Mastering Physics Assignment 2
Chapters 19, 20
due Monday, Feb 11 at 11 pm
Mastering Physics Assignment 3
Chapters 21, 22
Available Monday, Feb 11 at 8 am
due Friday, Feb 29 at 11 pm
Week of February 11 - 15
Experiment 3: e/m of electron
Wednesday, February 13, 2008
1
The next few weeks...
Ch. 18-22, 24, 25
Wednesday, February 13, 2008
2
Chapter 22: Electromagnetic Induction
• Induced emf and current
• Magnetic flux
• Faraday’s and Lenz’s laws
• Electric generators, back emf
• Omit 22.8, 22.9, (inductance and transformers)
Wednesday, February 13, 2008
3
Electromagnetic Induction
Charges inside the moving rod experience a
force due to the magnetic field...
–––
++
+
I
!v
I
I
I
Conductor
The moving conductor acts as a generator.
The basis of electromagnetic induction.
Wednesday, February 13, 2008
4
Moving coil relative to magnet
Motion of coil
toward the
magnet
I
!F
!v
x
!B⊥
!B
!F !B
⊥
A charge Δq inside the wire moves with the coil relative to the magnetic
field. A component of field, B!, is perpendicular to the velocity of the coil.
The magnetic forces induce a current to flow around the coil.
Wednesday, February 13, 2008
5
Moving magnet relative to coil
I
!v
!B⊥
!B
!B⊥
The charges in the coil are no longer moving as the coil is at rest, but
the induced current is the same as before...
There must be some more basic reason for the induced current.
→ Changing magnetic field at the position of the coil.
Wednesday, February 13, 2008
6
zero
Induced emf
When the magnet moves
relative to coil, a current is
induced in the coil.
Reversing the magnet N and S
poles reverses the deflection.
Moving the coil to the magnet
produces the same deflection
as moving the magnet to the
coil – only the relative motion
of coil and magnet matters.
Wednesday, February 13, 2008
Fm = !q v B
+!q
Fq = !q E
7
Motional emf
The magnetic force Fm separates
the + and – charges in the conductor.
The separated + and – charges
give rise to an electric field E in
the conductor.
At equilibrium, the electrostatic force:
!E
Fq = !q E
–!q
balances the magnetic force.
That is:
Fm
Fq
∆qE
= Fm
= ∆qvB
Also, E = V/L
The induced potential difference between the ends of the rod is: V = vLB
Wednesday, February 13, 2008
8
Induced emf
The emf induced between the ends of a conductor that is
moving in a magnetic field is:
V = vLB
(V = vLB sin ! when the angle between !B and !v is !)
The induced emf is the same whether the coil moves or the
magnet moves, only the relative motion matters.
Wednesday, February 13, 2008
9
Prob. 22.2/4: “Tethered Satellite Experiment”. A 20,000 m length
of wire is trailed behind the shuttle while in orbit around the
earth. The orbital speed of the shuttle is 7600 m/s.
If the earth’s magnetic field at the position of the shuttle is
5.1!10-5 T and the wire moves perpendicular to the field, what is
the induced emf between the ends of the wire?
–
Wire
v
!!
!B
+
V = vLB = 7600 ! 20,000 ! 5.1 ! 10-5 = 7752 V
Negative at the top.
Wednesday, February 13, 2008
10
Prob. 22.4/2: The drawing shows a type of blood flow meter. Blood is
conductive enough that it can be treated as a moving conductor. When it
flows perpendicular to a magnetic field, electrodes can be used to measure
the small voltage that develops across the vessel.
Suppose the speed of the blood is 0.3 m/s, the diameter of the vessel is 5.6
mm and B = 0.6 T. What is the magnitude of the voltage that is measured?
!B
+++
L
–––
Blood – a moving
conductor
!v
Wednesday, February 13, 2008
11
Prob. 22.5: Each rod of length L = 1.3 m moves at speed v = 2.7 m/s
in a magnetic field, B = 0.45 T. Find the motional emf for each.
+
–
Wednesday, February 13, 2008
12
Prob. 22.C6: Initially the rod is at rest. Describe the rod’s motion
after the switch is closed. Be sure to account for the effect of any
motional emf that develops.
V0
Wednesday, February 13, 2008
13
The rod experiences a magnetic force to the right and accelerates.
