Force on a current-carrying conductor
Force on a current-carrying conductor
Current is a flow of charge.
A charge $q flows at speed v along a length L
of wire in time $t.
Imagine + charges moving
along the conductor in the
direction of the current, I.
I=
!q
!t
and L = v$t
The force on the charge #q is:
! "
!q
F = !q v B sin " =
(v !t) B sin "
!t
There is a force F acting on
the charges moving along the
conductor that is due to the
magnetic field.
F = ILB sin !
I
!v
$q
L
The direction of the force is
given by the right hand rule.
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Friday, February 2, 2007
Ex 13:
Force per turn on lower end of loop:
By how much does the weight of the
coil of wire seem to increase when
the current is turned on in the coil?
18
B = 0.2 T, I = 8.5 A
125 turns
F = I L B, downwards
There are N = 125 turns in the coil, so
total force is:
F = N I L B = 125 I L B
I
F
B = 0.2 T, I = 8.5 A
F = 125 " (8.5 A) " (0.015 m) " (0.2 T) = 3.19 N
I
F
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Prob. 21.30: The triangular loop of wire carries a current, I = 4.7 A and
B = 1.8 T. Find the force acting on each side and the net force.
Prob. 21.29: A single turn coil is placed in a uniform magnetic field of
0.25 T. Each side of the coil is of length L = 0.32 m. I = 12 A.
Determine the magnetic force on each side of the coil.
"F
. F
0.32 m
F=0
l
F=0
L
h
L
. F
% = 550
I
"F
L = 0.32 m
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Force on a current-carrying conductor
21.34:
B = 0.05 T, vertical
m = 0.2 kg = mass of bar
I
L
v constant
A current I flows through the bar. The bar slides down the rails without
friction at constant speed.
What is I? In what direction does it flow?
An alternating current in the voice coil causes the cone of the
loudspeaker to move in and out and generate sound – an electrical signal
is converted into a sound wave.
• The component of the weight down the plane must be equal to the
# component of the magnetic force up the plane...
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Ex 5
23
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Force on a charge moving in a magnetic field
Ex. 5: The voice coil of a speaker has a diameter d = 0.025 m, contains
55 turns of wire and is placed in a magnetic field of 0.1 T. The current in
the coil is 2 A.
!F
F = qvB sin !
!B
!
Magnetic field
q
Force is at right angles to both !v and !B
a) Find the magnetic force that acts on the coil and cone.
b) The voice coil and cone have a combined mass of 0.02 kg. Find their
# acceleration.
!v
Force on a current in a
magnetic field
F = !q v B sin " =
!
"
!q
(v !t) B sin "
!t
I
!v
$q
L
F = ILB sin !
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Force on a charge moving in a magnetic field
Charges inside the moving rod experience a force
due to the magnetic field...
!F
F = qvB sin !
!B
!
Magnetic field
q
Force is at right angles to both !v and !B
–––
++
+
Force on a current in a
magnetic field
I
!v
I
I
!v
I
F = !q v B sin " =
!
"
!q
(v !t) B sin "
!t
Conductor
I
The moving conductor acts as a generator.
!v
$q
L
F = ILB sin !
The basis of electromagnetic induction (next chapter)
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The Torque on a Current-Carrying Coil
Magnetohydrodynamics
(MHD)
$ = angle
between normal
to coil and B
Ions within the water
allow a current to flow
when a potential
difference is applied.
w
L
The moving ions feel a
force due to the magnetic
field that pushes the water
out through the back of
the boat.
L
w
Area of coil, A = Lw
The force on each vertical arm of the coil is: F = ILB
Fw
Fw
The torque is: τ =
sin φ +
sin φ = I(Lw)B sin φ = IAB sin φ
2
2
(A = Lw)
If the coil has N turns: ! = NIAB sin "
The boat feels an equal
and opposite reaction
force to the front.
Propulsion with no moving parts!
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Variation of torque with $
If the coil has N turns: ! = NIAB sin "
Torque: ! = NIAB sin "
The formula is valid for any shape of coil, not just rectangular.
1
NIA is known as the “magnetic moment” of the coil.
! = 90◦
From above
sin(pi*phi/180)
0.8
Torque, &
$ = 0o
0.6
0.4
0.2
00
0
1800
3600
5400
!
