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Magnetism
The Magnetic Force
r
r
r r
F = qE + qv × B
B
x x x x x x
x x x x x x
v
x x x x x x
q
F
B
→→→→→
v
→→→→→
´ q
F
B
↑↑↑↑↑↑↑↑
v
↑↑↑↑↑↑↑↑
q
F=0
Magnetic Fields (History)
• Earliest mention (~800 BC) of magnetic effects are from
rocks, called lodestones, which are natural occurring
permanent magnets.
• Early seafarers used lodestones as compasses. The
word, lode, means “way or course” in middle english, so
lode+stone meant a “direction stone” for sailors.
• Lodestones contain magnetite. It is believed that iron
ore becomes magnetized when lightning struck the iron
ore with high current and magnetized the iron, creating
magnetite. There are many interesting legends about
lodestones
Magnetic Properties
• All magnets are found to have a PAIR of points of
attraction/repulsion,
called the NORTH and SOUTH poles.
• N & S poles attract
• N & N or S & S poles repel
• If the magnet is broken into
two pieces, there are still a pair
of North & South poles.
There are no magnetic
monopoles !!!! (none found
so far).
We live on a large magnet: planet Earth
• Earth has a magnetic field
indicated by the blue lines.
• Compass, or free floating
magnet, aligns its
N-pole direction along the
earth’s magnetic field lines.
• Note the Earth’s magnetic
South-pole is at the Earth’s
geographic North pole
Bearth~10-4T
Bar magnet sprinkled with iron filings
• Place a bar magnet
underneath paper
and then sprinkle iron
filings on the paper,
the filings behave like
tiny compasses and
align themselves along
magnetic field
lines.
Magnetic Field lines are similar in direction
to Electric Field lines from an electric dipole
Sources of Magnetic Fields, B
• In Chap 28 we will study the sources of magnetic fields, which
are created from permanent magnets, currents and moving charges
• For now (this week in Chap 27), let’s assume there exist vector
magnetic fields. They will look similar to electric fields, that is at
each position in space there are a field lines with direction and
magnitude. The unit of the magnetic field is given in units of Tesla
(“T”). We will mostly be concerned with constant magnetic field
examples.
Lorentz Force Law
• The force on a moving charged particle is given by,
r
r r
F = qv × B
where q is the charge, v is the vector velocity and B is the magnetic
Field in units of Tesla, T.
• NOTE if the particle is not moving, there is NO force from B field
• NOTE the force is perpendicular to the velocity, hence the
magnitude of the velocity does not change, but the direction does.
This is similar to centripetal acceleration studied in Phys 170.
If an E field is also present then
r
r
r r
F = qE + qv × B
Vector Cross Product
r r r
C = A× B
Use right hand rule to get vector C
direction and ABsinθ to get magnitude
r
r r
C = A B sin θ
r
C
θ
r
A
r
B
Rotate A into B with
your finger tips. Thumb
points into C direction
2nd Useful formula for vector cross product in component form
r
C = (Ay Bz − Az B y )xˆ + ( Az Bx − Ax Bz ) yˆ
+ (Ax B y − Ay Bx )zˆ
Divergence and Curl in spherical coordinates
More on the cross product
) )
)
Unit vectors:
i ×k = − j
) ) )
Check with right
i× j =k
hand rule
) ) )
j ×k = i
) )
) )
) )
i × i = 0; j × j = 0; k × k = 0
Example: A particle with q = -1.24 x 10-8 C and velocity
)
)
4
v = (4.19 ×10 m / sr)i + (−3.85 × 10 m / s ) j
)
enters a region of space with B = (1.40T )i
4
Find the force on it:
r
) )
) )
r v
F = qv × B = (−1.24 ×10−8 C )(1.40T )[(4.19 ×104 m / s )i × i − (3.85 ×104 m / s ) j × i ]
r
) )
r v
−8
4
F = qv × B = (−1.24 × 10 C )(1.40T )[(3.85 ×10 m / s ) j × i ]
r
)
r v
−4
F = qv × B = 6.68 ×10 N (−k )
Clicker:
Three points are arranged in a
uniform magnetic field. The B field
points into the screen.
r r r
Magnetic Force: F = qv × B
1) A positively charged particle is located at point A and is
stationary. The direction of the magnetic force on the particle is:
a) right
b) left
d) out of the screen
c) into the screen
If v = 0 ⇒ F = 0.
e) zero
2) The positive charge moves from point A toward B. The direction of
the magnetic force on the particle is:
r r
If
a) right
b) left
d) out of the screen
v⊥B
then F = qvB
c) into the screen
e) zero
If v is up, and B is into the
page, then F is to the left.
