Transcript Document 7240200

Magnetism
Magnetism:
Permanent and Temporary
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Assignments
 24/1-9,16-18,21-23,
and 669/1-9
 25/1-8,16,17,21
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General Properties of Magnets
 Like
magnetic poles repel; unlike
magnetic poles attract
 Magnetic field lines are directed from
north to south
 Magnetic field lines always form close
loops
 A magnetic field exists around any
wire that carries current
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Bar Magnets
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Gen’l Properties cont.
A
coil of wire (SOLENOID) that
carries a current has a magnetic field
about a permanent magnet
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Forces Caused by Magnetic
Fields
 When
a current-carrying wire is
placed in a magnetic field, a force
acts on the wire that is perpendicular
to both the field and the wire.
Meters operate on this principle.
 Magnetic field strength is measured
in tesla, T (one newton per ampere
per meter).
 B is the symbol for magnetic field
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Forces cont.
 An
electric motor consists of a coil of
wire (armature) placed in a
magnetic field. When current flows
in the coil, the coil rotates as a result
of the force on the wire in the
magnetic field.
 The force a magnetic field exerts on
a charged particle depends on the
velocity and charge of the particle
and the strength
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Forces cont.
 of
the magnetic field. The direction
of the force is perpendicular to both
the field and particle’s velocity.
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Key Equations
F
= BIL  Force on a current
carrying wire in a magnetic field.
Force = magnetic field strength x
current x length of wire. Newton =
tesla x amp x meter
 F = BqV  Force of a magnetic field
on a single charged particle. Force =
magnetic field strength x charge x
velocity of the charge. Newton =
tesla x coulomb x m/s
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Hand Rule #1- B field direction
around a current carrying wire
 Point
thumb in direction of current in
the wire
 Fingers of your hand circle the wire
and show the direction of the
magnetic field
– Knuckles, N
– Finger tips, S
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Hand Rule #2 – Determine the
polarity of an electromagnet
Wrap the fingers of your right hand
around the loops in the direction of the
current
 Extended thumb points toward the N pole
of the electromagnet

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Sample Problems
A
straight wire that carries a 5.0 amp
current is in a uniform magnetic field
oriented at right angles to the wire.
When 0.10 m of the wire is in the
field, the force on the wire is 0.20 n.
What is the strength of the magnetic
field, B?
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Solution
Known:
I = 5.0 amp
L = 0.10m
F = 0.20 N
F=BIL  B = F/IL
= 0.20N/5.0 amp(0.10m)
= 0.40 T
Unknown:
B =?
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Sample Problem
A beam of electrons travels at 3.0 x
106 m/s through a uniform
magnetic field of 4.0 x 10–1 T at
right angles to the field. How
strong is the force that acts on
each electron?
Known
Unknown
V = 3.0 x 106 m/s
F =?
B = 4.0 x 10–1 T
Q = - 1.6 x 10–19 c
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Solution
F = BqV
= 4.0 x 10–1 T (-1.6 x 10–19c)(3.0 x
106 m/s)
= -1.9 x 10–13 Tcm/s
= -1.9 x 10 -13 n
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The small picture – how
magnetism occurs
Domain theory – when enough
atoms of a substance line up in the
same direction
Strong magnets – iron and steel
Very strong – Alnico alloy
Weak – aluminum, platinum
Natural – magnetite or lodestodes
formed when rock was molten
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Magnetic field lines
Magnetic flux, (F) – number of field
lines passing through a surface
Unit: weber = 1 nm/amp
Magnetic flux density, B = F /A
Unit: wb/m2 = nm/a m2 = n/am
1 wb/m2 = 1 Tesla
Earth, 10–4 T
Humans, 10–11 T
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Electromagnetism-flowing
electrical current  magnetism
Ampere’s Rule for parallel, straight
conductors: F = 2k L I1 I2 / d
K = 10 –7 n/a2 = 10 –7 Tm/a
L, length, m
I, current, a
d, distance between wires
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Solenoid – conducting linear coil
which acts like a bar magnet
Increase B, magnetic flux density by
Increasing the current
Adding loops of wire
Inserting an iron core into solenoid – now
it is an electromagnet
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Electromagnetic Induction
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Michael Faraday and Joseph
Henry around the same time…
 Discovered
that when there is
relative motion between a magnetic
field and a complete circuit (and the
conductor cuts across the magnetic
field), that electricity will flow!!!
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Hand rule #3 – shows force
acting on wire in B field
 Lay
right hand flat, palm up
 Extend thumb 90 degrees to rest of
fingers
 Fingers point in direction of B field
 Thumb points in direction of current,
I
 Imaginary vector coming up
perpendicular out of the palm points
in the direction of force acting on
current carrying wire.
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If current flows, there must be
an EMF – this is EM induction
Faraday’s Law of Induction:
E = - N DF / D t
E, emf, volts
-N, # of turns of wire (- means the current
opposes the change that induced it)
DF, change in flux in weber, wb
D t, change in time, sec
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Sample Problem
 If
a coil of 200 turns is moved
perpendicularly in a magntic field at
a constant rate, find the induced
emf. The flux linkage change ( DF /
D t) is 4.00 x 10-6 wb in 0.0100 sec.
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Problem solution
E = - N DF / D t
E = (-200)(4.00 x 10 –6 wb)
1.00 x 10 –2 s
E = -8.00 x 10 –2 v
Imagine what thousands of turns
would produce!
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Generators at
Hoover Dam
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Electric Generators
 Convert
mechanical energy into
electrical energy by rotating a looped
conductor (armature) in a magnetic
field
 Alternating-Current electricity
produced is conducted by slip rings
and brushes to be used
*
 Direct current can be produced by
using split rings
*
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A coil with a wire is wound around a
2.0 m2 hollow tube 35 times. A
uniform magnetic field is applied
perpendicular to the plane of the coil.
If the field changes uniformly from
0.00 T to 0.55 T in 0.85 s, what is the
induced emf in the coil?
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A = 2.0 m2
N = 35
B = .55 T
T = 0.85 s
E = -N D F / D t = - NBA / D t
E = 35 (0.55 T) (2 m 2)
0.85s
E = 45.3 v
http://www.phys.ufl.edu/~phy3054/extras/contents/Welcome.html
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AC Generator Output
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Four sides of a loop depicted
in a magnetic field
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Cross-section of a rotating wire
loop and output current
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Lenz’s Law  The
direction of an induced current is
such that the magnetic field resulting
from the induced current opposes
the change in the field that caused
the induced current.
 When the N pole of a magnet is
moved toward the left end of a coil,
that end of the coil must become a
N, causing induced current flow in
opposition.
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 Lenz's
law: The induced emf
generates a current that sets up a
magnetic field which acts to oppose
the change in magnetic flux.
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Another way of stating Lenz's law is to say
that coils and loops like to maintain the
status quo (i.e., they don't like change).
If a coil has zero magnetic flux, when a
magnet is brought close then, while the
flux is changing, the coil will set up its
own magnetic field that points opposite
to the field from the magnet.
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On the other hand, a coil with a
particular flux from an external
magnetic field will set up its own
magnetic field in an attempt to
maintain the flux at a constant
level if the external field (and
therefore flux) is changed.
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Inductance
 The
property of an electric circuit by
which a varying current induces a
back emf in that circuit or a
neighboring circuit.
 Mutual Inductance, M
 Self Inductance, L
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Mutual Inductance
 Effect
that occurs in a transformer
when a varying magnetic field
created in the primary coil is carried
through the iron core to the
secondary coil, where the varying
field induces a varying emf.
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M = -Es / D Ip/ D t
Shows the ratio of induced emf in one
circuit to the rate of change of
current in the other circuit.
M, inductance, Henry
Es, average induced emf across
secondary
D Ip/ D t, time rate of change in
current in primary coil
- sign, induced v opposes D I
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Problem
 Find
the mutual inductance in an
electrical device in which the EMF in
the secondary is 200. V and the rate
of change of the current is 2.0 x 10-3
a/s.
 M = -Es / D Ip/ D t
 M = -200v / 2.0x10-3a/s
 M = -1x105 H
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Self Inductance
Ratio of induced emf across a coil to
the rate of change of current in the
coil
L = -E / D I / D t
L, henry
I, current, amp
T, time, sec
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Effective Value
Ieff = 0.707 Imax
Veff = 0.707 Vmax

