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MagnetismTuesday, February 22, 2011
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Announcements
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Magnetic Fields
s Magnets cause space to be modified in their vicinity, forming a “magnetic field”.
s
The magnetic field caused by magnetic“poles” is analogous to the electric fieldcaused by electric “poles” or “charges”.
s The north pole is where the magnetic field
lines leave the magnet, and the south pole iswhere they reenter.s Magnetic field lines differ from electric field
lines in that they are continuous loops with no
beginning or end.
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Magnetic
Field,B
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More Magnetic Fields
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Compasses
s If they are allowed to select their own
orientation, magnets align so that the north
pole points in the direction of the magneticfield.
s Compasses are magnets that can easily
rotate so that they can align themselves to a
magnetic field.s The north pole of the compass points in the
direction of the magnetic field.
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Sample Problem
s A compass points to
the Earth’s South
Magnetic Pole (whichis near the North
Pole).
s Is the North MagneticPole the north pole of
the Earth’s Magnetic
Field?
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Magnetic “Monopoles”
s Do not exist!
– This is another way that magnetic fields
differ from electric fields. – Magnetic poles cannot be separated from
each other in the same way that electric
poles (charges) can be.
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Units of Magnetic Field
s Tesla (SI)
– N/(C m/s)
– N/(A m)
s Gauss
– 1 Tesla = 104 gauss
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Magnetic Force on Particles
s Magnetic fields cause the existence of
magnetic forces.
s A magnetic force is exerted on aparticle within a magnetic field only if
– the particle has a charge.
– the charged particle is moving with at leasta portion of its velocity perpendicular to the
magnetic field.
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Magnetic Force on a Charged
Particles magnitude: F = qvBsinθ
– q: charge in Coulombs
– v: speed in meters/second – B: magnetic field in Tesla
θ: angle between v and B
s
direction: Right Hand Rules F
B= q v x B (This is a “vector cross
product”)
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The Right Hand rule to Determine a
Vector Cross Product1. Align your thumb along the velocity vector.
2. Orient your index finger so that it points in
the direction of the magnetic field.
3. Your palm faces the direction of the force
vector (which is the result).
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Sample Problem
Calculate the magnitude force exerted ona 3.0 µC charge moving north at 300,000
m/s in a magnetic field of 200 mT if thefield is directed
a) North.
b) South.c) East.
d) West.
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Sample Problem
s Calculate the magnitude and direction
of the magnetic force.
v = 300,000 m/s
B = 200 mT
q = 3.0µC
34o
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Wednesday, February 23, 2011
Motion of Charged Particles in
Magnetic Field
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Magnetic forces…
s are always orthogonal (at right angles)
to the plane established by the velocity
and magnetic field vectors.s can accelerate charged particles by
changing their direction.
s can cause charged particles to move incircular or helical paths.
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Magnetic forces cannot...
s change the speed or kinetic energy of
charged particles.
s do work on charged particles.
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Magnetic Forces…
s …are centripetal .
s Remember that centripetal acceleration
is v2/r.
s Remember centripetal force is therefore
mv2/r.
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Magnetic Forces are Centripetal
B
F
V
F
V
F
V
F
ΣF = ma
FB = Fc
qvBsinθ = mv2/rqB = mv/r
q/m = v/(rB)
V
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Sample Problem
What is the orbital radius of a proton moving at
20,000 m/s perpendicular to a 40 T magneticfield?
l bl
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Sample Problem
What must be the speed of an electron if it is to
have the same orbital radius as the proton inthe magnetic field described in the previous
problem?
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Thursday, February 24th, 2011
Electric and Magnetic Fields
Together At Last
S l P bl
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Sample Problem
An electric field of 2000 N/C is directed to the
south. A proton is traveling at 300,000 m/s tothe west. What is the magnitude and direction of
the force on the proton? Describe the path of
the proton? Ignore gravitational effects.
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Sample Problem A magnetic field of 2000 mT is directed to the
south. A proton is traveling at 300,000 m/s tothe west. What is the magnitude and direction of
the force on the proton? Describe the path of
the proton? Ignore gravitational effects.
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Sample Problem
s Calculate the force and describe thepath of this electron.
E = 2000 N/C
e-300,000 m/s
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Sample Problem
s Calculate the force and describe thepath of this electron.
e-
300,000 m/s
B = 2000 mT
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Sample problems How would you arrange a magnetic field and
an electric field so that a charged particle of velocity v would pass straight through without
deflection?