Clicker question:
3) The positive charge moves from point A toward C. The direction
of the magnetic force on the particle is:
a) up and right
b) up and left
c) down and right
d) down and left
r
r r
Magnetic Force: F = q v × B
If v is up and to the right, it is still perpendicular to B,
hence F = qvB then and F is up and to the left.
Clicker problem
• Two protons each move at speed v (as
shown in the diagram) in a region of
space which contains a constant B field
2A
in the -z-direction. Ignore the interaction
between the two protons.
– What is the relation between the
magnitudes of the forces on the two
protons?
(a) F1 < F2
2B
1
v
B
2
z
v
x
(c) F1 > F2
– What is F2x, the x-component of the force on the second proton?
(a) F2x < 0
2C
(b) F1 = F2
y
(b) F2x = 0
(c) F2x > 0
– Inside the B field, the speed of each proton:
(a) decreases
(b) increases
(c) stays the same
Clicker problem
• Two independent protons each move at
speed v (as shown in the diagram) in a
region of space which contains a
2A
constant B field in the -z-direction.
Ignore the interaction between the two
protons.
– What is the relation between the
magnitudes of the forces on the two
protons?
(a) F1 < F2
(b) F1 = F2
y
1
v
B
2
v
z
(c) F1 > F2
• The magnetic force is given by:
r
r
r
F = qv ´ B ⇒ F = qvB sin θ
• In both cases the angle between v and B is 90°!!
Therefore F1 = F2.
x
Clicker problem
• Two independent protons each move at
speed v (as shown in the diagram) in a
region of space which contains a
2B
constant B field in the -z-direction.
Ignore the interaction between the two
protons.
– What is F2x, the x-component of the
force on the second proton?
(a) F2x < 0
(b) F2x = 0
F1
F2
y
1
v
B
2
v
z
x
(c) F2x > 0
• To determine the direction of the force, we use the
right-hand rule.
r
r r
F = qv × B
• As shown in the diagram, F2x < 0.
Clicker problem
• Two protons each move at speed v (as
2C
shown in the diagram) in a region of
space which contains a constant B field
in the -z-direction. Ignore the interaction
between the two protons.
– Inside the B field, the speed of each
proton:
(a) decreases
(b) increases
y
1
v
B
2
z
v
x
(c) stays the same
Although the proton does experience arforce
(which deflects it), this is always ⊥ to v .
Therefore, there is no possibility to do work, so
r
kinetic energy is constant and v is constant.
Magnetic Flux
• Recall that we introduced the electric flux
r r
Φ E = ∫ E ⋅ dA
and used it in Gauss’ Law
E
dA
S
r r qin
Φ E = ∫ E ⋅ dA =
S
ε0
• Lets introduce the analog, the magnetic flux as the surface integral
of the normal component of the magnetic rfield r
Φ B = ∫ B ⋅ dA
S
The units are Tesla·m2, which are called Weber (Wb)’s
Magnetic Flux Law
There are no magnetic monopoles !
• We now state the magnetic flux law (analogous to Gauss’ Law)
r r
Φ M = ∫ B ⋅ dA = 0
• The net magnetic flux through any enclosed surface is always ZERO
This is equivalent to the non-existence of magnetic monopoles and the
fact that all magnetic lines form a loop (do not start/end on a charge).
S
S
Any enclosed surface will have net magnetic flux = zero
above are B fields from bar magnet and current loop
Example (not a clicker problem)
The magnetic flux through one face of a cube is +120Wb
A) What must the total magnetic flux through the other five
faces of the cube be?
B) Why did you not need to know the dimensions of the cube in
order to answer part A)?
C) Suppose the magnetic flux is due to a permanent magnet, show
Where the cube in part (A) might be be located
A) -120 Wb
??
+120 Wb
B) Flux is zero
for a closed
surface
C) Just outside the
block, with equal #
lines
leaving/entering
Comparison between bar magnet & E-dipole
The magnetic lines form closed loops. The electric fields start on the
+charges and end on the –charges. So if we place a closed Gaussian box
around a charge, we find non-zero electric flux. If we place a closed box
anywhere in the bar magnet case, we always obtain zero magnetic flux.
