Transcript 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, magnetic poles discussing 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 S Bar Magnet N N
Iron filings
Magnetic Poles
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 E Compass N
Magnetic Attraction-Repulsion
S S N N N S S N N S Magnetic Forces: Like Poles Repel 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 any point is the same as the direction indicated by this compass. at N S Field B is strong where lines are dense and weak where lines are sparse.
Field Lines Between Magnets
Unlike poles N S Attraction Leave N and enter S Like poles N N Repulsion
The Density of Field Lines
Electric field D A D N Magnetic field flux lines f
B
D D
A
D A Df S N Line density
E
D
N
D
A
Line density Magnetic Field B is sometimes called the flux density in Webers per square meter (Wb/m 2 ).
Magnetic Flux Density
• Magnetic flux lines are continuous and closed.
• Direction is that of the B vector at any point.
• Flux lines are NOT in direction of force but ^ .
B
A
D A Magnetic Flux density: When area A is perpendicular to flux:
B
A
BA
Df 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 of q n makes an angle with the B-field is:
BA
cos q a A q n 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 defined as the electric force per unit charge.
was 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.
+ + v v E
Magnetic Force on Moving Charge
Imagine a tube that projects charge with velocity v perpendicular B +q into field.
F B v 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 charge q , and a larger v , an increase in magnetic field B .
Direction of Magnetic Force
The right hand rule: With a flat right point thumb hand, in direction of velocity v , direction of B fingers in field. The flat hand pushes in the direction of force F .
F B v F B v S The force is greatest when the velocity decreases to zero for parallel motion.
v is perpendicular to the B field. The deflection
Force and Angle of Path
S S S Deflection force greatest when path perpendicular to field. Least at parallel.
F
v
sin q v sin q q v F B v
Definition of B-field
Experimental observations show the following:
F
qv F qv
sin q constant By choosing appropriate units for the constant of proportionality, we can now define the B-field as: Magnetic Field Intensity B:
B
F 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.
velocity 5 x 10 4 mT A 2-nC m/s charge is projected with at an angle of 30 0 with a magnetic field as shown. What are the magnitude and direction of the resulting force? 3 Draw a rough sketch.
q v = 2 x 10 -9 = 5 x 10 4 C m/s B = 3 x 10 -3 q = 30 0 T v sin f B 30 0 v F B v Using right-hand rule, the force is seen to be upward .
F
qvB
sin q -9 4 -3 (2 x 10 C)(5 x 10 m/s)(3 x 10 T) sin 30 0 Resultant Magnetic Force: F = 1.50 x 10 -7 N, upward
Forces on Negative Charges
Forces on requires a negative charges are opposite to those on positive charges. The force on the negative charge left-hand rule to show downward force F .
Right-hand rule for positive q F B v Left-hand rule for negative q F B v S S
Indicating Direction of B-fields
One way of indicating the directions of fields perpen dicular 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?
F v X X X X X X X X Up Left F v X X X X X X X X F v Up F Right
Crossed E and B Fields
The motion of charged particles, such as electrons, can be controlled by combined electric and magnetic fields.
Note: F E on electron is upward and opposite E-field.
But, F B down on electron is (left-hand rule).
Zero deflection when F B = F E + x x x x x x x x E F E e B v F B B e v v
The Velocity Selector
This device uses crossed fields to select only those velocities for which F B = F E . (Verify directions for +q) When F B
qvB v
E B
= F E :
qE
Source of +q + x x x x x x x x Velocity selector +q v By adjusting the E and/or B-fields, a person can select only those ions with the desired velocity.
Example 2.
B = 20 mT A lithium ion, q = +1.6 x 10 -16 C , is projected through a velocity selector where . The E-field is adjusted to select a velocity of 1.5 x 10 6 field E?
m/s Source of +q . What is the electric +
v
E B
x x x x x x x x +q v -
E = vB
V E = (1.5 x 10 6 m/s)(20 x 10 -3 T); E = 3.00 x 10 4 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.
F C F C
mv
2 ;
F B R mv
2
F B R
qvB
;
qvB
The radius of path is:
R
mv qB
+ Centripetal F c = F B X X X X X X R + F c + X X X X X X
Mass Spectrometer
+q x x x x x x x x +
v
E B
R Photographic plate slit
mv
2 x x x x x x x x x x x x x x x x x x x x x x x x x x x x m 2
qvB
m 1
R
Ions passed through a velocity selector at known velocity emerge into a magnetic field as shown. The radius is:
R
mv qB
The mass is found by measuring the radius R:
m
qBR 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 slit x x x x x x x x +
v
E B
Photographic R plate x x x x x x x x x x x x x x x x x x m
R v
E B
1000 V/m 0.5 T v = 2000 m/s
mv qB m
qBR v m
-19 (1.6 x 10 C)(0.5 T)(0.0728 m) 2000 m/s m = 2.91 x 10 -24 kg
The force on a current carrying wire in a magnetic field
F B
F
B
,
q
q
v
d
B
)
F
B
q
v
d
B
)
nAL I
v d qnA
F
B
I
L
B
BIL
sin q For a straight wire in a uniform field
d
F
B
Id
s
B F
B
I a
b d
s
B
For an arbitrary wire in an arbitrary field
F
B
I a
b d
s
B F
B
I
b a
d
s
B F
B
I
L
B F
B
I
)
B
d
s
0
F
B
0
Ex.
From the figure, the magnetic field ( and perpendicular to the plane. The conductor caries a current I B ) is uniform . Find the total magnetic force on these wire.
y I L I R I ( in ) I 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 F
B
0 0
For 2 and 4,
F
2
F
4
IaB
max max
F
2
b
2 )
b F
4
b
2 ) ) 2
b
2 max
IAB
If the field makes an angle with a line perpendicular to the plane of the loop:
F
1
a
2 sin q
F
3
a
2 sin q
IbB a
2 sin
IAB
sin q q
IbB a
2 sin q
IabB
sin q
I
A
B
Magnetic Dipole Moment
A (Amperes.m
2 )
μ
B If the wire makes N loops around A,
N
μ
loop
B
μ
coil
B The potential energy of the loop is:
U
μ
B
Hall Effect
qv d B E H
v d qE B H
D
V H
v d Bd
Hall constant or Hall coefficient
v d
I nqA
Δ V H IB nqt R H IB 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 F B v Left-hand rule for negative q F B v S S
Summary (Continued)
v sin q q v F B v For a charge moving in a B-field, the magnitude of the force is given by: F = qvB sin q
Summary (Continued)
The velocity selector:
v
E B
V + x x x x x x + v The mass spectrometer:
R
mv qB m
qBR v
slit x x x x x x x x +
v
R
E B
x x x x x x x x x x x 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
v 0
Quiz III
A
B 10.0
cm
An electron at point B A in Fig. has a speed v 0 of 2.94 10 6 m/s. Find a) the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from ; b) the time required for the electron to move from A to A B . to