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Chapter 21: Magnetic Forces and Fields
• Magnetic fields
• 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
• Omit 21.9, magnetic materials
Wednesday, February 6, 2008
1
Magnetic poles, north and south
Magnetic poles occur only as
N-S pairs, unlike electrical
charges that exist separately
as + and – entities.
Existence of a “magnetic
monopole”?
Only a theoretical possibility
at present.
Wednesday, February 6, 2008
2
The north pole (“north-seeking 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
Magnetic field lines form a pattern
similar to an electric dipole.
Wednesday, February 6, 2008
3
Magnetic field lines of the earth
the north pole of the earth is a magnetic south pole...
Wednesday, February 6, 2008
4
Force on a positive charge moving in a
magnetic field
!
q
F = qvB sin !
Magnetic field
Force is at right angles to both !v and !B
Right hand rule
!B
!v
F in Newtons
Palm (Push, i.e. force)
Fingers (Field)
q in Coulombs
v in m/s
thuMb (Motion of + charge)
B in Tesla (T)
Earth’s magnetic field ! 10-4 T
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5
21.C5: Determine whether each particle is positively or negatively
charged, or neutral.
#1: positive
!F
#2: neutral
#3: negative
!F
Wednesday, February 6, 2008
6
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.
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
Wednesday, February 6, 2008
7
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.
Wednesday, February 6, 2008
8
Motion of a charge in a uniform magnetic field
As B is at right angles to v:
F=qvB
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 –
Fc =
mv2
= qvB
r
So, r =
Wednesday, February 6, 2008
mv
qB
9
Prob. 21.11/12:
a) Is q positive or negative?
Right hand rule: q is negative
b) If the magnitude of q is 8.2"10-4 C,
what is the mass of the particle?
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
Wednesday, February 6, 2008
10
Mass spectrometer, to measure mass of ions
In the magnetic field:
qBr eBr
=
v
v
1 2
and mv = eV
2
!
so, v = 2eV /m
m=
KE = eV
!F
eB2r2
m=
2V
–V
m, q = e
v!0
Removes an electron from neutral atoms,
forming singly-charged positive ions
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11
Mass Spectrum
Isotopes of neon
eB2r2
m=
2V
⇐ atomic mass of isotopes of neon
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12
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.
Both species are accelerated through the same electric potential
difference and experience the same magnetic field, which causes
them to move on circular paths.
Find the ratio of the radii of the paths.
For a particle of charge q : m =
qB 2 r2
2V
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Velocity Selector
Magnetic Field
The charge +q feels an upward
force due to the magnetic field.
13
F = qvB
Electric Field
The charge +q feels a downward
force due to the electric field.
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
Wednesday, February 6, 2008
F = qE
14
Force on a current-carrying conductor
Current is a flow of charge.
Imagine + charges moving
along the conductor in the
direction of the current, I.
There is a force F acting on
the charges moving along the
conductor that is due to the
magnetic field.
The direction of the force is
given by the right hand rule.
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15
Force on a current-carrying conductor
A charge Δq flows at speed v along a length
L of wire in time Δt.
I=
!q
!t
and
L = vΔt
The force on the charge Δq is:
F = !q v B sin " =
F = ILB sin !
Wednesday, February 6, 2008
!
"
!q
(v !t) B sin "
!t
I
!v
Δq
L
16
By how much does the weight of the
coil of wire seem to increase when
the current is turned on in the coil?
B = 0.2 T, I = 8.5 A
I
F
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Force per turn on lower end of loop:
17
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
F = 125 " (8.5 A) " (0.015 m) " (0.2 T) = 3.19 N
Wednesday, February 6, 2008
18
MasteringPhysics
01/02/08 11:58 AM
Power Dissipation in Resistive Circuit Conceptual Question
A single resistor is wired to a battery as shown in the diagram below.
by this circuit as .
Define the total power dissipated
Mastering Physics printing bug!
Problem 1, assignment 2:
If you print out the problem with the usual
print command, you may get the wrong
diagram!!! On screen, diagram is correct.
Click print
icon at top
left instead!
Diagram for part B
Text for part A
Jim Birchall
Now, a second identical resistor is wired in series with the first resistor as shown in the second diagram to
the left
.
Part A
What is the power, in terms of , dissipated by this circuit?
Express your answer in terms of
.
Wednesday, February 6, 2008
19
http://session.masteringphysics.com/myct
Page 1 of 2
Force on a charge moving in a magnetic field
!F
F = qvB sin !