There are NO free “magnetic charges”.
These hypothetical particles are called “magnetic monopoles”.
r r Q
∫ E ⋅ dA =
ε0
r r
∫ B ⋅ dA = 0
The above two equations are TWO of the FOUR fundamental equations
from Maxwell, called Maxwell’s Eqns. (see Y&F pg.1216 for the other
two). Maxwell’s Eqns. predict and govern all of the physics of
electricity and magnetism. Congratulations on understanding
HALF of all electromagnetic science.
Charged particle motion in a constant B field
- velocity in plane
to B.
r
Suppose we have a magnetic field given by B = − B0 zˆ and
a particle starts out at the origin moving in the +x direction.
The particle will move in a circle with the radius R and angular
velocity ω:
r
r
v2
F = m a = qvB = m
R
y
mv
R=
qB
θ
magnetic force
provides
centripedal
force
R
v qB
ω= =
R m
v
x
independent
of r and v.
Review of Circular Motion
If the force F is perpendicular to the
velocity, the particle undergoes
circular motion:
r
r mv 2
F = ma =
R
v
R
F
Force F is always perpendicular to velocity.
F does no work.
Example from Y+F
In an experiment with cosmic rays, a vertical beam of particles that
have a charge magnitude of 3e and a mass 12 times the proton mass
enters a uniform horizontal Magnetic field of 0.250T and is bent in a
semicircle of 95cm. Find the speed of the particle.
mv
qB
q
v = BR
m
3(1.6 x 10 −19 C )
v=
(0.25T )(0.475m)
− 27
12(1.67 x 10 Kg )
R=
v = 2.84 x106 m / s
Example from Y+F
In an experiment with cosmic rays, a vertical beam of particles that
have a charge magnitude of 3e and a mass 12 times the proton mass
enters a uniform horizontal Magnetic field of 0.250T and is bent in a
semicircle of 95cm. Find the speed of the particle.
mv
R=
qB
q
v = BR
m
m
3(1.6 x 10 −19 C )
v=
(0.25T )(0.475m)
− 27
12(1.67 x 10 Kg )
v = 2.84 x106 m / s
Motion in a constant magnetic field
If the velocity of the charge particle is perpendicular to the B field
the motion is circle with radius R=mv/qB.
If the velocity is not perpendicular, the motion is a helix. In this
case we break up the velocity into components perpendicular v⊥
and parallel to the field v//.
r
r r
F = qv × B = 0
Therefore have uniform
motion in B direction.
Path looks like extended spring,
v// constant along the B
direction and v⊥ moves
in a circle.
R (of helix) = mv⊥ /qB
Circumference = 27 km = 2 π R
LHC@Geneva,
Switzerland
R=
mv
p
=
qB qB
The KEKB Collider (Tsukuba, Japan)
8 x 3.5 GeV beam energies
Belle detector
SCC RF(HER)
World record:
L = 2.1 x 1034/cm2/sec
ARES(LER)
Ares RF cavity
e+
source
3.1 km
around
Applications
Electric/Magnetic Velocity Selector
A charged particle enters a region with perpendicular
electric and magnetic fields. The electric and
magnetic forces will cancel if the velocity is
just right. Particles with this velocity will go through
undeflected. Others will be deflected.
r
r
r
r r
FE = qE = − Fm = −qv × B
r
r r
qE = qv B1
r
r E
v = r
B
Choosing a particular E/B ratio will select the desired
velocity. Example, E=10,000V/m and B=0.001T
selects v=10,000,000 m/s
Mass Spectrometer (Chemistry)
Measure m to identify substances:
1.) Ionize atoms by hitting them with
accelerated electrons.
2.) Acelerate ions through known potential V.
q=e if singly ionized.
U=qV
3.) Velocity select.
3.) Pass ions through known B field and
measure R.
R=
mv
qB
2 2
qB
q
B
v=
; v2 = 2 2
mR
m R
1 2
mv = qV
2
qB 2 R 2
m=
2V
Mass Spectrometer (Chemistry)
Applications:
Think CSI
Paleoceanography: Determine relative abundances of isotopes
(they decay at different rates geological age)
Space exploration: Determine what’s on the moon, Mars, etc.
Detect chemical and biol. weapons (nerve gas, anthrax, etc.).