V0
F = ILB
I
L
Wednesday, February 13, 2008
14
The moving rod now generates its own emf that opposes the emf of
the battery (a “back emf”). The current therefore decreases. The
rod continues to accelerate until the current is reduced to zero
(assuming no friction).
+
V0
F = ILB
V = vLB I
v
–
Speed constant when vLB = V0
Wednesday, February 13, 2008
15
Induced emf – equivalent circuit
I = 0 when V = V0
I
V = vLB
= “back emf”
V0
I
R
Resistance of rails and bar
Wednesday, February 13, 2008
16
60 W bulb, R = 240 !
22.7/6: How long do
the rails have to be
to light the 60 W
bulb for 0.5 s?
B = 0.4 T
!Fm
L = 0.6 m
Fapplied
Motional emf between ends of sliding rod, V = vLB
Power dissipated, W = VI = V2/R = 60 W, so (vLB)2/R = 60 W
!
√
60 × (240 !)
60R
Therefore, v =
=
= 500 m/s
LB
(0.6 m) × (0.4 T)
In 0.5 s, the rod slides 250 m!
Wednesday, February 13, 2008
60 W lamp
There is an induced
current I in the bar
when the bar is moving
in the magnetic field.
17
Work done to light the lamp
B = 0.4 T
Magnetic force on the
bar, Fm = ILB, opposes
the motion of the bar.
!Fm
L = 0.6 m
Fapplied
In 1 s, work done by the applied force in opposing Fm is W = Fm v
W = Fmv = (ILB) v = I (LBv) = I V = 60 W
That is, the power to light the bulb is supplied by doing work against
the magnetic force.
Wednesday, February 13, 2008
18
Work done to light the lamp
!
B
!Fm
Fapplied
• Sliding the bar along the rails generates an emf
• When the circuit is completed, a current flows and lights up the lamp
• A magnetic force acts on the current in the rod to oppose the motion
of the rod (a consequence of Lenz’s law – later)
• The work done in pushing the rod is equal to the electrical energy
dissipated in the lamp – mechanical energy is converted into electrical.
Wednesday, February 13, 2008
19
Motional emf
L
The emf induced between the ends of the
falling rod is:
V = vLB
!v
L
No current is flowing, so there is no magnetic
force on the rod.
The resistor R completes the circuit, so that
current flows and there is now a magnetic
force resisting the gravitational force that
accelerates the rod downwards.
The rod stops accelerating when the magnetic
force is equal to the weight of the rod.
!v
Wednesday, February 13, 2008
20
Prob. 22.9: A conducting rod 1.3 m long slides down between two
frictionless vertical copper tracks at a constant speed of 4 m/s
perpendicular to a 0.5 T magnetic field.
a) What is the mass of the rod?
b) Find the change in gravitational PE in 0.2 s.
c) Find the electrical energy dissipated in the resistor in 0.2 s.
R = 0.75 !
L = 1.3 m
Wednesday, February 13, 2008
21
Induced emf
Changing the area of the loop also
induces an emf.
An emf is induced in the coil
whenever the number of field lines
passing through the coil changes.
The number of field lines is a
measure of “magnetic flux”.
→ There is an induced emf
whenever there is a change of
magnetic flux passing through the
coil.
Wednesday, February 13, 2008
22
Mastering Physics Assignment 2
Chapters 19, 20
due today at 11 pm!
Mastering Physics Assignment 3
Chapters 21, 22
due Friday, Feb 29 at 11 pm
Week of February 11 - 15
Experiment 3: e/m of electron
Wednesday, February 13, 2008
23
Work done to light the lamp
!
B
!Fm
Fapplied
• Sliding the bar along the rails generates a motional emf: V = vLB
• When the circuit is completed and a current flows, the resulting
magnetic force opposes the motion of the rod.
• The work done in pushing the rod is equal to the electrical energy
dissipated in the lamp – mechanical energy is converted into electrical.
Wednesday, February 13, 2008
24
Induced emf
L
A0
L
A 0 = x0L
A
A = xL
The emf induced between the ends of the moving rod is: V = vLB
Between time t0 and t, the rod moves a distance x – x0 = v(t – t0), so
(x − x0 )
V = vLB =
LB =
(t − t0 )
!
