Download Magnetic Fields

General Physics II
By
Dr. Cherdsak Bootjomchai
(Dr.Per)
Chapter 7
Magnetic Field and
Magnetic Force
Objectives: After completing this
module, you should be able to:
• Define the magnetic field, discussing
magnetic poles and flux lines.
• Solve problems involving the
magnitude and direction of forces on
charges moving in a magnetic field.
• Solve problems involving the magnitude
and direction of forces on current
carrying conductors in a B-field.
Magnetism
Since ancient times, certain materials, called
magnets, have been known to have the property of
attracting tiny pieces of metal. This attractive
property is called magnetism.
S
Bar Magnet
S
N
N
Magnetic Poles
Iron
filings
N
The strength of a magnet is
concentrated at the ends,
called north and south
“poles” of the magnet.
S
A suspended magnet:
N-seeking end and
S-seeking end are N
and S poles.
W
N
S
N
Bar magnet
S
N
E
Compass
Magnetic Attraction-Repulsion
S
S
N
N
N
Magnetic Forces:
Like Poles Repel
S
S
N
N
S
Unlike Poles Attract
Magnetic Field Lines
We can describe
magnetic field lines
by imagining a tiny
compass placed at
nearby points.
The direction of the
magnetic field B at
any point is the same
as the direction
indicated by this
compass.
N
S
Field B is strong where
lines are dense and weak
where lines are sparse.
Field Lines Between Magnets
Unlike
poles
Attraction
N
S
Leave N
and enter S
N
Like poles
N
Repulsion
The Density of Field Lines
Electric field
DN
Magnetic field flux lines
D
B
DA
DA
S
Line density
DN
E
DA
DA
Df
N
Line density
Magnetic Field B is sometimes called the flux
density in Webers per square meter (Wb/m2).
f
Magnetic Flux Density
• Magnetic flux lines are
continuous and closed.

B
A
DA
Df
• Direction is that of the B
vector at any point.
• Flux lines are NOT in
direction of force but ^.
When area A is
perpendicular to flux:
Magnetic Flux
density:

B  ;  = BA
A
The unit of flux density is the Weber per square meter.
Calculating Flux Density When
Area is Not Perpendicular
The flux penetrating the
area A when the normal
vector n makes an angle
of q with the B-field is:
  BA cosq
n
A
q
a
B
The angle q is the complement of the angle a that the
plane of the area makes with the B field. (Cos q = Sin a)
Origin of Magnetic Fields
Recall that the strength of an electric field E was
defined as the electric force per unit charge.
Since no isolated magnetic pole has ever been
found, we can’t define the magnetic field B in
terms of the magnetic force per unit north pole.
We will see instead that
magnetic fields result from
charges in motion—not from
stationary charge or poles.
This fact will be covered later.
E
+
+
v
B^v
Magnetic Force on Moving Charge
Imagine a tube that
projects charge +q
with velocity v into
perpendicular B field.
F
B
v
N
S
Experiment shows:
F  qvB
Upward magnetic force F
on charge moving in B field.
Each of the following results in a greater magnetic
force F: an increase in velocity v, an increase in
charge q, and a larger magnetic field B.
Direction of Magnetic Force
The right hand rule:
With a flat right hand,
point thumb in direction
of velocity v, fingers in
direction of B field. The
flat hand pushes in the
direction of force F.
F
B
v
N
F
B
v
S
The force is greatest when the velocity v is
perpendicular to the B field. The deflection
decreases to zero for parallel motion.
Force and Angle of Path
N
N
N
S
S
S
Deflection force greatest
when path perpendicular
to field. Least at parallel.
F  v sinq
F
v sin q
q
v
B
v
Definition of B-field
Experimental observations show the following:
F  qv sin q
or
F
 constant
qv sin q
By choosing appropriate units for the constant of
proportionality, we can now define the B-field as:
Magnetic Field
Intensity B:
F
B
qv sin q
or
F  qvB sin q
A magnetic field intensity of one tesla (T) exists in a
region of space where a charge of one coulomb (C)
moving at 1 m/s perpendicular to the B-field will
experience a force of one newton (N).
Example 1. A 2-nC charge is projected with
velocity 5 x 104 m/s at an angle of 300 with a 3
mT magnetic field as shown. What are the
magnitude and direction of the resulting force?
Draw a rough sketch.
q = 2 x 10-9 C
v = 5 x 104 m/s
v sin f
B = 3 x 10-3 T
q = 300
F
B
300
v
B
v
Using right-hand rule, the force is seen to be upward.
F  qvB sin q  (2 x 10-9C)(5 x 104 m/s)(3 x 10-3T)sin 300
Resultant Magnetic Force: F = 1.50 x 10-7 N, upward
Forces on Negative Charges
Forces on negative charges are opposite to those on
positive charges. The force on the negative charge
requires a left-hand rule to show downward force F.
Right-hand
rule for
positive q
N
F
B
v
S
Left-hand
rule for
negative q
N
B
F
v
S
Indicating Direction of B-fields
One way of indicating the directions of fields perpendicular to a plane is to use crosses X and dots  :
A field directed into the paper
is denoted by a cross “X” like
the tail feathers of an arrow.
 
