Electromagnetic
Induction
Induced currents
From the demo we see:
S
S
N
Current induced when magnet moves through solenoid
Current induced when solenoid moves through magnet
Current not induced when magnet moves outside solenoid
No current induced if no motion
Amount of current induced increases if (all other things
being equal):
magnet is moved more rapidly
solenoid has more windings
solenoid’s corss section area is larger
magnet is stronger
Faraday’s Law
!
d
! · d!" = −
E
dt
∂S
n
B
!
S
! · dA
!
B
This describes all
the obser vations we made
Let’s understand all
the ingredients ...
S
C = ∂S
!
d
!
!
E · d" = −
dt
∂S
!
S
! · dA
!
B
n
B
S
Magnetic flux through surface S:
!
" · dA
"
φB ≡
B
C = ∂S
S
Similar to other fluxes we have encountered, eg, I =
- surface is open
- direction of A given by normal, n
- “counts” how many B-field lines cross A
!
S
!
J! · dA
!
d
!
!
E · d" = −
dt
∂S
!
S
! · dA
!
B
n
B
S
C = ∂S
Derivative:
dφB
d
=
dt
dt
!
S
" · dA
"
B
Gives the rate of change (with time) of flux.
Recall from demo: the current produced
depended on the rate of change of magnetic
flux through solenoid.
Magnetic Flux I
In order to change the
magnetic flux through
the loop, what would
you have to do?
1) drop the magnet
2) move the magnet upwards
3) move the magnet sideways
4) only (1) and (2)
5) any (all) of the above
Magnetic Flux I
In order to change the
magnetic flux through
the loop, what would
you have to do?
1) drop the magnet
2) move the magnet upwards
3) move the magnet sideways
4) only (1) and (2)
5) any (all) of the above
Moving the magnet in any direction would
change the magnetic field through the loop
and thus the magnetic flux.
Magnetic Flux II
In order to change the
magnetic flux through
the loop, what would
you have to do?
1) tilt the loop
2) change the loop area
3) use thicker wires
4) only (1) and (2)
5) all of the above
Magnetic Flux II
In order to change the
magnetic flux through
the loop, what would
you have to do?
1) tilt the loop
2) change the loop area
3) use thicker wires
4) only (1) and (2)
5) all of the above
Since Φ = B A cosθ , changing the
area or tilting the loop (which varies
the projected area) would change
the magnetic flux through the loop.
!
d
!
!
E · d" = −
dt
∂S
!
S
! · dA
!
B
n
S
Line integral of E is the “emf”:
E=
!
C
C = ∂S
! · d"
E
-C is the boundary of S
-Direction given by right hand rule
with respect to normal n to surface S
-The integral tells us that an electric
field E is generated, hence a current if a
conductor is present
B
The meaning of:
!
! · d!" = − d
E
dt
∂S
or
!
dφB
E =−
dt
S
! · dA
!
B
n
B
S
C = ∂S
-Somewhere in space there is a magnetic field
where it is not constant in time
an electric field is produced, or rather, “induced”
-Choose an arbitrary surface S
-Then the rate of change of the flux through that
surface S gives the line integral around C of the
induced electric field
-Minus sign needed to get correct direction of E
Moving Bar Magnet I
If a North pole moves toward the
loop from above the page, in what
direction is the induced current?
1) clockwise
2) counterclockwise
3) no induced current
Moving Bar Magnet I
If a North pole moves toward the
loop from above the page, in what
direction is the induced current?
1) clockwise
2) counterclockwise
3) no induced current
The magnetic field of the moving bar
magnet is pointing into the page and
getting larger as the magnet moves
closer to the loop. Thus the induced
magnetic field has to point out of the
page. A counterclockwise induced
current will give just such an induced
magnetic field.
Follow-up: What happens if the magnet is stationary but the loop moves?
Moving Bar Magnet II
If a North pole moves toward
the loop in the plane of the
page, in what direction is the
induced current?
1) clockwise
2) counterclockwise
3) no induced current
Moving Bar Magnet II
If a North pole moves toward
the loop in the plane of the
page, in what direction is the
induced current?
Since the magnet is moving parallel
to the loop, there is no magnetic
flux through the loop. Thus the
induced current is zero.
1) clockwise
2) counterclockwise
3) no induced current
Subtlety #1:
B
n
S
C
The surface S (and its boundary C) are arbitrary,
mathematical constructs
There is no need for a wire along C, an electric
field is created regardless of the presence of a
conductor (but see subtlety #3)
If a conductor is present, the electric field will
produce a current according to Ohm’s law
!
