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Physics 212
Lecture 17
Faraday’s Law
Physics 212 Lecture 17, Slide 1
Motional EMF
Change Area
of loop
Change magnetic field
through loop
Change orientation
of loop relative to B
In each case the flux of the magnetic field through the
circuit changes with time and an EMF is produced.
 B   B  dA
d B
dt
EMF
Physics 212 Lecture 17, Slide 2
Rotate the loop,change flux, generate emf.
B
Physics 212 Lecture 17, Slide 3
Move loop to a place wherethe B field is different,
change flux, generate emf.
B1
B2
v
Physics 212 Lecture 17, Slide 4
Checkpoint 1a
A copper loop is placed in a uniform magnetic field as shown. You are looking from the right.
Suppose the loop is moving to the right. The current induced in the loop is:
A. zero
B. clockwise
C. counterclockwise
• Motional emf is ZERO
•vXB=0
• no charge separation
• no E field
• no emf
• The flux is NOT changing
• B does not change
• the area does not change
• the orientation of B and A does not change
Physics 212 Lecture 17, Slide 5
Checkpoint 1c
Now suppose that the loop is spun around a vertical axis as shown, and that it makes one
complete revolution every second.
The current induced in the loop:
A. Is zero
B. Changes direction once per second
C. Changes direction twice per second
Current changes direction every time the loop
becomes perpendicular with the B field
emf ~ d/dt
(B  dA = max)  d/dt (B  dA ) = 0
O
X
B
dA
O
B
X
dA
Physics 212 Lecture 17, Slide 6
Faraday’s Discovery
d B
emf  
dt
True no matter how we change the flux. In fact,
the circuit may be stationary (no motional EMF) and
only the B-field changes with time. An EMF is still
produced. This implies that:
d B
emf   E  dl  
dt
Faraday’s Law
Change the B field in time so flux changes. Induce an emf
nnd therefore an Electric field. This emf tries to oppose
the change in flux. (Lenz’s Law)
 
d B
emf   E  d   
dt
B(t) decreasing
Induces an E field even if
there is no circuit there!
Physics 212 Lecture 17, Slide 8
Checkpoint 1b
A copper loop is placed in a uniform magnetic field as shown. You are looking from the right.
Looking from right
X X X X X X X X
X X X X X X X X
Now suppose the that loop is stationary and thatCheckpoint
the magnetic field
1b is
decreasing in time. The current induced in the loop is:
A. zero
B. clockwise
C. counterclockwise
• Motional emf is ZERO
• Circuit is stationary !
• HOWEVER: The flux is changing
• B decreases in time
• current induced to oppose the flux change
• clockwise current tries to restore B
that was removed
X X X X X X X X
X X X X X X X X
X X X X X X X X
Clockwise current
tries to restore B
Physics 212 Lecture 17, Slide 9
Checkpoint 2
A horizontal copper ring is dropped from rest directly above the north pole of a permanent magnet
F
O
X
B
B
Like poles repel
(copper is not
ferromagnetic)
Ftotal < mg
Will the acceleration a of the falling ring in the presence of the magnet
be any different than it would have been under the influence of just
gravity (i.e. g)?
A. a > g
B. a = g
C. a < g
a<g
This one is hard !
B field increases upward as loop falls
Clockwise current (viewed from top) is induced
Physics 212 Lecture 17, Slide 10
Checkpoint 2
A horizontal copper ring is dropped from rest directly above the north pole of a permanent magnet
HOW
IT
WORKS
Looking down
B
(copper is not
ferromagnetic)
Will the acceleration a of the falling ring in the presence of the magnet
be any different than it would have been under the influence of just
gravity (i.e. g)?
A. a > g
B. a = g
C. a < g
This one is hard !
B field increases upward as loop falls
Clockwise current (viewed from top) is induced
Main Field produces horizontal forces
“Fringe” Field produces vertical force
I
I
B
IL X B points UP
Ftotal < mg
a<g
Physics 212 Lecture 17, Slide 11
Calculation
A rectangular loop (height = a, length = b,
resistance = R, mass = m) coasts with a constant
velocity v0 in + x direction as shown. At t =0, the
loop enters a region of constant magnetic field B
directed in the –z direction.
y
a
v0
B
b x x x x x x x
x x x x x x x
x x x x x x x
x x x x x x x
x
What is the direction and the magnitude of the
force on the loop when half of it is in the field?
• Conceptual Analysis
–
–
Once loop enters B field region, flux will be changing in time
Faraday’s Law then says emf will be induced
• Strategic Analysis
–
–
–
Find the emf
Find the current in the loop
Find the force on the current
Physics 212 Lecture 17, Slide 12
Calculation
A rectangular loop (height = a, length = b,
resistance = R, mass = m) coasts with a constant
velocity v0 in + x direction as shown. At t =0, the
loop enters a region of constant magnetic field B
directed in the –z direction.
emf  
d B
dt
y
a
v0
B
b x x x x x x x
x x x x x x x
x x x x x x x
x x x x x x x
x
What is the magnitude of the emf induced
in the loop just after it enters the field?
(A) e = Babv02
(B) e = ½ Bav0 (C) e = ½ Bbv0 (D) e = Bav0 (E) e = Bbv0
y
a
v0
B
x
x
x
x x x x
b
x x x x x x x
a
x x x x x x x
x x x x x x x
Change in Flux = dB = BdA = Bav0dt
x
In a time dt
it moves by v0dt
The area in field
changes by dA = v0dt a
d B
 Bav o
dt
Physics 212 Lecture 17, Slide 13
Calculation
emf  
A rectangular loop (height = a, length = b,
resistance = R, mass = m) coasts with a constant
velocity v0 in + x direction as shown. At t =0, the
loop enters a region of constant magnetic field B
directed in the –z direction.
y
a
v0
d B
dt
B
b x x x x x x x
x x x x x x x
x x x x x x x
x x x x x x x
x
What is the direction of the current induced
in the loop just after it enters the field?
(A) clockwise
(B) counterclockwise
(C) no current is induced
emf is induced in direction to oppose the change in flux that produced it
y
a
v0
B
b x x x x x x x
x x x x x x x
x x x x x x x
x x x x x x x
Flux is increasing into the screen
Induced emf produces flux out of screen
x
Physics 212 Lecture 17, Slide 14
Calculation
A rectangular loop (height = a, length = b,
resistance = R, mass = m) coasts with a constant
velocity v0 in + x direction as shown. At t =0, the
loop enters a region of constant magnetic field B
directed in the –z direction.
emf  
y
a
v0
d B
dt
B
b x x x x x x x
x x x x x x x
x x x x x x x
x x x x x x x
x
What is the direction of the net force
on the loop just after it enters the field?
(A) +y
(B) -y
(C) +x
(D) -x