 
 
 








X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
A field directed out of the paper
is denoted by a dot “” like the
front tip end of an arrow.
Practice With Directions:
What is the direction of the force F on the charge in
each of the examples described below?
X
X
X
X
F
X X X Up
X+ X v X
X X X
X X X
F
   
Up
v
   
  -  
negative q
   
X
X
Left
X
X
v
X X
FX X
+
X X
X X
X
X
X
X
   
  F 
Right
   
v
   
Crossed E and B Fields
The motion of charged particles, such as electrons, can
be controlled by combined electric and magnetic fields.
Note: FE on electron
is upward and
opposite E-field.
+
But, FB on electron is
down (left-hand rule).
Zero deflection
when FB = FE
e-
x x x x
x x x x
v
-
FE
E
--
e
B
v
B
FB
v
The Velocity Selector
This device uses crossed fields to select only those
velocities for which FB = FE. (Verify directions for +q)
When FB = FE :
qvB  qE
E
v
B
Source
of +q
+
x x x x
x x x x
+q
v
-
Velocity selector
By adjusting the E and/or B-fields, a person can
select only those ions with the desired velocity.
Example 2. A lithium ion, q = +1.6 x 10-16 C,
is projected through a velocity selector where
B = 20 mT. The E-field is adjusted to select a
velocity of 1.5 x 106 m/s. What is the electric
field E?
E
v
B
E = vB
Source
of +q
+
x x x x
x x x x
+q
v
V
E = (1.5 x 106 m/s)(20 x 10-3 T);
E = 3.00 x 104 V/m
Circular Motion in B-field
The magnetic force F on a moving charge is always
perpendicular to its velocity v. Thus, a charge moving
in a B-field will experience a centripetal force.
mv 2
FC 
; FB  qvB;
R
FC  FB
Centripetal Fc = FB
X
mv
 qvB
R
X +X
X
X
X RX X
X
X
X
X FX
X
mv
R
qB
X
X
X
X
X
The radius of
path is:
2
+
X
+
X
c
X X
X
+
X
X
X
X
X
X
Mass Spectrometer
+q
-
slit
E
v
+ B
xx
Photographic
xx
plate
xx
R
xx
x x x x x x x x x
x x x x x x x x
x x x x x x x
m2
x x x x x x
x x x x
mv 2
 qvB
R
m1
Ions passed through a
velocity selector at
known velocity emerge
into a magnetic field as
shown. The radius is:
mv
R
qB
The mass is found by
measuring the radius R:
qBR
m
v
Example 3. A Neon ion, q = 1.6 x 10-19 C, follows
a path of radius 7.28 cm. Upper and lower B =
0.5 T and E = 1000 V/m. What is its mass?
+q
E
v
+ B
xx
Photographic
- xx
plate
xx
R
xx
x x x x x x x
slit x x x x x x x
x x x x x x x
m
x x x x x x
x x x x
E 1000 V/m
v 
B
0.5 T
v = 2000 m/s
mv
R
qB
(1.6 x 10-19C)(0.5 T)(0.0728 m)
m
2000 m/s
qBR
m
v
m = 2.91 x 10-24 kg
The force on a current carrying wire in
a magnetic field
FB ,q  qv d  B )
FB  qv d  B )nAL
I  vd qnA
FB  IL  B
FB  BIL sin q
For a straight wire
in a uniform field
dFB  Ids  B
b
FB  I  ds  B
a
For an arbitrary wire in
an arbitrary field
b
FB  I  ds  B
a
b 
FB  I   ds   B
a 
FB  IL  B
 )
FB  I  ds  B
 ds  0
FB  0
Ex. From the figure, the magnetic field (B) is uniform
and perpendicular to the plane. The conductor caries a
current I. Find the total magnetic force on these wire.
y
I
I
R
I
L
I (in )
x
Torque on a Current Loop in a Uniform
Magnetic Field
If the field is parallel to the plane of the loop
For 1 and 3,
LB  0
FB  0
For 2 and 4,
F2  F4  IaB
 max
 max
b
b
b
b
 F2  F4  IaB )  IaB )
2
2
2
2
 IaB )b
 max  IAB
If the field makes an angle with
a line perpendicular to the
plane of the loop:
a
a
  F1 sin q  F3 sin q
2
2
a
a
  IbB sin q  IbB sin q  IabB sin q
2
2
  IAB sin q
  IA B
Magnetic Dipole Moment
μ  IA
(Amperes.m2)
  μB
If the wire makes N loops around A,
  Nμ loop  B  μ coil  B
The potential energy of the loop is:
U  μ  B
Hall Effect
qvd B  qEH
E H  vd B
DVH  vd Bd
I
vd 
nqA
Hall constant or
Hall coefficient
IB R H IB
ΔVH 

nqt
t
Summary
The direction of forces on a charge moving in an electric
field can be determined by the right-hand rule for positive
charges and by the left-hand rule for negative charges.
Right-hand
rule for
positive q
N
F
B
v
S
Left-hand
rule for
negative q
N
B
F
v
S
Summary (Continued)
F
v sin q
q
For a charge moving in a
B-field, the magnitude of
the force is given by:
v
B
v
F = qvB sin q
Summary (Continued)
The velocity
selector:
E
v
B
The mass
spectrometer:
mv
R
qB
qBR
m
v
+
+
vq
x x x
x x x
x x
V
+q
-
xx
- xx +
xx
xx
x x x
slit x x x
x x x
x x
E
v
B
R
x
x
x
x
x
x
x
x
x x
x x
x x
x m
The end of Chapter 7
Magnetic Field and Magnetic Force
Quiz III
v0

A
10.0 cm
B
An electron at point A in Fig. has a speed v0 of 2.94  106 m/s.
Find a) the magnitude and direction of the magnetic field that
will cause the electron to follow the semicircular path from A to
B ; b) the time required for the electron to move from A to B.