www.sfu.ca/.../Graphics/Root_Mean_Square.gif
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Transformer
Two separate coils of wire placed near one
another that are used to increase or
decrease AC voltages with little loss of
energy.
 It contains a Primary coil and a Secondary
coil
 When the primary is connected to AC
voltage, the changing current creates a
varying magnetic field that is carried
through the core to the secondary coil.

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Transformer, cont.
In the secondary coil, the varying field
induces a varying emf. This is called
mutual inductance
Secondary voltage = secondary #turns
Primary voltage
primary # turns
Power = Voltage x Current
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Transformers lose no power
 Pp
= Ps  VpIp = VsIs
 Transformer Equation:
 Is = Vp = Np
 Ip
Vs
Ns
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Transformer Problem
A
step-up transformer has a primary
coil consisting of 200 turns and a
secondary coil that has 3000 turns.
The primary coil is supplied with an
effective AC voltage of 90.0v. A)What
is the Vs? B)If Is = 2.00a, find Ip.
C) What is the power in the primary
circuit?
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Solution to Transformer
Problem
 Vs
= NsVp/Np = 3000(90.0V)/200 =
1.35 kV
 Pp = Ps, VpIp = VsIs  Ip =
VsIs/Vp =
 Ip =1350v(2.00a)/90.0v = 30.0a
 Pp = VpIp = 90.0v(30.0a) = 2.70 kW
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Assignment: 488p1,5,7
1. Ns =
VsNp/Vp=(2400v)(75turns)/120v
1500 turns
5. A. Vs = VpNs/Np = 120v/5 =24.0 v
5. B. Is = Vs/Rs = 24v/15 W = 1.60 a
5. C. Ps=(Is)2 Rs = (1.60a)2 (15.0 W )
=
38.4 w
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Sources or for more information
http://physics.bu.edu/~duffy/py106.html
 http://sol.sci.uop.edu/~jfalward/magneticf
orcesfields/magneticforcesfields.html
 Most any high school or college physics
text
 www.physics.sjsu.edu/.../physics51/
mag_field.htm
 hyperphysics.phyastr.gsu.edu/.../elemag.html 52

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