For loop!of thin wire: !
L
!
!
!
!
ρJ · d " = ρ
Jd" = ρJL = ρ I = RI
E=
E · d" =
A
C
C
C
⇒
E
I=
R
×
×
×
×
×
×
×
×
×
×
10 cm
×
×
×
×
×
×
×
×B
×
×
×
×
×
×
×
×
×
×
×
×
Subtlety #2:
!
C
! · d!" != 0
E
n
S
C
The line integral of the electric field along a closed loop
is not zero!
Wait, didn’t we say the electric force is conser vative,
which means the line integral of the electric field
along any closed loop is zero?
And what about defining potential? It seems with
Faraday's Law when we follow potential changes
around a close loop we do no get zero, so how can I
define potential now?
Answers:
In the absence of magnetic field, or with constant
(non-time varying) magnetic field, E-field is indeed
conser vative, and electric potential can be defined
But electric potential cannot be defined in a region of
space with time-varying B-field
B
B-field into page increasing
Take S the plane surface bounded by circuit
Its boundary C is the circuit itself
Normal n into page
Faraday
R
!
d
!
!
E · d" = −
dt
∂S
!
S
+
! · dA
!
B
says an emf is produced
with tangential E field
counterclockwise (hence
current as indicated)
2IR
–
C
–
A
IR
× × ×
× × × ×
× × × × ×
× × × ×
× × ×
R
B
I
R
Reading of voltmeters across several resistances
indicates “voltage” is decreasing counterclockwise:
VA − VC
VC − VB
VB − VA
= IR
= IR
= IR
+
Give us some examples!
Example 31-3: A wire loop of radius 10 cm
has a resistance of 2.0 ohms. The loop is at
right angles to a uniform magnetic field B,
as shown in the figure. The field strength is
increasing at 0.10 tesla/second.
What is the magnitude of the induced
current in the loop?
!
"
(COB)
1
dB
2
Ans: I =
πr
R
1.6mA
dt
×
×
×
×
×
×
×
×
×
×
10 cm
×
×
×
×
×
×
×
×B
×
×
×
×
×
×
×
×
×
×
×
×
Example 31-2, modified: A long straight wire
carries a current I. A rectangular loop of
dimensions l by w, made of wire with cross
r
dr
I
section A and resistivity ρ, lies with its
closest edge a distance a away from the
wire as indicated in the figure.
(a) What is the magnetic flux through the
loop?
(b) Assuming I = I0 cos(ωt), what is the
current (magnitude and direction) induced in
the loop?
(COB)
Ans:
(a) φB =
µ0 I"
ln
2π
!
a+w
a
"
(b) Iloop
µ0 I0 ω sin(ωt)"A
=
ln
4πρ(" + w)
counterclokwise
a
B
!
a+w
a
w
"
Lenz’s Law: the sign in Faraday’s law, that is...
“The direction of the induced emf and current is such
that the magnetic field created by the induced current
opposes the change in magnetic flux.”
recall examples:
r
dr
×
×
×
×
×
×
×
×
×
×
10 cm
×
×
×
×
×
×
×
×B
×
×
×
×
×
×
×
×
×
×
×
×
I
a
B
w
Loop and Wire I
A wire loop is being pulled away
from a current-carrying wire.
What is the direction of the
induced current in the loop?
I
1) clockwise
2) counterclockwise
3) no induced current
Loop and Wire I
A wire loop is being pulled away
from a current-carrying wire.
What is the direction of the
induced current in the loop?
1) clockwise
2) counterclockwise
3) no induced current
The magnetic flux is into the page on the
right side of the wire and decreasing due
to the fact that the loop is being pulled
away. By Lenz’s Law, the induced B field
will oppose this decrease. Thus, the new
B field points into the page, which requires
an induced clockwise current to produce
such a B field.
I
Loop and Wire II
What is the induced current if
the wire loop moves in the
direction of the yellow arrow ?
I
1) clockwise
2) counterclockwise
3) no induced current
Loop and Wire II
What is the induced current if
the wire loop moves in the
direction of the yellow arrow ?
1) clockwise
2) counterclockwise
3) no induced current
The magnetic flux through the loop is
not changing as it moves parallel to
the wire. Therefore, there is no
induced current.
