WileyPLUS Assignment 2
Chapters 20, 21
Due Friday, February 12 at 11 pm
Week of Feb 9 – 11
Tutorial and Test 2
Chapters 19, 20, 21
Midterm Break
Week of Feb 15-19 !
Monday, February 8, 2010
1
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)
Monday, February 8, 2010
2
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.
Monday, February 8, 2010
3
Moving coil relative to magnet
Motion of coil
toward the
magnet
I
�v
�
F
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.
Monday, February 8, 2010
4
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.
Monday, February 8, 2010
5
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.
Monday, February 8, 2010
6
Fm = !q v B
Motional emf
+!q
Fq = !q E
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:
Fq
∆qE
=
=
Fm
∆qvB
Also, E = V/L
Fm
The induced potential difference between the ends of the rod is: V = vLB
Monday, February 8, 2010
7
Induced emf
When the angle between B and v is ", the emf induced between
the ends of the conductor is:
V = vLB sin"
The induced emf is the same whether the coil moves or the
magnet moves, only the relative motion matters.
Monday, February 8, 2010
8
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.
V = vLB sin"
– +
+
–
A: V = 0; B: V = 1.6 V; C) V = 0 along length
Monday, February 8, 2010
9
Prob. 22.3/2: “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.
Monday, February 8, 2010
10
Prob. 22.6/4: 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
Blood – a moving
conductor
+++
L E
�
–––
�v
L = 5.6 mm
1 mV
Monday, February 8, 2010
11
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
L
Monday, February 8, 2010
12
The rod experiences a magnetic force to the right and accelerates.
V0
Fm = ILB
I
L
Monday, February 8, 2010
13
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
Fm = ILB
V = vLB I
L
v
–
Speed constant when vLB = V0
Monday, February 8, 2010
14
Induced emf – equivalent circuit
I
V = vLB
= “back emf”
V0
I
R
Resistance of rails and bar
V0 – V = IR
I is reduced to zero when V = V0
then
v = V0/(LB)
Monday, February 8, 2010
15
60 W bulb, R = 240 #
22.9/7: 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 in lamp, 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!
Monday, February 8, 2010
16
Work done to light the lamp
60 W lamp
There is an induced
current I in the bar
when the bar is moving
in the magnetic field.
B = 0.4 T
�Fm
L = 0.6 m
Fapplied
Magnetic force on the
bar, Fm = ILB, opposes
the motion of the bar.
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.
Monday, February 8, 2010
17
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.
Monday, February 8, 2010
18
Clicker Question:
You have three light bulbs;
bulb A has a resistance of 240 #,
bulb B has a resistance of 192 #, and
bulb C has a resistance of 144 #.
Each of these bulbs is used for the same amount of time in a setup like that
in the drawing. In each case the speed of the rod and the magnetic field
strength are the same. Rank the setups in descending order, according to how
much work the hand in the drawing must do (largest amount of work first).
a) B, A, C
b) B, C, A
c) C, B, A
d) A, C, B
e) A, B, C
c) Rate of doing work = P = V2/R
V is the same in all three cases.
Monday, February 8, 2010
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
Monday, February 8, 2010
20
Prob. 22.10/9: 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 ΩΩ
a) 0.23 kg
b) -1.803 J
c) 1.803 J
L = 1.3 m
Monday, February 8, 2010
21
Are charge carriers positive or negative?
If q positive, I (conventional flow) and v in same direction
If q negative, I (conventional flow) and v in opposite directions
If q > 0
+V
++++++++++++++++++++++
q
I
�
xB
�v
$
F�
If q < 0
–V
–––––––––––––––––––––––––
I
F�
q
�v
�
xB
$
For conductors, charge carriers are negative
Results may differ for semiconductors
Monday, February 8, 2010
22