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Faraday’s Law and Lenz’s Law
B(t)
i
~
Physics 1304: Lecture 13, Pg 1
Overview of Lecture

Induction Effects

Faraday’s Law (Lenz’ Law)
• Energy Conservation with induced currents?

Faraday’s Law in terms of Electric Fields
Text Reference: Chapter 31.1-4
Physics 1304: Lecture 13, Pg 2
Induction Effects

Bar magnet moves through coil
S
N
N
S
N
S
v
 Current induced in coil
• Change pole that enters
 Induced current changes sign

Bar magnet stationary inside coil
 No current induced in coil

Coil moves past fixed bar magnet
 Current induced in coil
v
v
S
N
Physics 1304: Lecture 13, Pg 3
Induction Effects
from Currents
•
a
Switch closed (or opened)
b
 current induced in coil b
•
Steady state current in coil a
 no current induced in coil b

Conclusion:
A current is induced in a loop when:
• there is a change in magnetic field through it
• loop moves through a magnetic field

How can we quantify this?
Physics 1304: Lecture 13, Pg 4
Faraday's Law
•
Define the flux of the magnetic field through a surface (closed
or open) from:
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.
The minus sign indicates direction of induced current
(given by Lenz's Law).
Physics 1304: Lecture 13, Pg 5
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
Conservation of energy considerations:
Claim: Direction of induced current must be so as to oppose
the change; otherwise conservation of energy would be
violated.
» Why???
 If current reinforced the change, then the change
would get bigger and that would in turn induce a
larger current which would increase the change,
etc..
Physics 1304: Lecture 13, Pg 6
Lecture 18, CQ
y
A conducting rectangular loop moves with
constant velocity v in the +x direction
through a region of constant magnetic field B
in the -z direction as shown.
 What is the direction of the induced current in
1A
the loop?

(a) ccw
(b) cw
(b) cw
x
(c) no induced current
• A conducting rectangular loop moves with y
constant velocity v in the -y direction and a
constant current I flows in the +x direction as
shown.
• What is the direction of the induced
1B
current in the loop?
(a) ccw
XXXXXXXXXXXX
XXXXXXXXXXXX
X X X X X X X vX X X X X
XXXXXXXXXXXX
I
v
x
(c) no induced current
Physics 1304: Lecture 13, Pg 7
Lecture 16, ACT 1
A conducting rectangular loop moves with y
constant velocity v in the +x direction
through a region of constant magnetic field B
in the -z direction as shown.
 What is the direction of the induced current in
1A
the loop?

(a) ccw
(b) cw
XXXXXXXXXXXX
XXXXXXXXXXXX
X X X X X X X vX X X X X
XXXXXXXXXXXX
x
(c) no induced current
• There is a non-zero flux FB passing through the loop since B is
perpendicular to the area of the loop.
• Since the velocity of the loop and the magnetic field are CONSTANT,
however, this flux DOES NOT CHANGE IN TIME.
• Therefore, there is NO emf induced in the loop; NO current will flow!!
Physics 1304: Lecture 13, Pg 8
Lecture 16, ACT 1

y
A conducting rectangular loop moves with
constant velocity v in the +x direction through a
region of constant magnetic field B in the -z
direction as shown.
 What is the direction of the induced current in
the loop?
(a) ccw
(b) cw
(b) cw
x
(c) no induced current
• A conducting rectangular loop moves
with constant velocity v in the -y
direction and a constant current I
flows in the +x direction as shown.
• What is the direction of the induced
current in the loop?
(a) ccw
XXXXXXXXXXXX
XXXXXXXXXXXX
X X X X X X X vX X X X X
XXXXXXXXXXXX
y
I
i
v
x
(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.
• Therefore, 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.
Physics 1304: Lecture 13, Pg 9
Demo
E-M Cannon
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
Physics 1304: Lecture 13, Pg 10
Lecture 18, CQ