 
Force on a current in a magnetic field: F  IL  B
y
b
a
B
x x x x x x x
v0
I
x x x x x x x
• Force on top and bottom segments cancel (red arrows)
• Force on right segment is directed in –x direction.
x
Physics 212 Lecture 17, Slide 15
Calculation
A rectangular loop (height = a, length = b,
resistance = R, mass = m) coasts with a constant
velocity v0 in + x direction as shown. At t =0, the
loop enters a region of constant magnetic field B
directed in the –z direction.
What is the magnitude of the net force on
the loop just after it enters the field?
(A) F  4aBvo R (B) F  a 2 Bvo R
(C) F  a 2 B 2vo2 / R

 
F  IL  B
emf  
y
a
v0
d B
dt
B
b x x x x x x x
x x x x x x x
x x x x x x x
x x x x x x x
x
F  IL  B
e = Bav0
(D) F  a 2 B 2vo / R
 
F  ILB since L  B
y
b
a
F
B
x x x x x x x
v0
I
e
Bavo
I 
R
R
x x x x x x x
B 2 a 2vo
 Bav o 
F 
aB 
R
R


ILB
x
Physics 212 Lecture 17, Slide 16
Follow-Up
A rectangular loop (sides = a,b, resistance = R,
mass = m) coasts with a constant velocity v0 in + x
direction as shown. At t =0, the loop enters a
region of constant magnetic field B directed in a
the –z direction.
t = dt: e = Bav0
y
b
v0
B
x x x x x x x
x x x x x x x
x x x x x x x
x x x x x x x
x
What is the velocity of the loop when half of it is
in the field?
Which of these plots best represents the velocity as a function of
time as the loop moves form entering the field to halfway through ?
(A)
(B)
(C)
D)
(E)
X
This is not obvious,
but we know v must
decrease
Why?
X
b
a
Fright
B
x x x x x x x
v0
I
x x x x x x x
X
Fright points to left
Acceleration negative
Speed must decrease
Physics 212 Lecture 17, Slide 17
Follow-Up
A rectangular loop (sides = a,b, resistance = R,
mass = m) coasts with a constant velocity v0 in + x
direction as shown. At t =0, the loop enters a
region of constant magnetic field B directed in
the –z direction.
y
b
a
What is the velocity of the loop when half of it is
in the field?
v0
B
x x x x x x x
x x x x x x x
x x x x x x x
x x x x x x x
e = Bav0
x
Which of these plots best represents the velocity as a function of
time as the loop moves form entering the field to halfway through ?
dv
(A)
F  a B v / R (D)
m
• Why (D), not (A)?
2
2
dt
–
F is not constant, depends on v
a 2 B 2v
dv
F 
m
R
dt
Challenge: Look at energy
v  vo e t
2 2
a
where   B
mR
Claim: The decrease in kinetic energy of
loop is equal to the energy dissipated as
heat in the resistor. Can you verify??
Physics 212 Lecture 17, Slide 18