A − A0
t − t0
"
B=
(BA) − (BA)0
∆(BA)
=
t − t0
∆t
With B perpendicular to the loop, (BA) is the “magnetic flux” passing
through the loop. The induced emf is equal to the rate of change
of magnetic flux passing through the loop – Faraday’s Law.
Wednesday, February 13, 2008
25
Magnetic Flux
Unit of Magnetic Flux:
1 Weber (Wb) = 1 T.m2
Magnetic flux, ! = B cos " × A = BA cos "
Faraday’s Law: the induced emf is equal to the rate of change of magnetic flux
Wednesday, February 13, 2008
26
Magnetic Flux and Field Lines
! = BA cos 0◦ = BA
! = BA cos 60◦ = BA/2
! = BA cos 90◦ = 0
The magnetic flux is proportional to the number of magnetic field lines
passing through the coil.
Wednesday, February 13, 2008
R
27
Induced emf and rate of change of flux
!BI
B = 0.4 T
Faraday’s Law:
!Fm
L = 0.6 m
Fapplied
!BI
!BI
!"
= IR
!t
!A
=B
!t
V=
= BLv
As the rod is moved to the right, the area of the closed loop increases and
the magnetic flux passing through the loop increases. There is increasing
magnetic flux passing through the loop and pointing into the page.
The induced current, I, produces a magnetic field, BI, pointing out of
the page, that opposes the change in magnetic flux. This is Lenz’s law.
Wednesday, February 13, 2008
28
Faraday’s Law
The induced emf is equal to the
rate of change of magnetic flux.
Magnetic flux: ! = BA cos "
Lenz’s Law
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 in the first place.
Wednesday, February 13, 2008
29
Prob. 22.C11: Use Lenz’s law to verify
that the induced current is in the
direction in the diagram.
• The flux through the loop is into
the page and is decreasing as the
area of the loop decreases.
• The induced current produces a
magnetic field that opposes the
decreasing flux.
→ The magnetic field produced
by the induced current must point
into the page.
→ The current flows clockwise in
the loop.
Wednesday, February 13, 2008
30
Conducting ring falling through a magnetic field
I=0
The magnetic flux passing through the
ring is constant (zero), so there is no
induced emf or current.
I
The magnetic flux passing through
the ring is increasing and is directed
into the page.
!BI
!Fm
The induced current produces a
magnetic field, BI, that opposes the
increase of flux.
A magnetic force is generated that opposes the motion of the ring.
Wednesday, February 13, 2008
The magnetic flux passing
through the coil is constant, so
there is no induced emf or
current and no magnetic force.
31
!Fm = 0
!Fm
The magnetic flux passing
through the coil is decreasing
and points into the page.
The induced emf generates a current that
opposes the change in magnetic flux,
producing a magnetic field, BI, into the page.
x !BI
I
A magnetic force opposes the motion of the coil.
Wednesday, February 13, 2008
32
Eddy Currents
Move an electrical conductor into or out of a magnetic field. The
changing magnetic flux in the conductor produces a current that
circulates inside the conductor - an “Eddy current”.
Magnetic force
opposes motion
of metal !
electromagnetic
braking
Fm
Wednesday, February 13, 2008
33
Prob. 22.32/70: A bar magnet is falling through a metal ring. In part (a)
the ring is solid all the round around, but in part (b) it has been cut
through.
Explain why the motion of the magnet in part (a) is retarded, whereas it
is not in part (b).
Wednesday, February 13, 2008
34
Prob. 22.33: A circular loop of wire rests on a table. A long, straight
wire lies on this loop over its centre.
The current I in the straight wire is increasing. In what direction is
the induced current, if any, in the loop?
!
!
!
!
B
!
• What is the total magnetic flux through the loop?
• Does it change?
Wednesday, February 13, 2008
35
A wire is bent into a circular loop as shown. The radius of the circle is
2 cm. A constant magnetic field B = 0.55 T is directed perpendicular
to the plane of the loop. Someone grabs the ends of the wire and
pulls it taut, so the radius shrinks to zero in 0.25 s.
Find the magnitude of the average induced emf between the ends of
the wire.
B
I
Wednesday, February 13, 2008
36
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.
Induced magnetic field
I
BI
I
– – – –
BI
BI
++
++
E
I
Wednesday, February 13, 2008
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.