I
More examples.
Magnetic flux,
!
" · dA
"
φB ≡
B
S
can change in time by
changing magnitude of B field in time
changing area A in time
changing orientation of B field in time
changing orientation of S (direction of n) in
time
Moving Wire Loop I
A wire loop is being pulled
through a uniform magnetic
1) clockwise
field. What is the direction
2) counterclockwise
of the induced current?
3) no induced current
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 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 x x x x
Moving Wire Loop I
A wire loop is being pulled
through a uniform magnetic
1) clockwise
field. What is the direction
2) counterclockwise
of the induced current?
3) no induced current
x x x x x x x x x x x x
Since the magnetic field is uniform, the
magnetic flux through the loop is not
changing. Thus no current is induced.
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 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
Follow-up: What happens if the loop moves out of the page?
Moving Wire Loop II
A wire loop is being pulled
through a uniform magnetic
1) clockwise
field that suddenly ends.
2) counterclockwise
What is the direction of the
3) no induced current
induced current?
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
Moving Wire Loop II
A wire loop is being pulled
through a uniform magnetic
1) clockwise
field that suddenly ends.
2) counterclockwise
What is the direction of the
3) no induced current
induced current?
The B field into the page is disappearing in
the loop, so it must be compensated by an
induced flux also into the page. This can
x x x x x
x x x x x
x x x x x
x x x x x
be accomplished by an induced current in
x x x x x
the clockwise direction in the wire loop.
x x x x x
x x x x x
Follow-up: What happens when the loop is completely out of the field?
R
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
v
B
×
×
×
×
×
×
×
×
×
×
×
Changing area: Example 31-4. A circuit consists of t wo parallel conducting
rails a distance l apart connected at one end by a resistance R. A conducting
bar completes the circuit, joining the t wo bars electrically but being free to
slide along them. The whole circuit is in a constant, uniform magnetic field B
at right angles to the circuit, as shown in the figure. The bar is pulled to the
right with constant speed v, increasing the area of the circuit as it moves.
What is the current in the circuit? Assume the bar and rails are ideal
conductors, so the total circuit resistance is R.
(COB)
Ans:
B!v
I=
R
Rotating Wire Loop
If a coil is rotated as shown,
in a magnetic field pointing
to the left, in what direction
is the induced current?
1) clockwise
2) counterclockwise
3) no induced current
Rotating Wire Loop
If a coil is rotated as shown,
in a magnetic field pointing
to the left, in what direction
is the induced current?
1) clockwise
2) counterclockwise
3) no induced current
As the coil is rotated into the B field,
the magnetic flux through it increases.
According to Lenz’s Law, the induced B
field has to oppose this increase, thus
the new B field points to the right. An
induced counterclockwise current
produces just such a B field.
Changing surface orientation: Example 31-5.
A circular wire loop of radius a and resistance R is
initially perpendicular to a constant, uniform
magnetic field B. The loop rotates with angular
velocity ω about an axis through a diameter, as
shown in the figure. What is the current in the
loop?
(COB)
Ans:
πa2 Bω
I=
sin ωt
R
B
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
ω
N
This is an example of an
“Electric Generator”
Rotating
contacts
Stationary
brushes
(demo)
Electrical
load
peak
Emf,
S
Time
emf, and hence current,
have sinusoidal time dependence,
they are “alternating currents” (AC)
Note that the electric generator and the electric motor are the
same device:
motor: apply a current, get motion
generator: apply motion, get current
Generators
A generator has a coil of wire
rotating in a magnetic field.
If the rotation rate increases,
1) increases
how is the maximum output
voltage of the generator
3) stays the same
affected?
2) decreases
4) varies sinusoidally
Generators
A generator has a coil of wire
rotating in a magnetic field.
If the rotation rate increases,
1) increases
how is the maximum output
voltage of the generator
3) stays the same
affected?
The maximum voltage is the leading
term that multiplies sin(ωt) and is
given by ε0 = NBAω. Therefore, if ω
increases, then ε0 must increase as
well.
2) decreases
4) varies sinusoidally
Subtlety #3: Motional EMF
Recall we said the flux φB ≡
!
S
" · dA
" can change in time by
B
changing magnitude of B field in time
changing area A of surface S in time
changing orientation of B field in time
changing orientation of S (direction of n ) in time
When B changes in time there is always an emf produced (1st and 3rd bullets)
regardless of whether there is a conducting wire loop present.
