Chapter 19 Magnetic Fields
and Forces
Chapter 19 topics
 Magnetic Fields
 Magnetic Forces on a point charge
 Motion of a charged particle perpendicular
to a uniform B field
 Charged particles moving in crossed E and
B fields
 Magnetic Force on a current carrying wire
 Torque on a current loop
 Magnetic field due to an electric current
Magnetic Fields--similarities
with Electric Fields




North and South poles
Like poles repel
Opposite poles attract
Field lines outside the
material move from N to S
•Direction of the magnetic field is tangent to
the field lines
•The magnetic Field is the strongest where
the field lines are the closest
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Magnetic Fields: differences
from Electric Fields
 No Magnetic monopoles
 Magnetic Field lines
are closed loops
Earth’s Magnetic Field
 Why? Theory:
Flow of molten
iron and nickel in
earth’s outer
core
 Poles move around!
10-40km per year
 Polarity reversal has happened 100
times in the last 5 million years
Magnetic Force on a point charge
 Magnitude
(
F =q v"B
v"
v
)
θ
B
= q v # B = q (v sin $ )B
= q v(Bsin $ ) = q vB#
!
!
v
B"
θ
Units: Tesla
1T=1 N/(A•m)=1 N/(C•m/s)
B
!
2
Magnetic Force on a point charge
 Direction: RIGHT Hand Rule
 Perpendicular to both v and B
 Here Into
or Out of
the page
 Run fingers along v, curl them towards B,
 If q is positive, thumb points along F
 If q is negative, thumb points opposite F
If q is +
If q is -
v"
v
v
θ
B"
θ
B
B
!
!
Magnetic Force on a point charge
 If q is + find the
direction of F
F
v
B
F
B
F
v
B
v
 Right Hand Rule
 Perpendicular to both v and B
 Run fingers along v, curl them towards B,
 If q is positive, thumb points along F
 If q is negative, thumb points opposite F
Useful Coordinate systems
N
up
E
W
down
S
S
N
W
up
E
down
up
E
W
N
S
down
3
Example 19.2 Magnetic Force on an
Ion in the air
 At a certain place, the Earth’s magnetic field has
magnitude 0.5 mT. The field direction is 70.0°
below the horizontal. Its Horizontal component
points due north.
 (a) Find the magnetic force on an oxygen ion (O-2)
movingdue east at 250 m/s
 (b) compare the magnitude of the magnetic force
to the ion’s weight (5.2x10-25N) and the electric
force on it due to the Earth’s fair-weather
electric field (150 V/m downward).
Example Magnetic force on a
proton in a uniform magnetic field
 In a physics lab we generate a magnetic field of
magnitude 2T at an angle of 30° above the
horizontal. The horizontal component points due
West.
 (a) Find the magnetic force on proton moving due
east at 2.50x 106 m/s
 (b) Find the magnetic force on an electron moving
2.50x 106 m/s due north
Example 19.3 Electron in a
Magnetic Field
 An electron moves with a speed 2.0 x106 m/s in a
uniform magnetic field of 1.4 T directed due
north. At one instant the electron experiences an
upward force of 1.6 x 10-13 N. In what direction is
the electron moving at that instant?
 (There is more than one possibility, find all of the
solutions.)
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Charged Particle moving
Perpendicular to a uniform
magnetic field
 Bubble chamber
 Nuclear and Particle physics
 Mass spectrometer
 Used to identify what elements are
present in a sample
 Cyclotron
 Nuclear and particle physics
 Medicine
 Produce radioisotopes
 Proton beam radiosurgery (tumors)
Charged Particle moving Perpendicular
to a uniform magnetic field
 Bubble chamber
Magnetic field into the page
F
+
v
v
-
v
v
F
F
F
Charged Particle moving Perpendicular
to a uniform magnetic field
 Mass spectrometer
# F = qvBsin" = ma
But v is $ to B so sin" = 1
qvB = ma
If the velocity is $to the Force
we have circular motion
Material is ionized so it has
the same charge. But the
mass and velocity are
different. We will fix that
later with a velocity selector!
qvB = ma c = m
r=
v2
r
mv 2
qvB
!
5
Charged Particle moving Perpendicular
to a uniform magnetic field
 Cyclotron
Cyclotron
1
v
B
F
3
B
+V
v
E
-V
2
+V
-V
E
4
B
acceleration
r=
mv 2
qvB
Increased v
-->Increased r
-V
B
!