37
# = 15 rad/s
r = 0.5 m
BI I I
"θ
"
"
× #r2 = Br2!/2
The flux passing through the loop is: ! = BA = B
2#
!" Br2 !# Br2$
=
×
=
The induced emf is: V =
!t
2
!t
2
2
0.0038 × 0.5 × 15
and I = V/R = 0.00713/3
V=
= 7.13 × 10−3 V
2
= 2.4 mA
Wednesday, February 13, 2008
!
38
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
Wednesday, February 13, 2008
39
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
Wednesday, February 13, 2008
40
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
Wednesday, February 13, 2008
41
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.
Wednesday, February 13, 2008
42
Mastering Physics Assignment 3
Chapters 21, 22
due Friday, Feb 29 at 11 pm
Week of February 18 - 22
Midterm Break
Week of February 25 - 29
Tutorial and Test 2
Chapters 19, 20, 21
Wednesday, February 13, 2008
43
Faraday’s Law
The induced emf is equal to the
rate of change of magnetic flux.
Magnetic flux: ! = BA cos "
Lenz’s Law
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 in the first place.
Wednesday, February 13, 2008
44
Electric Generator
The magnetic flux passing
through the coil varies as the coil
rotates – an emf is generated.
Flux, ! = BA cos " = BA cos(#t)
Wednesday, February 13, 2008
45
EMF from electric generator,
using rate of change of flux
V = V0 sin(!t)
V0
Reminder of Lenz’s Law
Flux, ! = BA cos " = BA cos(#t)
Then, the induced emf is V = (−)
∆Φ
= BAω sin(ωt) (diff. calculus)
∆t
V = V0 sin(!t), V0 = BA!
Wednesday, February 13, 2008
46
Electric Generator
V = BLv sin !
EMF generated in moving conductors
V = BLv sin !
2V
Area, A = LW
+
–
V
–
+
V
W
Total emf generated: Vtot = 2BLv sin !
! "
W
v = r! =
!, ! = angular frequency of rotating coil
2
Therefore, Vtot = BLW ! sin " = BA! sin "
If the coil has N turns, then: Vtot = NBA! sin " = V0 sin " = V0 sin ωt
Wednesday, February 13, 2008
47
Alternating Current Electric Generator
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.
V0 = 28 V
The period is T = 0.42 s
Peak voltage, V0 = 28 V = NBA#.
V = NBAω sin ωt
(Could also be V = N BAω cos ωt)
14.96
! = 2" f , so f =
= 2.38 Hz
2"
2"
2"
=
= 14.96 rad/s
T
0.42 s
V0
28
Therefore, B =
=
= 0.624 T
NA! 150 × 0.02 × 14.96
!=
f = 1/T = 2.38 Hz
Wednesday, February 13, 2008
48
Root Mean Square (rms)
V = V0 sin #t
V0
V0 = NBA#
Time
Time
Power $ V2
V2 = V02 sin2 #t
V02
Mean = V02/2
Vrms = V0/!2 = rms voltage
Time
Mean power $ Vrms2 = V02/2
Wednesday, February 13, 2008
49
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...
Wednesday, February 13, 2008
50
Electric Generator
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.
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.”
Wednesday, February 13, 2008
51
Back emf
Vback
Symbol for
AC
generator
A power supply drives the motor
Kirchhoff’s loop law:
Wednesday, February 13, 2008
The motor acts as a generator
V −Vback = IR
52
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
Vback
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.
Wednesday, February 13, 2008
53
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?
Wednesday, February 13, 2008
54
22.43/74: A motor is designed to operate on 117 V and draws a
current of 12.2 A when it first starts up. At its normal operating
speed, the motor draws a current of 2.30 A.
Obtain:
a) the resistance of the armature coil,
b) the back emf developed at normal speed,
c) the current drawn by the motor at one-third normal speed.
Wednesday, February 13, 2008
55
Prob. 22.25/71: A conducting coil of 1850 turns is connected to a
galvanometer. The total resistance of the circuit is 45 !. The area
of each turn is 4.7 ! 10-4 m2.
The coil is moved into a magnetic field, the normal to the coil being
kept parallel to the magnetic field. The amount of charge that is
induced to flow around the circuit is 8.87 ! 10-3 C.
Find the magnitude of the magnetic field.
Wednesday, February 13, 2008
56
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
Wednesday, February 13, 2008
57