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Physics 212
Lecture 16
Motional EMF
Conductors moving in B field
Induced emf !!
Physics 212 Lecture 16, Slide 1
Music
Who is the Artist?
A)
B)
C)
D)
E)
Gram Parsons
Tom Waits
Elvis Costello
Townes Van Zandt
John Prine
Why?
Theme of the week?
Guys who sing with Emmylou Harris !!
I do like these albums..
Giving you a break from Jazz this week…
Electromotive Force (EMF):


f  dl
where
Units of EMF:
work = force x distance
EMF = force/q x distance
= work/q
=
Closed loop integral, so …
F
f 
q
Force per
unit charge
of ANY sort
e.g.:
fE  E
fB  v  B
To get a non-zero EMF,
need a force that can
push charges around a loop,
i.e.
Physics 212 Lecture 16, Slide 5
The Big Idea
B
When a conductor moves through a region
containg a magnetic field:
XXXXXXXXX
F
L + -
Magnetic forces may be exerted on the charge
carriers in the conductor:
F
v
XXXXXXXXX
FB  qv  B
Creates an
-> potential difference
across the ends of the bar:
Vdue to B   f B  dl  (v  B)  L
This is called MOTIONAL EMF:
 vBL
in above
example
Move a conductor in a magnetic field in the right way,
and you’ve got a magnetic battery
05
Physics 212 Lecture 16, Slide 8
Putting it together
Change Area
of loop
Change magnetic field
through loop
Change orientation
of loop relative to B
Faraday’s Law
  B A
d
 
dt
We will study this law in detail & in another context next time !
Physics 212 Lecture 16, Slide 9
Two identical conducting bars (shown in end view) are moving through a vertical magnetic
field. Bar (a) is moving vertically and bar (b) is moving horizontally.
Checkpoint 1
B
Which of the following is true?
A.
A A motional emf exists in the bar for case (a), but not (b)
B A motional emf exists in the bar for case (b), but not (a)
B.
C
C.
D A motional emf exists in the bar for both cases (a) and (b)
D. A motional emf exists in the bar for neither case (a) nor case (b)
Rotate picture by 90o
(top view)
XXXXXXXXX
a
vX + - B
b
F
+ F
XXXXXXXXX
v || B  Fa = 0
:22
Fb = qvB
Bar a
No magnetic force
on charges
No charge separation
No emf
v
Bar b
Magnetic force on charges
Charge separation
emf = (F/q)L = vBL
Physics 212 Lecture 16, Slide 10
A conducting bar (green) rests on two frictionless wires connected by a resistor as shown.
Checkpoint 2a
Changing Area
A
B
The
C entire apparatus is placed in a uniform magnetic field pointing
into
D the screen, and the bar is given an initial velocity to the right.
The motion
of the
green bar
Rotate
picture
bycreates
90o a current through the bar
A. going up
A
B.
B going down
:22
Physics 212 Lecture 16, Slide 11
A conducting bar (green) rests on two frictionless wires connected by a resistor as shown.
Checkpoint 2a
Equivalent circuit
Changing Area
A
B
The
C entire apparatus is placed in a uniform magnetic field pointing
into
D the screen, and the bar is given an initial velocity to the right.
R
I
V
The motion
of the
green bar
Rotate
picture
bycreates
90o a current through the bar
A. going up
A
B.
B going down
XXXXXXXXX
B
Bar
F
+F
v0
Opposite forces on charges
Charge separation
emf = (F/q)L = v0BL
XXXXXXXXX
:22
Fb = qv0B
Physics 212 Lecture 16, Slide 12
A conducting bar (green) rests on two frictionless wires connected by a resistor as shown.
Checkpoint 2b
F
Energy
I
External agent must
exert force F to the
right to maintain
constant v0
Changing Area
A
B
The
entire apparatus is placed in a uniform magnetic field pointing
C
into
the screen, and the bar is given an initial velocity to the right.
D
The current through this bar results in a force on the bar
A.
A down
B up
B.
C right
C.
D
D.
E left
E. into the screen
Preflight 5
This energy is
dissipated in the
resistor!
Counterclockwise Current

