Chapter 31
Faraday’s law
Using Ampere’s law in chapter 30 we investigated how an
electric current I can generate a magnetic field B.
In this chapter we shall study the fourth (and last) of
Maxwell’s equations known as Faraday’s law.
Faraday’s law tells us how a magnetic flux ΦB that changes
with time can generate an electric field E
All electric power generation is based on Faraday’s law.
(31-1)
A
I
ΦB(t)
A
I
+
Fig.a
-
E
Fig.b
In 1831 Faraday discovered that if the magnetic flux ΦB
through a closed circuit changes with time as shown in fig.a,
then a voltage E appears, as if there were a battery in the
circuit, as shown in fig.b. This emf is called an “induced”
emf . The resulting current I can be measured by the ammeter
in the circuit and is known as “induced” current. The whole
phenomenon is called “magnetic induction”
(31-2)
n θ
B
A
Faraday’s law
dΦB
E =−
dt
The induced emf in a circuit is equal to rate of change (with
a negative sign) of the magnetic flux through the circuit.
Magnetic flux ΦB = BAcosθ (for uniform B)
The magnetic flux ΦB depends on B, A, and θ
We can have a time-varying magnetic flux if:
• B changes with t
• θ changes with t
• A changes with t
(31-3)
Michael Faraday
1791-1867
(31-4)
In fig.(d) we we change the
angle θ between B
(generated by the upper
loop) and the normal n to the
lower loop. The magnetic
flux through the lower loop
varies with t
B
n
θ
B
In fig.(a) we change B
with time as we approach
the bar magnet closer to
the wire loop.
Example 31-2 (page 841)
Φ = Bo A cos 0 = Bo A
A = LL1
L1 = L − vt →
A = L ( L − vt ) → Φ B = Bo L ( L − vt )
dΦ B
E =−
= Bo Lv
dt
n
L1
vt
E Bo Lv
I= =
R
R
= 0.15 A
Bo = 1 T
L = 0.1 m
R = 0.065 Ω
v = 0.1 m/s
I=?
(31-5)
Faraday’s Law
n θ
B
A
dΦB
E =−
dt
Φ B = BA cosθ
The minus sign in Faraday’s law indicates the direction in
which the current I will flow in the loop of the figure above.
The current direction is given by Lenz’s rule that can be
summarized as follows:
Lenz’s Rule: The induced current Iind will flow in such a
direction so that the resulting magnetic field Bind tends to
oppose the magnetic flux change that produced Iind in the first
place
(31-6)
Example 31-1
Φ B = ∫ BdA cos θ
(B non-uniform)
As the bar magnet gets closer to the ring B increases and
therefore ΦB increases. This is the first step in the application
of Lenz’s rule. We must determine whether the magnetic flux
increases or decreases.
(31-7)
In the previous slide we saw that as the magnet approaches
the ring the magnetic flux ΦB increases
ΦB = Bnet Acosθ
Bnet = Bmagnet + B ind
Scenario 1
Bind
Scenario 2
Bind
Scenario 1.
Bind and Bmagnet are
antiparallel and
subtract. Thus ΦB
decreases with time
Scenario 2.
Bind and Bmagnet are
parallel and add up.
Thus ΦB increases
with time
(31-8)
Revisit Example 31-2 Magnetic flux ΦB= BnetL(L – vt)
From the equation above, we conclude that as time t increases,
the magnetic flux decreases
Bnet = Bo + Bind
Scenario 1
Scenario 2
Bind and Bo are parallel and Bind and Bo are antiparallel and
add up. Thus ΦB increases subtract. Thus ΦB decreases
Bind
Bind
(31-9)
Example 31-4
Page 842
The magnetic field Bw
generated by the wire at a
distance r is given by the
expression:
µo I
Bw =
2π r
As the wire approaches the loop, the average distance r
decreases. Thus the flux ΦB through the loop increases
(31-10)
Bind
Bind
Bw
Bw
As the wire approaches the loop the flux ΦB
Scenario 1
Bw and Bind are antiparallel
and they subtract. Thus the
flux ΦB through the loop
decreases
increases
Scenario 2
Bw and Bind are parallel and
they add up. Thus the flux
ΦB through the loop
increases
(31-11)
Example 31-5, page 843
n
Φ B = Bo A cos 0 = Bo A
A=Dvt → Φ B = Bo Dvt
The magnetic flux ΦB through the loop increases with time t
d ΦB
E =−
= − Bo Dv
dt
ρ l ρ ( 2 D + 2vt )
R=
=
A
Dvt
E
I=
R
(31-12)
The magnetic flux ΦB through the
loop increases with t
Scenario 1:
Bo and B ind are antiparallel
and they subtract. Thus ΦB
decreases
Bo and B ind are parallel and
they add up. Thus ΦB
increases
(31-13)
Iind
Bind
Bind
Iind
FL
E
F+
FB = q(v × B) F = qvB →
E = F/q = vB
Voltage across rod V = LE = vBL
If we move a conductor in a magnetic field an emf
develops across its length as shown in the figure above.