+V
E
This works because the time to complete a cycle stays constant
A charged particle in
crossed E and B fields
 Velocity selector
 Mass spectrometer
 Electromagnetic flow meter
 Blood flow
 The Hall effect
 Moving charges in a solid (wire or
sample)
 Used to measure magnetic fields
6
E
Velocity Selector
Crossed E and B fields
Means E perpendicular to B
B
F = FE + FB
FB
= qE + q(v " B)
+
v
If a charged particle is not
deflected, F=0
FE
0 = qE + q(v " B)
!
= E + (v " B)
v=
v
B
E
B
!
!
Mass Spectrometer
mv 2
qvB
mv
E
=
and since v = 1
qB
B1
mE1
=
qB1B
r=
!
Electromagnetic Flowmeter
 Measures the speed of blood flow in
arteries during surgery
3. Ions separate
1. Apply B field
+
-
S
v
N
2. Ions experience a force
+ v
-
-
FBB
+ FB+
- - - - - - + v
B
E
+ + + + + + +
B
4. Produces an E field
5. Force E field
opposite
Force B field
6. Equilibrium is
reached and forces
are equal
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Electromagnetic Flowmeter
- - - - - - + v
B
E
+ + + + + + +
B
V
We know B since we applied it.
E is determined by measuring ΔV
with a voltmeter
E=ΔV/d
Hall Effect
-
There is no natural charge flow1) apply a current I
2) In the presence of a B field
3) Charges are deflected
4) Charge separation
creates an Electric
field that we can F = F
E
B
measure
qE = qvB
E
B
d
v = E/B
E=ΔVH/d where ΔVH
is the Hall Voltage
- - - - - - -
Same thing as flow meter only we have a solid
material (eg a wire) rather than an artery.
I
ΔV H
Hall Effect--Why?
We can use this to measure the magnetic field.
FE = FB
qE = qvB
v = E/B
-
E B
•E is determined by measuring
ΔV H with a voltmeter E=ΔV H/d;
The velocity can be determined
from the current
I=neAvD
Where n is the electron density
e is the electron charge
A is the cross sectional area of
the wire
B
- - - - - - B
FE = FB
qE = qvB
v = E/B
d
I
d
ΔV H
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Example:crossed E,B fields
 You are going to build a velocity
selector that will select negative ions
moving to the right at 9000 m/s.
 You have a magnet which produces a
0.70T B-field into the page.
 What is the magnitude and direction
of the electric field?
Magnetic Force
on a current carrying wire
A
I
Here the current is just a bunch of moving charges
FB = q(v " B)
N = number of charge carriers
vD = drift velocity of charges
= Nq(v D " B)
n = number of charges per volume
I=nqAvD
A = cross sectional area
L = length of wire
F = IL B
!
FB = nALq(v D " B)
= I(L " B)
"
= ILB"
= ILBsin #
!
!
Magnetic Force
on a current carrying wire
F
L
I
F = IL" B
= ILB"
= ILBsin #
B
!
Direction: run fingers along I,
Curl them into B
Thumb points along F
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Example: Force on a current
carrying wire
We have a 2 T magnet whose poles each have
a diameter of 10 cm as shown below
B
Between the poles we suspend
S
a wire with a current of 10
N
mA running out of the
page.
(a) What direction is the
force?
(b)What is the magnitude
of the force
I
N
Torque on a current loop
IL
F4 = ILB
= IaB = F2
3
4
B
1
2
" = Fr sin #
b
= F4 sin #
2
!
" = IABsin #
A= loop area
!
!
Electric Motor
DC motor applet
http://www.walter-fendt.de/ph14e/electricmotor.htm
10
Audio Speaker
s
I
s
s
B
N
s
s
s
s
F
s
Moving Charges Create
Magnetic fields
 Moving charges experience magnetic
forces -->Moving charges create magnetic
fields
 Charges at rest feel no magnetic force -->
Charges at rest create no magnetic fields
 Charges feel electric forces --> Charges
create electric fields (moving or not)
Magnetic field
due to a long straight
current carrying wire
Direction:
Point your thumb along I
Curl your fingers in the
direction of B
I
r
B
Magnitude:
B=
!
µ0 I
T•m
, µ0 = 4 "x10#7
2"r
A
r is the radial distance from
the current carrying wire
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Magnetic field
due to a loop or solenoid
Magnetic field strength
Direction:
µ NI
B= 0
Point your thumb along I inside a solenoid
L
N=number of turns of wire
Curl your fingers in the
L=solenoid length
direction of B
!L
Example
r
d
X
θ
P
B=?
 Consider two wires a distance
2.0 cm apart and carrying
currents of magnitude, I=5.0 A,
in opposite directions. Find the
magnetic field at point, P, a
distance r=5.0 cm from the
center of the wires.
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