 
F  IL  B
:22
 vBL 
F 
 LB
 R 
F points to left
 vBL 
2
P  Fv  
 LBv  I R
 R 
Physics 212 Lecture 16, Slide 13
A wire loop travels to the right at a constant velocity. Which plot best represents the induced
current in the loop as it travels from left of the region of magnetic field, through the magnetic
field, and then entirely out on the right side?
B=0
v
B=5T
Out of Screen
Checkpoint 5
20
Physics 212 Lecture 16, Slide 14
A wire loop travels to the right at a constant velocity. Which plot best represents the induced
current in the loop as it travels from left of the region of magnetic field, through the magnetic
field, and then entirely out on the right side?
B=0
v
B=5T
Out of Screen
Let’s step
through this one
Checkpoint 5
20
Physics 212 Lecture 16, Slide 15
A wire loop travels to the right at a constant velocity. Which plot best represents the induced current in
the loop as it travels from left of the region of magnetic field, through the magnetic field, and then entirely
out on the right side?
L
fB
+
v
t
Only leading side has charge separation
emf = BLv (cw current)
A wire loop travels to the right at a constant velocity. Which plot best represents the induced current in
the loop as it travels from left of the region of magnetic field, through the magnetic field, and then entirely
out on the right side?
-
L fB fB
+
-
+
t
v
Leading and trailing sides have charge separation
emf = BLv – BLv = 0 (no current)
A wire loop travels to the right at a constant velocity. Which plot best represents the induced current in
the loop as it travels from left of the region of magnetic field, through the magnetic field, and then entirely
out on the right side?
-
L fB
+
20
Only trailing side has charge separation
emf = BLv (ccw current)
v
t
Physics 212 Lecture 16, Slide 16
A wire loop travels to the right at a constant velocity. Which plot best represents the induced
current in the loop as it travels from left of the region of magnetic field, through the magnetic
field, and then entirely out on the right side?
B=0
v
B=5T
Out of Screen
Checkpoint 5
20
Physics 212 Lecture 16, Slide 17
Changing B field
A conducting rectangular loop moves with velocity v toward an infinite straight wire carrying
current as shown.
Checkpoint 3
What is the direction of the induced current in the loop?
A. clockwise
B. counter-clockwise
C. there is no induced current in the loop
20
Physics 212 Lecture 16, Slide 18
Changing B field
A conducting rectangular loop moves with velocity v toward an infinite straight wire carrying
current as shown.
Checkpoint 3
What is the direction of the induced current in the loop?
A. clockwise
B. counter-clockwise
C. there is no induced current in the loop
B 
-
 +
B  
+
v
-
 B
  B
I
20
Physics 212 Lecture 16, Slide 19
Generator: Changing Orientation
On which legs of the loop is charge separated?
A)
B)
C)
D)
22
Top and Bottom legs only
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 20
Generator: Changing Orientation
L
w
At what angle q is emf the largest?
(A) q = 0
(B) q = 45o
(C) q = 90o
(D) emf is same at all angles
 
vB
Largest for q = 0 (v perp to B)
 
F
w
  2 EL  2 L  2 L v  B  2 L ( ) B cos q   AB cos( t )
q
2
22
=v
Physics 212 Lecture 16, Slide 21
Generator: Changing Orientation
A rectangular loop rotates in a region containing a constant magnetic field as shown.
The side view of the loop is shown at a particular time during the rotation. At this time, what
is the direction of the induced (positive) current in segment ab?
A. from b to a
B. from a to b
C. there is no induced current in the loop at this time
20
Checkpoint 4
Physics 212 Lecture 16, Slide 22
Generator: Changing Orientation
A rectangular loop rotates in a region containing a constant magnetic field as shown.
The side view of the loop is shown at a particular time during the rotation. At this time, what
is the direction of the induced (positive) current in segment ab?
A. from b to a
B. from a to b
C. there is no induced current in the loop at this time
v
 
vB

+
v
20
Checkpoint 4
I
X 
B
vB
Physics 212 Lecture 16, Slide 23
Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total
resistance of 5W 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 24
Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total
resistance of 5W 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
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 25
Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total
resistance of 5W 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?
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)
Zero
 
vB
Physics 212 Lecture 16, Slide 26
Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total
resistance of 5W 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?
B into page
I
v
h
x
L
What is the emf induced on the top segment?
A) left is positive
B)
left is negative
C)
Zero
 
vB
perpendicular to wire
Physics 212 Lecture 16, Slide 27
Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total
resistance of 5W 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?
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)
Zero
 
vB
Physics 212 Lecture 16, Slide 28
Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total
resistance of 5W 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?
B into page
I
v
h
x
L
Which expression represents the emf induced in the left wire
(A)
(B)
(C)
 left
 I
 o Lv
2x
 left
 I
 o hv
2x
 left 
qv B  qE
 I
B o
2x
  Eh  vBh
E  vB

o I
hv
2x
o I
Lv
2 ( L  x )
Physics 212 Lecture 16, Slide 29
Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total
resistance of 5W 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?
B into page
I
v
h
x
L
Which expression represents the emf induced in the right wire
(A)  right
o I

hv
2 ( L  x )
 I
(B)  right  o hv
2x
(C)  right 
qv B  qE
B
o I
2 ( L  x )
  Eh  vBh
E  vB

o I
hv
2 ( L  x )
o I
Lv
2 ( h  x )
Physics 212 Lecture 16, Slide 30
Example Problem
A rectangular loop (h=0.3m L=1.2 m) with total
resistance of 5W 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?
B into page
I
v
h
x
L
Which expression represents the total emf in the loop?
(A)
 loop 
o I
o I
hv 
hv
2x
2 ( L  x )
(B)
 loop 
o I
o I
hv 
hv
2x
2 ( L  x )
(C)
 loop  0
I loop 
 loop
R
 I 1
1 
I loop  o hv  

2R  x L  x 
Physics 212 Lecture 16, Slide 31
Follow-Up
A rectangular loop (h=0.3m L=1.2 m) with total
resistance of 5W is moving away from a long straight
wire carrying total current 8 amps.
I
B into page
+
+
v
I
h
x
L
What is the direction of the induced current?
(A)
Clockwise
(B)
Counterclockwise
Left > Right
Clockwise current
Physics 212 Lecture 16, Slide 32
Follow-Up
A rectangular loop (h=0.3m L=1.2 m) with total
resistance of 5W is moving away from a long straight
wire carrying total current 8 amps.
B into page
I
What is the direction of the force exerted by
the magnetic field on the loop?
(A)
UP
(B)
DOWN
(C)
LEFT
(B)
RIGHT
(B)
F = 0
h
x
I
v
L
B into page
I
Total force from B
Points to the left !!
Physics 212 Lecture 16, Slide 33