When S changes in time (2nd and 4th bullets) there is a current induced if
there is a conducting loop at C, the boundary of S.
So if S and C are just mathematical constructs and B is constant in time
there is no E field produced.
Motional EMF: the inside scoop
×
×
B
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
v
I
It is just like “E x B” fields (eg, Hall effect)!
Magnetic force on moving charges in
conductor (moving with whole wire)
!
F!B = q!v × B
points up in figure. This is force on
individual charge.
Along this stretch of wire, charges move, build up
charge differential that results in electric field
that counters magnetic force. Build up continues
until forces cancel each other: F!E = −F!B
This gives E = E! = vB!
If we close the loop this will results in a current
E
vB!
I=
=
R
R
×
×
B
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
Now we check we get the same result
from Faraday’s law
v
I
This is as in the example above of area
changing with time.
The rate of change of the enclosed area
through which there is a B field is
dA
dx
=!
= !v
dt
dt
So the rate of change of magnetic flux through square is
dφB
dA
=B
= B"v
dt
dt
Faraday’s law gives then that the emf and current are
as before:
E = B!v
B!v
I=
R
Moral:
there is no induced electric field for constant
magnetic field
for wires (loops included) moving in a constant
magnetic field there are currents produced by
the magnetic force on charges
these currents are precisely what you would
get by applying Faraday’s law to compute an
emf around the loop and then using Ohm ’s law
Where did the energy come from ?
In our example, energy is dissipated by the
resistance (the wire is resistive) at a rate
Pdiss
2
(B!v)
= I 2R =
R
B × F
mag
×
×
×
I
Fmag
×
×
× Fmag
×
x
×
×
B
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
The velocity is maintained by applying an
external force,
F!applied = −F!mag
The force does work at a rate
Papplied = F!applied · !v
Fapplied
Put these together with
B!v
!
!
Fmag = |I " × B| and I =
Papplied
(B!v)2
=
R
R
The electric energy dissipated comes from the external work done!
This is true of electric generators too, of course.
v
I
Example/demo: Eddy currents
Pivot
v
B
S
I
Mechanical energy
is converted to electrical
energy that is dissipated
as heat (thermal energy)
N
(a)
Falling Magnet I
A bar magnet is held above the floor
and dropped. In 1, there is nothing
between the magnet and the floor.
In 2, the magnet falls through a
copper loop. How will the magnet in
1) it will fall slower
2) it will fall faster
3) it will fall the same
case 2 fall in comparison to case 1?
S
N
S
1
copper
loop
N
2
Falling Magnet I
A bar magnet is held above the floor
and dropped. In 1, there is nothing
between the magnet and the floor.
In 2, the magnet falls through a
copper loop. How will the magnet in
1) it will fall slower
2) it will fall faster
3) it will fall the same
case 2 fall in comparison to case 1?
When the magnet is falling from above
the loop in 2, the induced current will
produce a North pole on top of the loop,
which repels the magnet.
When the magnet is below the loop, the
induced current will produce a North
pole on the bottom of the loop, which
attracts the South pole of the magnet.
S
N
1
copper
loop
Follow-up: What happens in case 2 if you flip the magnet so
that the South pole is on the bottom as the magnet falls?
S
N
2
Falling Magnet II
If there is induced
current, doesn’t
that cost energy?
Where would that
energy come from
in case 2?
1) induced current doesn’t need any energy
2) energy conservation is violated in this case
3) there is less KE in case 2
4) there is more gravitational PE in case 2
S
N
S
1
copper
loop
N
2
Falling Magnet II
If there is induced
current, doesn’t
that cost energy?
Where would that
energy come from
in case 2?
1) induced current doesn’t need any energy
2) energy conservation is violated in this case
3) there is less KE in case 2
4) there is more gravitational PE in case 2
In both cases, the magnet starts with
the same initial gravitational PE.
In case 1, all the gravitational PE has
been converted into kinetic energy.
In case 2, we know the magnet falls
slower, thus there is less KE. The
difference in energy goes into making
the induced current.
S
N
S
1
copper
loop
N
2
Induced electric fields in solenoids
Consider a solenoid with time-dependent current
The magnetic field inside the solenoid is uniform and
B(t) = µ0 nI(t)
(COB)
1
dI(t)
E(r) = µ0 rn
2
dt
×
×
×
B
×
×
×
×
R
×
×
×
×
×
×
×
r
E