Let us predict the results of variants of the
electromagnetic cannon demo which you just
observed.
Ring 1
 Suppose two aluminum rings are used in the
demo; Ring 2 is identical to Ring 1 except that it
has a small slit as shown. Let F1 be the force on
Ring 1; F2 be the force on Ring 2.
Ring 2
(a) F2 < F1
(b) F2 = F1
(c) F2 > F1
– Suppose two identically shaped rings are used in the demo.
Ring 1 is made of copper (resistivity = 1.7X10-8 W-m); Ring 2 is
made of aluminum (resistivity = 2.8X10-8 W-m). Let F1 be the
force on Ring 1; F2 be the force on Ring 2.
(a) F2 < F1
(b) F2 = F1
(c) F2 > F1
Physics 1304: Lecture 13, Pg 11
Lecture 18, CQ

For this ACT, we will predict the results of variants of
the electromagnetic cannon demo which you just
observed.
Ring 1
 Suppose two aluminum rings are used in the
demo; Ring 2 is identical to Ring 1 except that it
has a small slit as shown. Let F1 be the force on
Ring 1; F2 be the force on Ring 2.
Ring 2
(a) F2 < F1
(b) F2 = F1
(c) F2 > F1
• The key here is to realize exactly how the force on the ring is
produced.
• A force is exerted on the ring because a current is flowing in the ring
and the ring is located in a magnetic field with a component
perpendicular to the current.
• An emf is induced in Ring 2 equal to that of Ring 1, but NO CURRENT is
induced in Ring 2 because of the slit!
• Therefore, there is NO force on Ring 2!
Physics 1304: Lecture 13, Pg 12
Lecture 18, CQ

For this ACT, we will predict the results of variants of
the electromagnetic cannon demo which you just
observed.
Ring 1
 Suppose two aluminum rings are used in the
demo; Ring 2 is identical to Ring 1 except that it
has a small slit as shown. Let F1 be the force on
Ring 1; F2 be the force on Ring 2.
Ring 2
(a) F2 < F1
(b) F2 = F1
(c) F2 > F1
– Suppose two identically shaped rings are used in the demo.
Ring 1 is made of copper (resistivity = 1.7X10-8 W-m); Ring 2 is
made of aluminum (resistivity = 2.8X10-8 W-m). Let F1 be the force
on Ring 1; F2 be the force on Ring 2.
(a) F2 < F1
(b) F2 = F1
(c) F2 > F1
• The emf’s induced in each ring are equal.
• The currents induced in each ring are NOT equal because of the
different resistivities.
• The copper ring will have a larger current induced (smaller resistance)
and therefore will experience a larger force (F proportional
current).
Physics 1304:to
Lecture
13, Pg 13
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?

xxxxxx
I
xxxxxx
xxxxxx
w
Lenz’ Law  clockwise!! x x x x x x
x

What is the magnitude?
» Magnetic Flux:
» Faraday’s Law:
\

Physics 1304: Lecture 13, Pg 14
v
Energy Conservation?
•
The induced current gives rise to
a net magnetic force ( F
) on
the loop which opposes the
motion.
F'
xxxxxx
F
•
Agent must exert equal but
I
xxxxxx
xxxxxx
w
xxxxxx
F'
x
opposite force to move the loop with velocity v; \ agent does work at rate
P, where
•
Energy is dissipated in circuit at rate P'

P = P' !
Physics 1304: Lecture 13, Pg 15
v
Faraday's Law
•
Define the flux of the magnetic field through a surface (closed
or open) from:
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.
The minus sign indicates direction of induced current
(given by Lenz's Law).
Physics 1304: Lecture 13, Pg 16
Conceptual Question
In figure (a), a solenoid produces a magnetic field whose strength
increases into the plane of the page. An induced emf is established in
a conducting loop surrounding the solenoid, and this emf lights bulbs
A and B. In figure (b), points P and Q are shorted. After the short is
inserted,
1.
2.
3.
4.
5.
6.
bulb A goes out; bulb
bulb B goes out; bulb
bulb A goes out; bulb
bulb B goes out; bulb
both bulbs go out.
none of the above
B
A
B
A
gets
gets
gets
gets
brighter.
brighter.
dimmer.
dimmer.
Physics 1304: Lecture 13, Pg 17
Physics 1304: Lecture 13, Pg 18
DB  E



