Post on 25-Apr-2023
WileyPLUS Assignment 2Chapters 20, 21
Due: Wednesday, Feb 11 at 6 pm
Next WeekMidterm Break!
Midterm Test Ch 18-22, 24, 25
20 multiple-choice questions
Thursday, March 5, 7-9 pm,
41Wednesday, February 11, 2009
Faraday’s Law
The induced emf is equal to the
rate of change of magnetic flux.
V = !"/!t
The direction of the induced current is such that the
magnetic field produced by the current opposes the change in
magnetic flux that generated the current.
Lenz’s Law
Magnetic flux: != BAcos"
42Wednesday, February 11, 2009
Prob. 22.70/32: Indicate the direction of the electric field between
the plates of the capacitor if the magnetic field is decreasing in
time. Charge can flow around the circuit and pile up on the capacitor
plates.
++++
– – – –E
I
I
I
BI
BI
BI
Induced magnetic field
43Wednesday, February 11, 2009
#
Prob. 22.26: A 0.5 m copper bar, AC, sweeps around a conducting circular track at 15 rad/s.
A uniform magnetic field points into the page, B = 0.0038 T.
Find the current in the loop ABC.
The loop forms a closed circuit of increasing area, so the magnetic flux passing through the loop increases and an emf is generated.
and I = V/R = 0.00713/3
= 2.4 mA
IBI
$ = 15 rad/s
r = 0.5 m
The flux passing through the loop is: != BA= B
!"
2#
"!#r2 = Br
2!/2
The induced emf is: V =!"
!t=Br
2
2! !#
!t=Br
2$
2
V =0.0038!0.52!15
2= 7.13!10"3 V
θI
$ = 15 rad/s
44Wednesday, February 11, 2009
Guitar pickup
• The strings are magnetizable
• A permanent magnet magnetizes them
• The vibration of a string changes the magnetic flux through the
coil at the frequency of vibration of the string
• An emf is induced in the coil at that frequency
45Wednesday, February 11, 2009
Playback of tape recording
• Recorded tape is magnetized in N-S patches
• The tape passes by the playback head which channels and concentrates
the magnetic field through an iron core
• The changing magnetic flux induces an emf in the coil
46Wednesday, February 11, 2009
Moving coil microphone
• Sound waves cause the diaphragm of the microphone to move in/out
• A coil moves with the diaphragm relative to a permanent magnet,
causing the magnetic flux through the coil to change in step with the
pressure variations of the sound wave
• An emf is set up in the coil at the frequency of the sound wave
47Wednesday, February 11, 2009
Ground fault detector
If the return current (green) is equal to the supply current (red), the
magnetic fluxes around the iron ring are equal and opposite and cancel.
If the currents differ, the fluxes do
not cancel and there is a net flux
varying at 60 Hz, which induces a
current in the sensing coil.
48Wednesday, February 11, 2009
Electric Generator
The magnetic flux passing
through the coil varies as the coil
rotates – an emf is generated.
Flux, != BAcos"= BAcos(#t)
49Wednesday, February 11, 2009
V0
V =V0 sin(!t)
EMF from electric generator,
using rate of change of flux
Flux, != BAcos"= BAcos(#t)
V = V0sin($t), V0 = NBA$
(diff. calculus)
50Wednesday, February 11, 2009
V = BLvsin!
V = BLvsin!
Electric GeneratorEMF generated in moving conductors
Total emf generated: Vtot = 2BLvsin!
Therefore, Vtot = BLW!sin"= BA!sin"
If the coil has N turns, then: Vtot = NBA!sin"=V0 sin"
W
Area, A = LW
v= r!=!W
2
"!, != angular frequency of rotating coil
VV ++–
–
2V
= V0 sin!t
51Wednesday, February 11, 2009
Clicker Question
You have a fixed length of wire and need to design a generator.
You have only two options:
1) a one turn square coil, or
2) a two turn square coil.
Which option should you choose so that your generator will
produce the greatest peak emf for a given frequency and
magnetic field strength?
A) One turn square
B) Two turn square
C) It doesn’t matter
Answer A): V0 ~ NA, with A = a2, and a ~ 1/N, so A ~ 1/N2
So, V0 ~ N/N2 - if you increase the turns you lose on area rapidly.
a
a
a
a
A = a2
L = 4aNV0 = NBA$
52Wednesday, February 11, 2009
Prob. 22.63/39: The cross-
sectional area of the coil is 0.02
m2 and the coil has 150 turns.
Find the rotation frequency of
the coil and the magnetic field.
The period is T = 0.42 s
Therefore, B=V0
NA!=
28
150!0.02!14.96 = 0.624 T
!=2"
T=
2"
0.42 s= 14.96 rad/s
Peak voltage, V0 = 28 V = NBA$.
Alternating Current Electric Generator
V0 = 28 V
f = 1/T = 2.38 Hz V = NBA!sin!t(Could also be V = NBA! cos !t)
53Wednesday, February 11, 2009
TimeTime
Root Mean Square (rms)
Power % V2
V0 = NBA$V0
V2 = V02 sin2 $t
Time
V02
Mean = V02/2
Average power % Vrms2 = V0
2/2
Vrms = V0/!2 = rms voltage
V = V0 sin $t
54Wednesday, February 11, 2009
Prob. 22.40: A generator uses a coil that has 100 turns and a 0.5 T
magnetic field. The frequency of the generator is 60 Hz and its
emf has an rms value of 120 V.
Assuming that each turn of the coil is a square, determine the
length of the wire from which the coil is made.
• What is the peak voltage generated?
• What is the area of the coil?
A = a2...
55Wednesday, February 11, 2009
Electric Generator
Electric Motor
Just like a generator, but use a current to cause the coil to rotate.
Once the coil is rotating, it acts as a generator, producing an emf that opposes the rotation of the coil! – the “back emf.”
L
L As the coil rotates and a current is induced in
it, a magnetic force is generated that opposes
the rotation of the coil.
! Work has to be done to rotate the coil
against this torque.
56Wednesday, February 11, 2009
Back emf
Vback
A power supply drives the motor The motor acts as a generator
Kirchhoff’s loop law: V !Vback = IR
Symbol for
AC
generator
57Wednesday, February 11, 2009
Vback
Prob. 22.38/36: A vacuum cleaner is plugged into a 120 V socket and
draws 3 A of current in normal operation (motor running at full speed,
back emf at maximum value) and the back emf is then 72 V.
Find the coil resistance of the motor.
V !Vback = IR
So, 120 – 72 = 3R
R = (120 – 72)/3 = 16 "
When the motor is first switched on, while it’s still spinning slowly, the
back emf is small and: V – 0 = IR,
Then, I = (120 V)/(16 !) = 7.5 A ! the motor draws extra current while it is speeding up. This is the time it’s most likely to trip the breaker.
58Wednesday, February 11, 2009
Prob. 22.36/38: A 120 volt motor draws a current of 7 A when
running at normal speed. The resistance of the armature wire is
0.72 &.
a) Determine the back emf generated by the motor.
b) What is the current at the instant the motor is turned on and
has not begun to rotate?
c) What series resistance must be added to limit the starting
current to 15 A?
59Wednesday, February 11, 2009
Summary of Chapter 22
• EMF induced in a moving conductor: V = vLB sin#
• Mechanical work has to be done to produce
electrical energy – you don’t get something for
nothing
• Magnetic flux, " = BA cos', proportional to
number of field lines.
• Faraday’s law: V = (–)N !"/!t,
Lenz’s law: the magnetic field
produced by the induced current opposes
the changing magnetic flux.
• Electric generator, back emf
60Wednesday, February 11, 2009