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Unformatted text preview: Physics 212
Lecture 16 Motional EMF
Conductors moving in B field Induced emf !!
Induced emf 50
40
30
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0 Confused Avg = 2.7 Confident Physics 212 Lecture 16, Slide 1 Music
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A)
B)
C)
D)
E) Susan Tedeschi
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Marcia Ball
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mood Wonderful albums Physics 212 Lecture 16, Slide 2 Physics 212
Lecture 16 Motional EMF
Conductors moving in B field Induced emf !!
Induced emf 50
40
30
20
10
0 Confused Avg = 2.7 Confident Physics 212 Lecture 16, Slide 3 The Big Idea
B When a conductor moves through a region
containg a magnetic field:
Magnetic forces may be exerted on the charge
carriers in the conductor XXXXXXXXX
F L + F = qv × B v F
XXXXXXXXX + These forces produce a charge separation in the
conductor  + This charge distribution creates an electric field
in the conductor
The equilibrium distribution is reached when the
forces from the electric and magnetic fields
cancel
The equilibrium electric field produces a
potential difference (emf) in the conductor
05 E  qvB = qE FB + E = vB
V = EL FE +
E  V = vBL Physics 212 Lecture 16, Slide 4 Preflight 2
BB A
B
C
D Rotate picture by 90o Bar a XXXXXXXXX No force on charges
No charge separation
No E field
No emf a b
F vX +  B +F XXXXXXXXX Fa = 0
:22 Fb = qvB
qvB v Bar b
Opposite forces on charges
Charge separation
E = vB
emf = EL = vBL 80
60
40
20
0 Physics 212 Lecture 16, Slide 5 BB Equivalent circuit
A
B
C
D Changing Area
Rotate picture by 90o A
B V Preflight 4
70
60 XXXXXXXXX 50 Bar F B +F XXXXXXXXX
:22 I R Fb = qv0B v0 40 Opposite forces on charges
Charge separation
E = v0B
emf = EL = v0BL 30
20
10
0 Physics 212 Lecture 16, Slide 6 BB F Energy
External agent must
External
exert force F to the
right to maintain
constant v
constant I
A
B
C
D Changing Area A
B
C
D
E Preflight 50
5
Preflight 5
40 Counterclockwise Current F = IL × B vBL F = LB
R
:22 This energy is
This
dissipated in the
resistor!
resistor! F points to left vBL 2
P = Fv = LBv = I R
R 30
20
10
0 Physics 212 Lecture 16, Slide 7 BB Let’s step
step
through this one
through 60
50
40
30
20
10 Preflight 11
20 0 Physics 212 Lecture 16, Slide 8 L E + v t Only leading side has charge separation
emf = BLv (cw current)
BLv  LE + E + t
v Leading and trailing sides have charge separation
emf = BLv – BLv = 0 (no current)
BLv  LE +
20 Only trailing side has charge separation
emf = BLv (ccw current)
BLv v t Physics 212 Lecture 16, Slide 9 Preflight 11
20 Physics 212 Lecture 16, Slide 10
10 Changing B field
BB Preflight 6
50 ∏ ∏+ B ∏ B ∏∏∏ + v  ∏B ∏∏∏ B 40
30
20
10 I
20 0 Physics 212 Lecture 16, Slide 11
11 Generator: Changing Orientation
BB On which legs of the loop is charge separated?
A)
B)
B)
C)
D) 22 Top and Bottom legs only
Top
Front and Back legs only
All legs
None of the legs v×B
parallel to top and bottom legs
Perpendicular to front and back legs Physics 212 Lecture 16, Slide 12
12 Generator: Changing Orientation
BB
L w At what angle θ is emf the largest?
