Physics 212 Lecture 12 Today's Concept: Magnetic Force on moving charges.
-
Upload
tanya-sloman -
Category
Documents
-
view
225 -
download
0
Transcript of Physics 212 Lecture 12 Today's Concept: Magnetic Force on moving charges.
Physics 212Physics 212Lecture 12Lecture 12
Today's Concept:Today's Concept:
Magnetic Force on moving chargesMagnetic Force on moving charges
F qv B
How about the fact that soon our planet's magnetic field will turn on
it's end and switch polarity, if it doesn't disappear in the process that
is, like it did for Mars. This flip will confuse animals that rely on internal compasses for migration. Also, due to the standard anomalies in polarity
prior to the flip (see graphic at http://www.pbs.org/wgbh/nova/magn
etic/reversals.html) our field will weaken enough to make cancer
even more prevalent that it is currently from increased exposure to radiation. Also these anomalies will
create multiple north and south poles covering much of the upper
northernand lower southern hemisperes in auroras at night. Also, the weakening of the magnetic field may allow radiation to interfere with satellite transmissions. Hello cable TV! And most importantly, after the flip our compasses will point south.
Today’s Plan:Today’s Plan:
1)1) Review of magnetismReview of magnetism2)2) Review of cross Review of cross
productproduct3)3) Example problemExample problem
Key Concepts:Key Concepts:
1)1) The force on moving charges due to a The force on moving charges due to a magnetic field.magnetic field.
2)2) The cross product.The cross product.
Magnetism
N S
F qv B
I
Magnetic Observations
N S
• Bar Magnets
• Compass Needles
•Magnetic Charge?
S N N S
N S
N S
N S N S N Scut in half
The earth is a big (weak) magnet
• Compass needle deflected by electric current
I
• Magnetic fields created by electric currents• Magnetic fields exert forces on electric currents (charges in
motion)V
V
I
IF F
V
V
I
I
F
F
Magnetic Observations
Magnetism & Moving Charges
All observations are explained by two simple equations:
F qv B
02
ˆ4
I ds r
dBr
Today
Next Week
Cross Product Review• Cross Product different from Dot Product
– A●B is a scalar; A x B is a vector
– A●B proportional to the component of B parallel to A– A x B proportional to the component of B perpendicular to A
• Definition of A x B– Magnitude: ABsin– Direction: perpendicular to plane defined by A and B with sense
given by right-hand-rule
Remembering Directions: The Right Hand Rule
F qv B x
y
z
BF
v
Magnetic Force
F qv B
xy
zB
F
screen
x
y
zB
v
+q
V
V
I
IF F
Magnetic Force
Three points are arranged in a uniform magnetic field. The B field points into the screen.
1) A positively charged particle is located at point A and is stationary. The direction of the magnetic force on the particle is:
F qv B
Since it has no velocity, there is no Since it has no velocity, there is no magnetic force.magnetic force.
Three points are arranged in a uniform magnetic field. The B field points into the screen.
1) A positively charged particle is located at point A and is stationary. The direction of the magnetic force on the particle is:
F qv B
If my finger point in the direction of If my finger point in the direction of velocity, and I curl my fingers toward velocity, and I curl my fingers toward the electric field, my thumb points left.the electric field, my thumb points left.
Motion of Charge q in Uniform B Field
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
x x x x x x x
x x x x x x xUniform B into page
• Force is perpendicular to v– Speed does not change– Uniform Circular Motion
• Solve for R:
2
v
aR
2
v
qvB mR qB
mvR
qv
Fq
v
F
q v
F
q
v
F
R
F qv B F qvB
mv pR
qB qB
The drawing below shows the top view of two interconnected chambers. Each chamber has a unique magnetic field. A positively charged particle is fired into chamber 1, and observed to follow the dashed path shown in the figure.
Right hand rule. Put thumb in same Right hand rule. Put thumb in same direction of v and palm in direction of direction of v and palm in direction of force (right). This shows the magnetic force (right). This shows the magnetic field points out of the page.field points out of the page.
