Faraday's Law
dS
B
B
• A Practice Exam is posted on the web page
• The Equation sheet is posted on the web page
• Exam material:
• From RC circuits through Faraday’s Law discussion
• Does not include RL circuits
• We are near the end of the course
• A tip: Don’t get behind on the FlipItPhysics and HW’s
Physics 122
Lecture 22
Magnetization (problem 5 in HW)
Dipoles in paramagnets or ferromagnets will align in the direction
of external field contributing a net  (magnetic moment). The per
unit volume is called the magnetization:
/
The magnetization contributes to the field in addition to the field
due to currents (in the textbook :
)
Gauss Law for B fields
Remember that the sources of E field flux were charges:
.
/
There are no monopoles for magnetic fields
.
Lines of B field
do not start from
points in space
as E (where lines
started at
charges).
A Quick Reminder to get started today
We will use Lenz’ Law:
EMF directed to
oppose the change in
flux through the loop
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?
1.
2.
3.
4.
As loop rotates up, flux “enters” from left to right
The loop produces an EMF to oppose this.
EMF drives a current around the loop
Direction of current in loop must create magnetic field
(or flux through loop) that points from right to left
5. Curl right hand along a to b and around top of loop into
page; your thumb points to the left
Wednesday’s Observations
EMF Observed when …
Change Area of loop
Change magnetic field
through loop (could also
be B(t)
Change orientation of
loop relative to B
Can be also understood with Faraday’s Law
 
  B A
d
 
dt
Faraday's Law
• Define the flux of the magnetic field through a surface
(closed or open) from:


 B   B  dS
dS
B

B
• Faraday's Law:
The emf induced in a circuit is determined by the time rate
of change of the magnetic flux through that circuit.
 
d B
emf   E  d   
dt
We will also talk
about this
The minus sign indicates
direction of induced current
(given by Lenz's Law).
Quick Flux Clickers
Suppose you double the magnetic field in a given region and
quadruple the area through which this magnetic field exists.
The effect on the flux through this area would be to
A.
B.
C.
D.
E.
Leave it unchanged
Double it
Quadruple it
Increase by factor of 6
Increase it by factor of 8


 B   B  dS
B’  2B
A’  4A
’B  8 B
Flux Clicker


 B   B  dS
A 3.0-cm by 5.0-cm rectangular coil has 100 turns. Its axis makes an
angle of 55º with a uniform magnetic field of 0.35 T. What is the
magnetic flux through this coil?
A.
B.
C.
D.
E.
3.0 x 10–4 Wb
4.3 x 10–4 Wb
3.0 x 10–2 Wb
4.3 x 10–2 Wb
5.3 x 10–2 Wb
Area = .03 x 0.05 = .0015 m2
x 100 turns
cos(55) = 0.57
B = (0.35 T)(100)(.0015)(.57) = 0.03 Wb
=55
Reminder: Lenz's Law
• Lenz's Law:
The induced current will appear in such a direction that it
opposes the change in flux that produced it.
B
B
S
N
v
N
S
v
Clicker - checkup
A conducting rectangular loop moves with
constant velocity v in the -y direction and a
constant current I flows in the +x direction
y
as shown.
I
i
What is the direction of the induced current
in the loop?
v
x
(a) ccw
(b) cw
(c) no induced current
The flux through this loop DOES change in time since the
loop is moving from a region of higher magnetic field to a
region of lower field.
By Lenz’ Law, an EMF will be induced which will oppose
the change of flux.
The current i is induced in the clockwise direction to
restore the flux.
Flux Clicker (look closely)
A loop rests in the xy plane. The z axis is normal to the plane. The
direction of the changing flux is indicated by the arrow on the z axis.
The diagram that correctly shows the direction of the resultant
induced current in the loop is
B
increasing
B
decreasing
No
Flux is
increasing
in +z;
Current
adds flux
in +z
direction
B
increasing
B
decreasing
B
B
B
B
B
A)
B
increasing
B)
No
Flux is
reduced
in –z;
need to
add flux
in –z
C)
D)
E)
No
No
YES !!
Flux is
Flux is
Flux is
increasing increasing getting
smaller;
in –z;
in –z;
in +z;
need to
Current
add flux adds flux need to
add flux
in +z
in +z
in +z
Demo E&M Jumping Rings
v
• Connect solenoid to a source of
alternating voltage.
• The flux through the area  to axis of
solenoid therefore changes in time.
~
side view
• A conducting ring placed on top of the
solenoid will have a current induced in it
opposing this change.
F B