-0.2
-0.4
-0.6
-0.8
-1
0
100
200
300
400
500
600
700
Torque changes sign! – rotation reverses, no good for an electric motor
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Reverse the current at the right times...
Electric motor
! = 0◦
...the torque is always in the same direction and the motor continues to rotate
1
sin(pi*phi/180)
1
0.8
2
1
2
Torque, &0.6
0.4
0.2
00
0
1800
3600
5400
!
-0.2
-0.4
-0.6
-0.8
-1
0
100
200
300
400
500
600
700
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Commutator reverses direction of current when $ = 0o, 180o...
Torque continues in the same direction
33
Prob. 21.41: The loop is free to rotate about the z-axis.
Find the torque on the loop and whether the 35º angle will increase or
decrease.
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Prob. 21.44: A square coil and a rectangular coil are made from the same
length of wire. Each contains a single turn. The long sides in the
rectangle are twice as long as the short sides.
Find the ratio of torques the coils experience in the same magnetic field.
I = 4.4 A
B = 1.8 T
• Find the relation between the lengths of the sides of the square and the
# rectangle, given that the perimeters are of the same length.
• Work out the areas of the two shapes and relate to the torques.
a = 2b
l
b
l
$ = 90º – 35º
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36
Magnetic field produced by currents
Magnetic field produced by currents
At a distance r from an infinitely long, straight wire:
I
B=
µ0I
2!r
Can be proved using
Ampere’s law (later)
µ0 = “permeability of free space”
= 4! × 10−7 T.m/A
B
Right hand rule: the magnetic field wraps around the
wire in the direction indicated by the fingers of the
right hand.
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21.C18: Currents of the same magnitude flow in each of the wires that
run perpendicular to the plane of the page. Choose the direction of the
current for each wire, so that when any single current is turned off, the
total magnetic field at point P at the centre is directed toward a corner
of the square.
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Prob. 21.49/47: In a lightning bolt, 15 C of charge flows in 1.5 ms.
Assuming the lightning bolt can be represented as a long, straight line of
current, what is the magnitude of the magnetic field at a distance of 25 m
from the bolt?
X
• What is the current?
!B4
!B1 + !B3 = 0
!B3
!2 + B
!4 = 0
B
!B1
!2
B
X
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Currents at opposite
corners should be in the
same direction
39
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40
Prob. 21.59: The picture shows the end view of three wires that are
perpendicular to the screen. The currents in the wires are I1 = I2 = I.
Prob. 21.72: Two long, straight wires are separated by 0.12 m. They
carry currents of 8 A in opposite directions. Find the magnetic field at A
and at B.
What current I3 is needed in wire 3 to make the magnetic field at the
empty corner equal to zero?
!B3
a
I1
45º
45º
!B2
45º
45º
l
a
!B1
!B1 + !B2
a
|!B3| = |!B1 + !B2|
I3
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Magnetic field produced by currents
a
I2
I
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42 1
Force on a charge moving parallel to a conductor
At a distance r from an infinitely long, straight wire:
B=
µ0I
2!r
µ0 = “permeability of free space”
= 4! × 10−7 T.m/A
I
B
Fingers show
direction in
which the
magnetic field
wraps around
the wire
F = q0vB
Direction of
force on the
charge
Friday, February 2, 2007
µ0 I
2πr
Direction in
which the
B magnetic field
wraps around
the wire
I
B
B=
43
Friday, February 2, 2007
Replace the moving
charge by a current
! attractive force
between currents in
same direction
44
Force on a moving charge
21.C14: The two outer wires are held in place.
! attractive force
between currents in
same direction
I=3A
q0 = 6.5"10-6 C,
Which way will the middle wire move?
!F23
!F12
The charge is moving parallel to
the wire at a distance r = 0.05 m
at speed v = 280 m/s.
r
1
F = q0vB sin% = q0vB
µI
At position of qo: B = o
2!r
µ0I
So, F = q0v ×
2!r
F = (6.5 × 10−6 C) × (280 m/s) ×
F = 2.2 × 10−8 N
2
3
4! × 10−7 × (3 A)
2! × (0.05 m)
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Force between long current-carrying wires
!B1
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46
Prob. 21.54: Find the net force acting on the wire loop that is beside the
long, straight wire.