Force is at right angles to both !v and !B
!B
!
q
Magnetic field
!v
Force on a current in a
magnetic field
F = !q v B sin " =
!
"
!q
(v !t) B sin "
!t
I
!v
Δq
L
F = ILB sin !
Wednesday, February 6, 2008
20
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
0.32 m
F=0
F=0
L
L
I
L = 0.32 m
" F
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21
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.
" F
L
h
900
Wednesday, February 6, 2008
. F
θ = 550
22
21.34: B = 0.05 T, vertical
m = 0.2 kg = mass of bar
m
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?
• 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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23
Charges inside the moving rod experience a
force due to the magnetic field...
–––
++
+
I
!v
I
I
I
Conductor
The moving conductor acts as a generator.
The basis of electromagnetic induction (next chapter)
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24
Force on a current-carrying conductor
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.
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25
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.
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.
B = 0.1 T
m = 0.02 kg
d = 0.025 m
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26
Magnetohydrodynamics
(MHD)
Ions within the water
allow a current to flow
when a potential
difference is applied.
The moving ions feel a
force due to the
magnetic field that
pushes the water out
through the back of the
boat.
The boat feels an equal
and opposite reaction
force to the front.
Propulsion with no moving parts.
Wednesday, February 6, 2008
27
The Torque on a Current-Carrying Coil
! = angle
between normal
to coil and B
L
From above
w
w
L
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
If the coil has N turns: ! = NIAB sin "
Wednesday, February 6, 2008
(A = Lw)
28
If the coil has N turns: ! = NIAB sin "
The formula is valid for any shape of coil, not just rectangular.
NIA is known as the “magnetic moment” of the coil.
! = 90◦
! = 0o
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29
Variation of torque with φ
Torque: ! = NIAB sin "
1
sin(pi*phi/180)
0.8
Torque, τ 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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30
Reverse the current at the right times...
...the torque is always in the same direction and the motor continues to rotate
1
sin(pi*phi/180)
0.8
Torque, τ0.6
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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Electric motor
! = 0◦
1
2
1
2
Commutator reverses direction of current when φ = 0o, 180o...
Torque continues in the same direction
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32
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.
I = 4.4 A
B = 1.8 T
! = 90º – 35º
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33
The next few weeks...
Ch. 18-22, 24, 25
Wednesday, February 6, 2008
34
The Torque on a Current-Carrying Coil
! = angle
between normal
to coil and B
L
From above
w
w
L
Area of coil, A = Lw
The force on each vertical arm of the coil is: F = ILB
Torque exerted on the coil: τ = N IAB sin φ
N = number of turns, A = area of coil
Wednesday, February 6, 2008
35
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.
• 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
Wednesday, February 6, 2008
L
36
Magnetic field produced by currents
I
Right hand rule #2: the magnetic field wraps
around the wire in the direction indicated by the
fingers of the right hand.
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37
Magnetic field produced by currents
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
Wednesday, February 6, 2008
B
Fingers show
direction in
which the
magnetic field
wraps around
the wire
38
21.C18: Currents of the same magnitude flow in each of the wires
that run perpendicular to the plane of the screen. 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.
X
!B4
!B1 + !B3 = 0
!B3
!2 + B
!4 = 0
B
!B1
!2
B
X
Currents at opposite
corners should be in
the same direction
Wednesday, February 6, 2008
39
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?
• What is the current?
Wednesday, February 6, 2008
40
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.
Wednesday, February 6, 2008
41
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.
What current I3 is needed in wire 3 to make the magnetic field at
the empty corner equal to zero?
!B3
I1
a
45º
45º
L
a
45º
45º
!B1
!B2
!B1 + !B2
a
|!B3| = |!B1 + !B2|
I3
Wednesday, February 6, 2008
a
I2
42I1
Force on a charge moving parallel to a conductor
I
B=
B
F = q0vB
F = q0 vB = q0 v
Direction of
force on the
charge
µ0 I
2πr
µ0 I
2πr
Direction in
which the
magnetic
field wraps
around the
wire
Replace the moving
charge by a current
→ attractive force
between currents in
same direction
Wednesday, February 6, 2008
43
Force on a charge moving in a magnetic field
!F
F = qvB sin !
Force is at right angles to both !v and !B
!B
!
q
Magnetic field
!v
Force on a current in a
magnetic field
F = !q v B sin " =
!
"
!q
(v !t) B sin "
!t
I
!v
Δq
L
F = ILB sin !