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Electric and Magnetic Fields
Together
E
B
v = E/B
e-
S l P bl
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Sample Problem
It is found that protons traveling at 20,000 m/s
pass undeflected through the velocity filter below. What is the magnitude and direction of
the magnetic field between the plates?
400 V
e20,000 m/s
0.02 m
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Friday, February 25, 2011
Magnetic Force on Current
Carrying Wires
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Magnetic Force on Current-
Carrying Wires F = I L B sinθ
– I: current in Amps
– L: length in meters
– B: magnetic field in Tesla
θ: angle between current and field
S l P bl
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Sample Problem
What is the force on a 100 m long wire bearing
a 30 A current flowing north if the wire is in a
magnetic field (directed into the page) of 400
mT?
S l P bl
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Sample Problem
What is the magnetic field strength if the currentin the wire is 15 A and the force is downwardand has a magnitude of 40 N/m? What is thedirection of the current?
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Magnetic Fields…
s Affect moving charge
– F = qvBsinθ
– F = ILBsinθ
– Hand rule is used to determine direction of
this force.
s Caused by moving charge!
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Magnetic Field for
Long Straight Wires B = µoI / (2πr)
µo: 4π × 10-7 T m / A
• magnetic permeability of free space
– I: current (A)
– r: radial distance from center of wire (m)
Ri h H d R l
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Right Hand Rule
for straight currents
1. Curve your fingers
2. Place your thumb (which ispresumably pretty straight)in direction of current.
3. Curved fingers representcurve of magnetic field.
4. Field vector at any point is
tangent to field line.
i
r • ×
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For
straightcurrents
I
Sample Problem
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Sample Problems What is the magnitude and direction of the
magnetic field at point P, which is 3.0 m away
from a wire bearing a 13.0 Amp current?
I = 13.0 A
P
3.0 m
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Monday, February 28, 2011
Superposition in Magnetic Fields
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Principle of Superposition
s When there are two or more currents
forming a magnetic field, calculate B
due to each current separately and thenadd them together using vector
addition.
Sample Problem
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Sample Problems What is the magnitude and direction of the
magnetic field at point P if there are two wires
producing a magnetic field at this point?
I = 13.0 A
P
3.0 m
I = 10.0 A
4.0 m
Sample Problem
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Sample Problem
s Where is the magnetic field zero?
I = 13.0 A
7.0 m
I = 10.0 A
Sample Problem not in notes
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Sample Problem – not in notess What is the magnitude and direction of the
force exerted on a 100 m long wire that
passes through point P which bears a currentof 50 amps in the same direction?
I1 = 13.0 A
P
3.0 m
I2 = 50.0 A
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Tuesday, March 1, 2011
Loops of Wire / Solenoids
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In the 4th Grade
s You learned that
coils with current
in them makemagnetic fields.
s The iron nail was
not necessary tocause the field; it
merely intensified
it.
I
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Solenoid
s A solenoid is a coil of wire.s When current runs through the wire, it
causes the coil to become an“electromagnet”.
s Air-core solenoids have nothing insideof them.
s Iron-core solenoids are filled with iron tointensify the magnetic field.
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Right Hand Rule for magnetic fields
around curved wires
1. Curve your fingers.
2. Place them along wireloop so that your fingerspoint in direction of current.
3. Your thumb gives thedirection of the magneticfield in the center of theloop, where it is straight.
4. Field lines curve around
and make completeloo s.
I
B
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Magnetic Field around Curved
Current
B
Sample Problem
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Sample ProblemWhat is the direction of the magnetic field
produced by the current I at A? At B?
I
AB
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Magnetic Flux
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Magnetic Flux
s The product of magnetic field and area.s Can be thought of as a total magnetic
“effect” on a coil of wire of a given area.
