Faraday's Law
dS
B
B
G G
Φ B ≡ ∫ B • dS
dΦ B
ε=−
dt
Global Review
• Electrostatics
» motion of “q” in external E-field
» E-field generated by Σqi
• Magnetostatics
» motion of “q” and “i” in external B-field
» B-field generated by “I”
• Electrodynamics
» time dependent B-field generates E-field ac
circuits, inductors, transformers, etc
» time dependent E-field generates B-field
• electromagnetic radiation - light
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 33: 1 - 5
Induction Effects
•
Bar magnet moves through coil
S
N
N
S
N
S
⇒ 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
S
N
v
v
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
• this can happen many different ways
•
How can we quantify this?
An example of induction
A wire loop falling into an increasing magnetic field
Force acting on
moving charges
time
magnetic
field
N
N
downward
velocity
N
Faraday's Law
•
Define the flux of the magnetic field through an open surface
as:
dS
G G
Φ B ≡ ∫ B • 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.
ε=−
dΦ B
dt
So what is
this emf??
The minus sign indicates direction of induced current
(given by Lenz's Law).
electro-motive force or emf
time
A magnetic field, increasing in time, passes through the blue loop
An electric field is generated “ringing” the increasing magnetic field
Ringing electric field will drive currents, just like a voltage difference
Loop integral of E-field is the “emf”
G G
ε = ∫ E • dl
•
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,
1
etc..
ACT 1
• 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.
1A – What is the direction of the induced
current in the loop?
(a) ccw
(b) cw
y
(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
(b) cw
XXXXXXXXXXXX
XXXXXXXXXXXX
X X X X X X X vX X X X X
XXXXXXXXXXXX
x
I
v
(c) no induced current
x
ACT 1
• 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.
1A – What is the direction of the induced
current in the loop?
(a) ccw
(b) cw
y
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 ΦB 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!!
ACT 1
• 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.
1A – What is the direction of the induced
current in the loop?
(a) ccw
(b) cw
y
(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
(b) cw
XXXXXXXXXXXX
XXXXXXXXXXXX
X X X X X X X vX X X X X
XXXXXXXXXXXX
x
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.
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
ε=−
dΦ B
dt
dΦ B
dx
= wB = wBv
dt
dt
⇒
I=
ε 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
w B v
wBv ⎞
F = IwB = ⎛⎜
⎟ wB =
⎝ R ⎠
R
•
F'
xxxxxx
I=
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
•
wBv
R
Energy is dissipated in circuit at rate P'
2
w 2 B2 v 2
wBv ⎞
⎛
P′ = I R = ⎜
⎟ R=
⎝ R ⎠
R
2
⇒
P = P' !
∆B/∆t → E
•
Faraday's law ⇒ a changing B induces
an emf which can produce a current in a
loop.
•
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 =
G G
W = ∫ qE • d l
•
This work can also be calculated from ε = W/q.
x x xEx x x x x x x
E
xxxxxxxxxx
r
• In order for charges to move (i.e., the
xxxxxxxxxx
current) there must be an electric field.
B
x
x
x
x
xxxxxx
∴ we can state Faraday's law more
E
generally in terms of the E field which is x x x x x x x x x x
E
produced by a changing B field.
∆B/∆t → E
•
Putting these 2 eqns together:
G G
W = ∫ qE • d l
G G
ε = ∫ E • dl
W
⇒
ε=
q
• Therefore, Faraday's law can be
rewritten in terms of the fields as:
Line integral
around loop
•
G
G
dΦ B
E
•
d
l
=
−
∫
dt
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
Rate of change of
flux through loop
G G
∫ E • d l = 0 for E fields generated by charges at
3
Note that
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.
y
ACT 2
• 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.
3A
– What is sign of the induced emf in a
ring of radius R at time t=t1?
(a) ε < 0
( E ccw)
3B
(b) ε = 0
(c) ε > 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
XXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
R
XXXXXXXXX
XXXXXXXX
XXXXXXX
XXXX
Bz
t1
(c) E2R = 4ER
x
t
ACT 2
• 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.
3A
– What is sign of the induced emf in a
ring of radius R at time t=t1?
(a) ε < 0
( E ccw)
(b) ε = 0
(c) ε > 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 clockwise sense
so that the current it would drive would
create a magnetic field in the -z direction.
y
XXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
R
XXXXXXXXX
XXXXXXXX
XXXXXXX
XXXX
Bz
t1
x
t
Note: the signs work this
way:
• E cw > 0 means that
dS points in -z direction.
• dB/dt > 0 ⇒ dΦ/dt < 0
• Faraday’s Law
⇒ ε = - dΦ/dt > 0
y
ACT 2
• 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.
3A
– What is sign of the induced emf in a
ring of radius R at time t=t1?
(a) ε < 0
( E ccw)
3B
(b) ε = 0
(c) ε > 0
( E cw)
XXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
R
XXXXXXXXX
XXXXXXXX
XXXXXXX
XXXX
Bz
– 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
• 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.
t
t1
(c) E2R = 4ER
dΦ B
dB
= − πR 2
dt
dt
G
G
∫ E • d l = E ( 2π R )
x
E∝R
Demo
E-M Cannon
•
Connect solenoid to a source of
alternating voltage.
•
The flux through the area ⊥ to axis of
solenoid therefore changes in time.
v
~
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
2
top view
ACT 3
• For this ACT, we will predict the results of
variants of the electromagnetic cannon demo .
2A
– 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 1
Ring 2
(a) F2 < F1
2B
(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 Ω-m); Ring 2 is
made of aluminum (resistivity = 2.8X10-8 Ω-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
ACT 3
• For this ACT, we will predict the results of
variants of the electromagnetic cannon demo
which you just observed.
2A
– 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.
(a) F2 < F1
(b) F2 = F1
Ring 1
Ring 2
(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!
ACT 3
• For this ACT, we will predict the results of
variants of the electromagnetic cannon demo
which you just observed.
2A
– 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.
(a) F2 < F1
2B
(b) F2 = F1
Ring 1
Ring 2
(c) F2 > F1
– Suppose two identically shaped rings are used in the demo.
Ring 1 is made of copper (resistivity = 1.7X10-8 Ω-m); Ring 2 is
made of aluminum (resistivity = 2.8X10-8 Ω-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 to current).