Wednesday, February 6, 2008
44
Force between long current-carrying wires
!B1
!B1
L
d
L
d
At wire 2: F = I2LB1 = I2L
µ0I1 µ0I1I2
=
L
2!d
2!d
(Due to I1 in wire 1, at position of wire 2)
Force per unit length of wire 2 is: µ0I1I2
2!d
Equal and opposite force on wire 1 due to current in wire 2
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45
21.C14: The two outer wires are held in place.
Which way will the middle wire move?
!F23
!F12
1
Wednesday, February 6, 2008
2
3
46
Force between current-carrying conductors
I1
I2
F!
−F!
!
B1 =
x B1
µo I1
2πd
F = I2 LB1
d
Force per unit length of wire:
F =
µ0 I1 I2
2πd
Force is attractive when the
currents are in the same direction.
Wednesday, February 6, 2008
47
21.C12: A conducting wire is wound into a helical shape. The helix acts
as a spring and expands back toward its original shape after the coils
are squeezed together and released. The bottom end of the wire
just barely touches the mercury (a good electrical conductor) in the
cup. After the switch is closed, current in the circuit causes the
light to glow. Does the bulb glow continually, glow briefly and then go
out, or repeatedly turn on and off like a turn signal on a car?
Wednesday, February 6, 2008
48
Prob. 21.54: Find the net force acting on the wire loop that is beside
the long, straight wire.
Find the forces acting on wire loop due to current I1.
I1 = 12 A
!F1
(1)
I2 = 25 A
"!
B1
(2)
B1 is due to I1
!F2
I2 = 25 A
"!
B1
Wednesday, February 6, 2008
49
Magnetic field at centre of current loop of radius R
B=
µ0I
N
2R
at centre of a coil of radius R with N turns
Wednesday, February 6, 2008
50
Magnetic field at centre of current loop of radius R
N
S
B=
µ0I
N
2R
at centre of a coil of radius R with N turns
Wednesday, February 6, 2008
51
Magnetic field produced by currents
B=
µ0I
2!r
B=
µ0I
N
2R
I
B
Wednesday, February 6, 2008
52
21.-/53:
!B2 !B1
"
Need
B1 = B2
I2 = 6.6 I1
At the centre of the loop, B = 0. What is H?
Due to the loop: B1 =
For B1 = B2 :
µ0I1
2R
Due to the wire: B2 =
I1
I2
=
R !H
So, H = R
µ0I2
2!H
I2
6.6
=
R = 2.1R
!I1
!
Wednesday, February 6, 2008
53
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 P at the common centre is zero?
I1 = 7.2 A
P
r1 = 0.015 m
I2 = ?
r2 = 0.023 m
N2 = 180
N1 = 140
Wednesday, February 6, 2008
54
Magnetic field from a solenoid
Inside the solenoid
B = µ0nI
n = number of turns per unit length of the solenoid
Magnetic field is very uniform at centre of solenoid
Wednesday, February 6, 2008
21.C15:
Attractive
or repulsive
forces?
55
S
S
21.C16:
Attractive
or repulsive
forces?
S
N
Wednesday, February 6, 2008
N
N
NS
N
S S
N
56
Ampère’s Law
Construct a closed path around the
two wires.
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
Then:
! B||"l = µ0I
I is the net current passing
through the surface,
I = I 1 + I 2.
Wednesday, February 6, 2008
57
Field from a long, straight wire, using Ampère’s Law
Choose a circular path of radius r
around the wire.
B is the same all the way around.
so, ! B||"l = B × 2#r = µ0I
B=
µ0I
2!r
As before!
Wednesday, February 6, 2008
58
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
"l
!B
B|| = 0 ! I = 0, zero net
current passes
through the
surface
Wednesday, February 6, 2008
59
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=
µ0I
2!r
b) Show that the field is zero inside the cylinder.
Wednesday, February 6, 2008
60
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 θ?
(Hint: the magnetic force can be taken to act at the centre of
gravity of the rod).
F = ILB
There are
forces at the
hinge
F
L/2
90º
I
Centre of gravity
L/2
Floor
mg
Wednesday, February 6, 2008
61
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.
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 #s = 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?
Brake shoe
• Find out the maximum torque
exerted by the coil
• Equate it to the torque exerted
on the shaft by the brake shoe
Wednesday, February 6, 2008
62
Summary of chapter 21
v, I
Force on a moving charge:
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: B = #o I/2!r
Magnetic field at centre of a circular loop: B = #o NI/2R
Magnetic field at centre of a solenoid:
B = #o nI
Ampere’s law:
! B||"l = µ0I
Wednesday, February 6, 2008
63