This emf is known as “motional emf ”
(31-14)
F =q(v×B)
(31-15)
F-
n
L
vt
F+
Consider the area A swept by the moving rod
Magnetic flux Φ B = BA cos 0 = BA = BLvt
dΦB
Induced emf E = −
= − BLv
dt
A = Lvt
Example 31-6 , page 846
Consider the magnetic flux through the area A swept by the rod
2
2
θ
θ
L
ω
t
ω
tL
A=π L2
=
→ θ =ω t → A = π L2
=
2π
2
2π
2
Φ B = BA
Bω tL
ΦB =
2
dΦB
Bω L2
E =−
=−
dt
2
2
n
θ
(31-16)
Polarity of induced emf. By considering the magnetic forces
for positive and negative charges we can determine that
VA > V B
FB = q(v × B)
-
B
FF+
+
A
v
(31-17)
When a metal plate moves
in a magnetic field the
induced emf results in
currents known as
“eddy currents”
(31-18)
Eddy currents result in Joule
heating of the metal plate
The energy W dissipated as heat
is given by the equation:
W = Pt = IindR2t
where R is the resistance of the
eddy current path
We try to avoid this heating by
cutting slots in the metal plate.
The slots act like open switches
and prevent the flow of eddy
currents
(31-19)
Revisit of example 31-2. In this example we calculated the
induced current I = Bo vL/R We now consider the magnetic
forces that act on the wire loop. F B = I(l×B).
Fa = 0
Fd = Fc = Fb = ILBosin(90°) = ILBo
Fz = Fb - Fd = 0
Fext
Fx = -Fc = -ILBo
Fx = - Bo2vL2 /R
To move the loop at
constant speed we must
exert an external force
Fext = -Fc on the loop so
that Fxnet = 0
Fext = Bo2vL2 /R
(31-20)
The induced current in example 31-2 was I = BovL/R
The external force necessary to keep the loop moving with a
constant velocity Fext = Bo2vL2 /R
The external force
produces work at the rate
Pext = Fext . v = Bo2v2 L2 /R
The induced current
dissipates heat on the loop
at the rate:
Pjoule = I2R = B o2 v2 L2 /R
Fext
(31-21)
Note: Pext = Pjoule
The rate at which the
force Fext is doing work
equals the rate at which
heat is generated
AC-genarator
An AC-generator consists of N wire loops (each of area A)
rotating in a uniform magnetic field B with an angular
velocity ω
(31-22)
(31-23)
AC - generator
θ= ωt
Magnetic flux through one loop Φ1 = BA cos θ
Net magnetic flux Φ B = N Φ1 = NBA cos ω t
dΦB
Induced emf E = −
= NBAω sin ω t
dt
NBAω ≡ Vmax → E = Vmax sin ω t
R
(31-24)
E
I
An AC generator has an emf
E = Vmax sinωt , Vmax = NBAω
We connect the generator to a
resistor R. The current I is given
by Ohm’s law.
E NBAω
I= =
sin ωt
Power P dissipated on R:
R
R
2 2 2 2
N
B Aω
2
P=I R=
sin 2 ω t
R
Power P dissipated on R
(31-25)
N 2 B 2 A 2ω 2
P=
sin 2 ω t
R
Pav
P varies with time and thus it is
not convenient to use. Instead
we use the concept of average
power Pav which is defined as
T
1
follows:
Pv =
P (t )dt
∫
T
0
N 2 B 2 A 2ω 2 1
2
Pav =
sin
? tdt
∫
R
T0
T
2
N 2 B 2 A2ω 2 Vmax
→ Pav =
=
2R
2R
T
1
1
2
sin ω tdt =
∫
T 0
2
Consider a magnetic field B that varies with time according
to the equation: B = Bo +αt
Using Faraday’s law
we see that there is
an electric field E
whose lines are
forming circles
around the magnetic
field lines. We will
determine the
magnitude of E
(31-26)
B = Bo +αt
B
E
ur ur
Induced emf E = ∫ E ⋅ d E
E
γ
E
E
γ = circle of radius r
ur ur
∫ E ⋅ d E = ∫ EdS cos(180°) = −2π rE
γ
B
E
2π rE = −π r α
2
B
γ
dΦ B
E =−
Φ B = π r 2 ( Bo + α t )
dt
→ E = −π r 2α
rα
→ E=
2
(31-27)