Transcript Magnetism 1415 edition

Magnetism
Magnetism:
Permanent and Temporary
See FSU’s site for much more information!!
http://micro.magnet.fsu.edu/electromag/index.
html
http://micro.magnet.fsu.edu/electromag/java/i
ndex.html
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Magnetic Topics
• Magnetic Fields
• Electromagnetic Induction
• Electromagnetism
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Assignments
• Read the handout and the Powerpoint on
the website.
• These are in the handout:
• 456/1,6
• 465/1-2,4
• 488/ 1,2b,5,7
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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 http://www.walterfendt.de/ph11e/mfbar.htm
• A magnetic field exists around any wire
that carries current
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Gen’l Properties cont.
• A coil of wire that carries a current has a
magnetic field about a permanent magnet
• http://micro.magnet.fsu.edu/electromag/jav
a/faraday/index.html
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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.
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Forces cont.
• The force a magnetic field exerts on a
charged particle depends on the velocity
and charge of the particle and the strength
of the magnetic field. The direction of the
force is perpendicular to both the field and
particle’s velocity.
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• E = - N DF / D t
• Ampere’s Rule for parallel, straight
conductors: F = 2k L I1 I2 / d
Transformer Equations
• Pp = Ps  VpIp = VsIs
• Is = Vp = Np
• Ip Vs
Ns
Induction
• M = -Es / D Ip/ D t
• L = -E / D I / D t
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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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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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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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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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Fingers point
in direction
of b field
Thumb direction of
current flow
Imaginary
vector
coming from
palm is
direction that
conductor is
forced out of
the field
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Moving Charges in a Magnetic Field
Right-Hand Rule for Moving Charges
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Force between 2 current carrying wires.
Ampere’s Rule for parallel, straight
conductors: F = 2k L I1I2
d
F, force, newtons, n
k, constant, 10-7 n/a2
L, length of wires, meters, m
I, current in wires, amp, a
d, distance between wires, meters, m
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Ampere’s Rule Problem
Calculate the strength of the two equal currents
that must flow through two parallel conductors that
are 1.5 m long and are 0.25 m apart if a force of
attraction of 4.9x10-6 n is present.
F = 2k L I1I2
I = Fd
F = 4.9x10-6 n
d
√ 2kL
d = 0.25m L = 1.5 m
I=
4.9x10-6 n (0.25m) a2
1.22x10-6
√ 2 x10-7n (1.5m)
√ 3x10-7
I = 2.02 a
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20-6 Force Between 2 Parallel
Wires
Fig 20-23
Fig 20-24
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20-3 Force on Electric Current in Magnetic Field
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Sources
• Other physics information:
• http://www.phys.ufl.edu/courses/phy2054/s08/lec
tures/2054_ch19A.pdf
• Great diagrams of magnetic rules:
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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!!! An induced EMF causes
electricity to flow.
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If current flows, there must be
an EMF – this is EM induction
Here’s the formula:
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 linkage 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 magnetic field at a
constant rate, find the induced emf. The
flux linkage change ( DF / D t) is 4.00 x 106 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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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
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Generator Output
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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 he
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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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
http://www.powertransformer.us/prima
ryvoltage.png
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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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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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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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