•
Ex x x x x x x
x
x
x
Faraday's law  a changing B induces an
E
emf which can produce a current in a loop.
xxxxxxxxxx
In order for charges to move (i.e., the
r
xxxxxxxxxx
current) there must be an electric field.
\ we can state Faraday's law more generally x x x xBx x x x x x
in terms of the E field which is produced by E
a changing B field.
x x x x x x x xEx x
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 e = W/q.
Physics 1304: Lecture 13, Pg 19
DB  E
•
Putting these 2 eqns together:

•

Therefore, Faraday's law can be
rewritten in terms of the fields as:
x x xEx x x x x x x
E
xxxxxxxxxx
r
xxxxxxxxxx
B
xxxxxxxxxx
E
x x x x x x x xEx x
Note that
for E fields generated by charges at rest
(electrostatics) since this would correspond to the potential difference
between a point and itself. Consequently, there can be no "potential
function" corresponding to these induced E fields.
Physics 1304: Lecture 13, Pg 20
Lecture 18, CQ

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) e < 0
( E ccw)
(b) e = 0
(c) e > 0
( E cw)
– 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
y
XXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
R
XXXXXXXXX
XXXXXXXX
XXXXXXX
XXXX
Bz
t1
x
t
(c) E2R = 4ER
Physics 1304: Lecture 13, Pg 21
y
Lecture 18, CQ

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) e < 0
( E ccw)
(b) e = 0
(c) e > 0
( E cw)
• There will be an induced emf at t=t1
because the magnetic field (and therefore
the magnetic flux) is changing. It makes NO
DIFFERENCE that at t=t1 the magnetic field
happens to be equal to ZERO!
• The magnetic field is increasing at t=t1
(actually at all times shown!) which induces
an emf which opposes the corresponding
change in flux. ie electric field must be
induced in a counter clockwise sense so
that the current it would drive would create
a magnetic field in the z direction.
XXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
R
XXXXXXXXX
XXXXXXXX
XXXXXXX
XXXX
x
B
t1
t
Physics 1304: Lecture 13, Pg 22
y
Lecture 18, CQ

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) e < 0
( E ccw)
(b) e = 0
(c) e > 0
( E cw)
XXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
R
XXXXXXXXX
XXXXXXXX
XXXXXXX
XXXX
B
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
t1
x
t
(c) E2R = 4ER
• The rate of change of the flux is
proportional to the area:
• The path integral of the induced
electric field is proportional to the
radius.
NOTE: This result is important for the operation
of the Betatron.
Physics 1304: Lecture 13, Pg 23
Textbook Problems
Problems 31-33, 31-39, 31-46, 31-67
Physics 1304: Lecture 13, Pg 24
Induced EMF by motion
Physics 1304: Lecture 13, Pg 25
Rod Moving in B-Field
Consider a metal rod moving in a
B-field. The free charges in the
rod will experience a force given X X X X X X X X X X X X
- - -by,
X
X X X X X X X - -
F  qv  B
the top the force would be
upwards and negative charge
would thus accumulate on end.
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 X X X X X X X X X X X
The work done in separating the
charges is:
W  F l
 qvBl
By definition the EMF is then,
e
W

q
i.e.
e  vBl
Physics 1304: Lecture 13, Pg 26
The previous result is connected with Faraday’s law. To see this we re-interpret
Faradays law as the induced EMF along the path of a moving conductor in the
presence of a constant of changing B-field. This induced EMF would be equal to
the rate at which magnetic flux sweeps across the path.
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
Ds

dFB
dt
e  vBL

Ds
BL
Dt
Aswept  DsL
e
d
d
( sBL)  ( BA)
dt
dt
e
dFB

dt
: rate at which flux is swept.
Physics 1304: Lecture 13, Pg 27
Electric Generators and Motors
n̂
Consider a loop of wire in
a constant magnetic
field. If we rotate the
loop the flux through
the cross sectional
area will change.
By Faradays Law an
Induced EMF will be
generated.
B
FB  BA
e
B

i
i
FB (t )  BA cos (t )
i
The converse is also true if we run a current through the
wire then a torque will be excerted on the loop making it
turn.
Physics 1304: Lecture 13, Pg 28
Textbook Problems
Physics 1304: Lecture 13, Pg 29
Textbook Problems
Physics 1304: Lecture 13, Pg 30
Electric Motors and Generators
Physics 1304: Lecture 13, Pg 31
Textbook Problems
Physics 1304: Lecture 13, Pg 32