Download PowerPoint

Physics 212
Lecture 16
Motional EMF
Conductors moving in B field
Induced emf !!
Physics 212 Lecture 16, Slide 1
The Big Idea
B
When a conductor moves through a region
containg a magnetic field:





05
Magnetic forces may be exerted on the charge


carriers in the conductor
XXXXXXXXX
F

F  qv  B
L + -
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
E
-
qvB  qE
E  vB
V  EL
FB
+
FE
+
E
-
V  vBL
Physics 212 Lecture 16, Slide 2
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)
Bar a
No force on charges
No charge separation
No E field
No emf
XXXXXXXXX
a
vX + - B
b
F
+ F
XXXXXXXXX
Fa = 0
:22
Fb = qvB
v
Bar b
Opposite forces on charges
Charge separation
E = vB
emf = EL = vBL
Physics 212 Lecture 16, Slide 3
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
+F
XXXXXXXXX
:22
Bar
F
Fb = qv0B
v0
Opposite forces on charges
Charge separation
E = v0B
emf = EL = v0BL
Physics 212 Lecture 16, Slide 4
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 5
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 6
L
E
+
v
t
Only leading side has charge separation
emf = BLv (cw current)
-
LE
+
E
-
+
t
v
Leading and trailing sides have charge separation
emf = BLv – BLv = 0 (no current)
-
LE
+
20
Only trailing side has charge separation
emf = BLv (ccw current)
v
t
Physics 212 Lecture 16, Slide 7
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 8
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 9
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 10
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 Lv  B  2 L( )B cos q  AB cos(t )
q
2
22
=v
Physics 212 Lecture 16, Slide 11
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 12
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 next time !
Physics 212 Lecture 16, Slide 13
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 14
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 15
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 16
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 17
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 18
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 
qvB  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 19
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 
qvB  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 20
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 21
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 22
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 23