BA
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Maximum Flux
s The area is aligned so that aperpendicular to the area points parallel
to the field
B
A
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Minimum Flux
s The area is aligned so that aperpendicular to the area points
perpendicular to the field
B
A
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Intermediate Flux
s The area is neither perpendicular nor isit parallel
B
A
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Magnetic Flux
s ΦB = B A cosθ
– ΦB: magnetic flux in Webers (Teslameters2)
– B: magnetic field in Tesla – A: area in meters2. θ: the angle between the area and the
magnetic field.
s ΦB = B⋅ A
Sample Problem
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Sample Problem
Calculate the magnetic flux through a rectangular wire frame 3.0 m
long and 2.0 m wide if the magnetic field through the frame is4.2 mT.
a) Assume that the magnetic field is perpendicular to the areavector.
b) Assume that the magnetic field is parallel to the area vector.
c) Assume that the angle between the magnetic field and thearea vector is 30o.
Sample Problem
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Sample Problems Assume the angle is 40o, the magnetic field is 50 mT, and
the flux is 250 mWb. What is the radius of the loop?
B
A
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Wednesday, March 2, 2011
Faraday’s Law of Induction
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Induced Electric Potential
s A system will respond so as to opposechanges in magnetic flux.
s A change in magnetic flux will be partiallyoffset by an induced magnetic field whenever possible.
s Changing the magnetic flux through a wireloop causes current to flow in the loop.
s This is because changing magnetic fluxinduces an electric potential.
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Faraday’s Law of Induction
s ε = -N∆ΦB/∆t
– ε: induced potential (V)
– N: # loops
– ΦB: magnetic flux (Webers, Wb)
– t: time (s)
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A closer look …
s ε = -∆ΦB/∆t
s ε = -∆(BAcosθ)/∆t
– To generate voltage
• Change B
• Change A
• Change θ
S l P bl
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Sample Problem
s A coil of radius 0.5 m consisting of 1000 loops is
placed in a 500 mT magnetic field such that the flux ismaximum. The field then drops to zero in 10 ms. What
is the induced potential in the coil?
S l P bl
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Sample Problems A single coil of radius 0.25 m is in a 100 mT magnetic
field such that the flux is maximum. At time t = 1.0
seconds, field increases at a uniform rate so that at 11seconds, it has a value of 600 mT. At time t = 11seconds, the field stops increasing. What is theinduced potential
s
A) at t = 0.5 seconds?s B) at t = 3.0 seconds?s C) at t = 12 seconds?
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Lenz’s Law
s The current will flow in a direction so as
to oppose the change in flux.
s Use in combination with hand rule topredict current direction.
Sample Problem
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ps The magnetic field is increasing at a rate of
4.0 mT/s. What is the direction of the
current in the wire loop?
Sample Problem
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ps The magnetic field is increasing at a rate of
4.0 mT/s. What is the direction of the
current in the wire loop?
Sample Problem
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ps The magnetic field is decreasing at a rate of
4.0 mT/s. The radius of the loop is 3.0 m,
and the resistance is 4 Ω. What is themagnitude and direction of the current?
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Monday, March 6, 2011
Motional EMF
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Motional emf
s ε = BLv – B: magnetic field (T)
– L: length of bar moving through field – v: speed of bar moving through field.
s Bar must be “cutting through” field lines.It cannot be moving parallel to the field.
s This formula is easily derivable fromFaraday’s Law of Induction
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Motional emf - derivation
s ε = ∆ΦB/∆t
s ε = ∆(BA) /∆t (assume cosθ = 1)
s ε = ∆(BLx) /∆t
s ε = BL∆x /∆t
s ε = BLv
Sample Problem
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Sample Problems How much current flows through the resistor?
How much power is dissipated by the
resistor?
× × × × × × × × × × ×
× × × × × × × × × × ×
× × × × × × × × × × ×
× × × × × × × × × × ×
× × × × × × × × × × ×
× × × × × × × × × × ×
B = 0.15 T
v = 2 m/s
50 cm
3 Ω
Sample Problem
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Sample Problems In which direction is the induced current
through the resistor (up or down)?
× × × × × × × × × × ×
× × × × × × × × × × ×
× × × × × × × × × × ×
× × × × × × × × × × ×
× × × × × × × × × × ×
× × × × × × × × × × ×
B = 0.15 T
v = 2 m/s
50 cm
3 Ω
Sample Problem
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Sample Problems Assume the rod is being pulled so that it is traveling at
a constant 2 m/s. How much force must be applied to
keep it moving at this constant speed?
× × × × × × × × × × ×
× × × × × × × × × × ×
× × × × × × × × × × ×
× × × × × × × × × × ×
× × × × × × × × × × ×
B = 0.15 T
v = 2 m/s
50 cm
3 Ω
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