• There will then be a force on the ring
since it contains a current which is
circulating in the presence of a magnetic
field.
B
F
B
top view
Think about this for a minute …
A horizontal conducting ring is dropped from rest above
the north pole of a permanent magnet
F
O
X
B
B
Like poles repel Ftotal  mg
ag
Will the acceleration a of the falling ring 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
B field increases upward as loop falls
Clockwise current (viewed from top) is induced Think about this for a minute …
A horizontal conducting ring is dropped from rest above
the north pole of a permanent magnet
HOW IT WORKS
Looking down
B
B
I
I
IL X B points UP
Ftotal  mg
Will the acceleration a of the falling ring 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
Summary:
Faraday’s Law:
 
d B
emf   E  d   
dt
where
 
 B   B  dA
emf → current → field a) induced only when flux is changing
b) opposes the change
Flux in loop can change when …
• Magnitude of magnetic field varies in time
• Loop orientation with respect to field varies
• Size of loop varies
Checkpoint
Suppose a current flows in a horizontal conducting loop in such a way
that the magnetic flux produced by this current points upward.
1) As viewed from above, in which direction is this current flowing?
This is just the RHR
Put fingers in direction of current;
Thumb points in the field direction
Checkpoint
A magnet makes the vertical magnetic field shown by the red arrows.
A horizontal conducting loop passes though the field from left to right
as shown.
The upward flux through the loop as a function of time is shown by the blue trace. Which of
the red traces below best represents the current induced in the loop as a function of time as
it passes over the magnet? (Positive means counter-clockwise as viewed from above):
1. Enters, induced flux must point down 
negative pulse
2. Middle no change
3. Leaving; flux must point up, positive pulse
A Change in B makes an electric field
• Faraday's law: a changing B induces an
emf which can produce a current in a x
loop.
x
x xEx x x x x x x
E
xxxxxxxxx
r
• For charges to move, there must be an
xxxxxxxxxx
electric field (and a path)
B
xxxxxxxxxx
• Faraday's law in terms of E field
E
produced by a changing B field.
x x x x x x x xEx x
 
d B
emf   E  d   
dt
• Suppose B is increasing into the screen as shown above. An E
field is induced in the direction shown. To move a charge q
 
around the circle would require an amount of work = W   qE  dl
• This work can also be calculated from  = W/q.
Clicker
The magnetic field in a region of space of
radius 2R is aligned with the z-direction and
changes in time as shown in the plot.
What is sign of the induced emf in a ring
of radius R at time t=t1?
(a)  < 0
( E ccw)
(b)  = 0
(c)  > 0
( E cw)
y
XXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
R
XXXXXXXXX
XXXXXXXX
XXXXXXX
XXXX
Bz
The magnetic field is increasing at t = t1.
It started into the board, and reverses at t = t1 .
It does not matter that it is actually zero, d/dt >0
The induced EMF make a field in the negative z
direction to compensate this change
That’s into the page:  clockwise current
Field at t = 0
t1
x
t
Clicker
What is the relation between the
magnitudes of the induced electric fields ER
at radius R and E2R at radius 2R ?
(a) E2R = ER (b) E2R = 2ER (c) E2R =
y
XXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
R
XXXXXXXXX
XXXXXXXX
XXXXXXX
4ER
XXXX
Bz
x
The rate of change of the flux is dΦ B   πR 2 dB
proportional to the area:
dt
dt
t1
The path integral of the induced electric
 
field is proportional to the radius.  E
 dl  E (2R )
Therefore:
ER
t
Calculation
• Suppose we pull with velocity v a
coil of resistance R through a
region of constant magnetic field
B.
– What will be the induced current?
» What direction?
• Lenz’ Law  clockwise!!
xxxxxx
xxxxxx
xxxxxx
I
w
xxxxxx
x
– What is the magnitude?
» Magnetic Flux:
» Faraday’s Law:

Φ B  xwB

d B
dt
dx
d B
 wB  wBv
dt
dt

I
 wBv

R
R
v
Energy Conservation?
• The induced current gives rise to
a net magnetic force ( F
) on
the loop which opposes the
motion.
F
2 2
wBv 
w B v
F  IwB  
 wB 
 R 
R
F'
xxxxxx
I
wBv
R
I
xxxxxx
xxxxxx
w
xxxxxx
F'
x
v
• Agent must exert equal but
opposite force to move the loop with velocity v;  agent does
work at rate P, where
w 2 B2 v 2
P  Fv 
R
• Energy is dissipated in circuit at rate P'
2
wBv 
w 2 B2 v 2

P  I R  
 R
 R 
R
2

P = P' !
A verbal clicker … just think about it
A copper loop is placed in a uniform field. You are looking from the right
Suppose the loop is moving to the right. The current induced in the loop is
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