At
is emf
(A) θ = 0
(B) θ = 45o
(C) θ = 90o
(C)
(D) emf is same at all angles
(D) emf v×B Largest for θ = 0 (v perp to B)
Largest
(v perp F
w
ε = 2 EL = 2 L = 2 Lv × B = 2 L( )ωB cosθ = ωAB cos(ωt )
q
2
22 Physics 212 Lecture 16, Slide 13
13 Changing Orientation
BB Preflight 8
50 v v×B ∏
+ v I X v×B 40 B 30
20
10
0 20 Physics 212 Lecture 16, Slide 14
14 Putting it together Change Area
of loop Change magnetic field
through loop Change orientation
of loop relative to B Faraday’s Law Φ ≡ B⋅ A dΦ
ε =−
dt
Physics 212 Lecture 16, Slide 15
15 Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total resistance of
5Ω is moving away from a long straight wire carrying total
current 8 amps. What is the induced current in the loop when
it is a distance x=0.7 m from the wire? I v h
x L •Conceptual Analysis:
•Long straight current creates magnetic field in region of the loop.
•Vertical sides develop emf due to motion through B field
•Net emf produces current •Strategic Analysis:
•Calculate B field due to wire.
•Calculate motional emf for each segment
•Use net emf and Ohm’s law to get current Physics 212 Lecture 16, Slide 16
16 Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total resistance of
5Ω is moving away from a long straight wire carrying total
current 8 amps. What is the induced current in the loop when
it is a distance x=0.7 m from the wire? BB I v h
x L What is the direction of the B field produced by the wire in the
region of the loop? A) Into the page
B) Out of the page
C) Left D) Right
E) Up
Physics 212 Lecture 16, Slide 17
17 Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total resistance of
5Ω is moving away from a long straight wire carrying total
current 8 amps. What is the induced current in the loop when
it is a distance x=0.7 m from the wire? BB B into page
I v h
x L What is the emf induced on the left segment? A) Top is positive
B) Top is negative
C) v×B Zero Physics 212 Lecture 16, Slide 18
18 Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total resistance of
5Ω is moving away from a long straight wire carrying total
current 8 amps. What is the induced current in the loop when
it is a distance x=0.7 m from the wire? BB B into page
I v h
x L What is the emf induced on the top segment? A) left is positive v×B B) left is negative perpendicular to wire C) Zero Physics 212 Lecture 16, Slide 19
19 Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total resistance of
5Ω is moving away from a long straight wire carrying total
current 8 amps. What is the induced current in the loop when
it is a distance x=0.7 m from the wire? BB B into page
I v h
x L What is the emf induced on the right segment? A) top is positive
B) top is negative
C) v×B Zero Physics 212 Lecture 16, Slide 20
20 Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total resistance of
5Ω is moving away from a long straight wire carrying total
current 8 amps. What is the induced current in the loop when
it is a distance x=0.7 m from the wire? BB B into page
I v h
x L Which expression represents the emf induced in the left wire
(A) ε left = µ0 I
Lv
2π x (B) ε left = µ0 I
hv
2π x (C) ε left = µ0 I 2π ( L + x ) ε = Bhv
Lv µ0 I
B=
2πx µ0 I
ε=
hv
2πx
Physics 212 Lecture 16, Slide 21
21 Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total resistance of
5Ω is moving away from a long straight wire carrying total
current 8 amps. What is the induced current in the loop when
it is a distance x=0.7 m from the wire? BB B into page
I v h
x L Which expression represents the emf induced in the right wire
(A) ε right = 2π ( L + x ) hv µ0 I
hv
2π x
µ0 I
=
Lv
2π (h + x ) (B) ε right =
(C) ε right µ0 I ε = Bhv
B= µ0 I 2π ( L + x) ε= µ0 I 2π ( L + x) hv Physics 212 Lecture 16, Slide 22
22 Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total resistance of
5Ω is moving away from a long straight wire carrying total
current 8 amps. What is the induced current in the loop when
it is a distance x=0.7 m from the wire? BB B into page
I v h
x L Which expression represents the total emf in the loop?
(A) ε loop = µ0 I
µ0 I
hv +
hv
2π x
2π ( L + x ) Iloop = (B) µI
µ0 I
ε loop = 0 hv −
hv
2π x
2π ( L + x ) Iloop = (C) ε loop = 0 ε loop
R µ0 I 1
1
hv − 2π R x x + L Physics 212 Lecture 16, Slide 23
23 ...
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