!B1
L
d
d
L
I1 = 12 A
!F1
µ0I1 µ0I1I2
At wire 2: F = I2LB1 = I2L
=
L
2!d
2!d
(1)
I2 = 25 A
" !B1
(2)
B1 is due to I1
(Due to I1 in wire 1, at position of wire 2)
!F2
Force per unit length of wire 2 is: µ0I1I2
2!d
I2 = 25 A
" !B1
Equal and opposite force on wire 1 due to current in wire 2
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48
Magnetic field from a current loop of radius R
21.-/53:
!B2 !B1
"
B1 = B2
I2 = 6.6 I1
At the centre of the loop, B = 0. What is H?
Due to the loop: B1 =
µ0I
B=
N
2R
for a coil of N turns
For B1 = B2 :
µ0I1
2R
I1
I2
=
R !H
Due to the wire: B2 =
So, H = R
µ0I2
2!H
I2
6.6
=
R = 2.1R
!I1
!
21.55
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50
Magnetic field from a solenoid
21.55: Two circular coils are concentric and lie in the same plane. The
inner coil contains 140 turns or wire, has a radius r1 = 0.015 m and
carries a current I1 = 7.2 A. The outer coil contains 180 turns and had a
radius r2 = 0.023 m.
What must be the current in the outer coil so that the net magnetic field
at the common centre is zero?
I1 = 7.2 A
P
I2 = ?
r2 = 0.023 m
N2 = 180
r1 = 0.015 m
Inside the solenoid
B = µ0nI
n = number of turns per unit length of the solenoid
N1 = 140
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52
21.C15:
Attractive or
repulsive
forces?
S
Construct a closed path around the
two wires.
S
21.C16:
Attractive or
repulsive
forces?
Ampère’s Law
N
S
N
NS
Split the path into short segments of
length #l, measure the magnetic field,
B|| in the direction of each segment.
Add up all of the B|| #l
N
Then: ! B||"l = µ0I
N
S S
Friday, February 2, 2007
I is the net current passing
through the surface, (I1 + I2).
N
53
Field from a long, straight wire, using Ampère’s Law
Choose a circular path of radius r around
the wire.
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Prob. 21.61/60: A uniform magnetic field is everywhere perpendicular to
the page. A circular path is drawn on the page. Use Ampere’s law to
show there can be no net current passing through the circular surface.
! B||"l = µ0I around the circle
B is the same all the way around.
!B
so, ! B||"l = B × 2#r = µ0I
B=
#l
µ0I
2!r
!B
B|| = 0
I = 0, zero net
current passes
through the surface
As before!
21.60
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Prob. 21.34: A thin, uniform rod 0.45 m long and of mass 94 g is attached
to the floor by a hinge at point P. A uniform magnetic field B = 0.36 T is
directed into the screen. The current in the rod is I = 4.1 A. The rod is in
equilibrium. What is the angle %?
(Hint: the magnetic force can be taken to act at the centre of gravity of the
rod).
Prob. 21.62: A very long, hollow cylinder is formed by rolling up a thin
sheet of copper. Electric charges flow along the copper sheet parallel to the
axis of the cylinder, forming a hollow tube carrying current I.
a) Use Ampere’s law to show that the magnetic field outside the cylinder at
# a distance r from the axis is:
B=
F = ILB
µ0I
2!r
b) Show that the field is zero inside the cylinder.
F
There are
forces at the
hinge
L/2
90º
I
Centre of gravity
L/2
Floor
mg
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58
Summary of chapter 21
Prob. 21.43: Coil: N = 410 turns; area per turn, A = 3.1 " 10-3 m2; current
I = 0.26 A. Magnetic field, B = 0.23 T.
Force on a moving charge: # #
A brake shoe is pressed against the shaft to keep the coil from turning.
The coefficient of static friction between the shaft and the brake shoe is
µs = 0.76. The radius of the shaft is 0.012 m. What is the magnitude of
the minimum normal force that the brake shoe exerts on the shaft?
#
F = qvB sin%
Force on a current-carrying conductor:# #
F = ILB sin%
Torque on a coil:# #
& = NIAB sin$
#
#
#
#
#
#
#
#
Magnetic field from a long, straight wire:#
Brake shoe
v, I
B = µo I/2!r
Magnetic field at centre of a circular loop:# B = µo NI/2R
Magnetic field at centre of a solenoid:#
• Find out the maximum torque
exerted by the coil
Ampere’s law:
#
B = µo nI
! B||"l = µ0I
• Equate it to the torque exerted on
the shaft by the brake shoe
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