The drawing below shows the top view of two interconnected chambers. Each chamber has a unique magnetic field. A positively charged particle is fired into chamber 1, and observed to follow the dashed path shown in the figure.
It turns sharper...the stronger the field It turns sharper...the stronger the field the sharper the turnsthe sharper the turns
CalculationCalculationA particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known.
What is B?• Conceptual Analysis
– What do we need to know to solve this problem?
(A) Lorentz Force Law (B) E field definition (C) V definition
(D) Conservation of Energy/Newton’s Laws (E) All of the above
X X X X X X X X XX X X X X X X X X X X X X X X X X XX X X X X X X X X
BBB
x0
x0
exits here
enters here
d
q, mq, m
– Absolutely ! We need to use the definitions of V and E and either conservation of energy or Newton’s Laws to understand the motion of the particle before it enters the B field.
– We need to use the Lorentz Force Law (and Newton’s Laws) to determine what happens in the magnetic field.
E
CalculationCalculation
• Strategic Analysis– Calculate v, the velocity of the particle as it enters the magnetic
field– Use Lorentz Force equation to determine the path in the field as
a function of B– Apply the entrance-exit information to determine B
A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known.
What is B?
X X X X X X X X XX X X X X X X X X X X X X X X X X XX X X X X X X X X
BBB
x0
x0
exits here
enters here
d
q, mq, m
E
• What is v0, the speed of the particle as it enters the magnetic field ?
0
2qEdv
m0
2Ev
m 0
2qEv
md 0
qEdv
m
(A) (B) (C) (D)(A) (B) (C) (D)
• Why??– Conservation of Energy
• Initial:Initial: Energy =Energy = U = qV = qEdU = qV = qEd• Final:Final: Energy = KE = ½ mv Energy = KE = ½ mv00
220
2qEdv
m
– Newton’s Laws• a = F/m = qE/ma = F/m = qE/m• vv00
22 = 2ad = 2ad2
0 2qE
v dm
0
2qEdv
m
20
1
2mv qEd
X X X X X X X X XX X X X X X X X X X X X X X X X X XX X X X X X X X X
BBB
x0
x0
exits here
enters here
d
q, mq, m
E
• Why??– Path is circle !
• Force is perpendicular to the velocityForce is perpendicular to the velocity• Force produces centripetal accelerationForce produces centripetal acceleration• Particle moves with uniform circular Particle moves with uniform circular
motionmotion
X X X X X X X X XX X X X X X X X X X X X X X X X X XX X X X X X X X X
• What is the path of the particle as it moves through the magnetic field?
(A) (B) (C)(A) (B) (C)
X X X X X X X X XX X X X X X X X X X X X X X X X X XX X X X X X X X X
X X X X X X X X XX X X X X X X X X X X X X X X X X XX X X X X X X X X
X X X X X X X X XX X X X X X X X X X X X X X X X X XX X X X X X X X X
BBB
x0
x0
exits here
enters here
d
q, mq, m
E
• What is the radius of path of particle?
(A) (B) (C) (A) (B) (C)
0R x 02R x0
1
2R x
• Why?? X X X X X X X X XX X X X X X X X X X X X X X X X X XX X X X X X X X X
RRxo
X X X X X X X X XX X X X X X X X X X X X X X X X X XX X X X X X X X X
BBB
x0
x0
exits here
enters here
d
q, mq, m
E
• Why??
(A) (B) (C) (D) (A) (B) (C) (D)
F ma 2v
qvB mR
m v
Bq R
0
2m qEd
Bqx m
What is B in terms of v, x0, q and m?
X X X X X X X X XX X X X X X X X X X X X X X X X X XX X X X X X X X X
BBB
x0
exits here
enters here
d
q, mq, m
E
0
mv
Bqx
2
qEdv
mWe already found that
0mx
Bqv
0qx
Bmv 0
qv
Bmx