25. Faraday’s Law of Induction
Concept 25.1: A conductor moving in a region
with a magnetic field has a potential difference
x
x
built up across it.
+x x
Fe
The magnetic force on the mobile electrons is
r
r r
Fm = qv × B (down).
• motional emf
• motional emf in a closed circuit
• Faraday’s Law
Since the electrons in the bar cannot go in
a circle, negative charge accumulates at
the bottom of the bar. This
an
r creates
r
electric field and force Fe = qE (up).
There is no net force
r and
r more charge
r no
separation once E = −v × B .
Serway and Beichner
Section 31.2, 31.1
top
This creates emf ε = −
r
r r
-
l
r xr
x
x
x
x
E
Fm
x
rx
∫ E • ds = − E • l = l • v × B
B
x
x
v
x
x
-x
x
This emf acts like a “battery” which separates positive and
negative charge, but this “battery” is distributed throughout the
Physics 1E03 Lecture 25
bar itself.
2
1
Quiz of concept 25.1
a) LvB cosθ sinφ
c) LvB sinθ sinφ
r
x
x
bottom
Physics 1E03 Lecture 25
A conducting bar moves
with velocity v, at an
angle φ (in the plane of
the screen) to a constant
magnetic field. The bar
has length L and is
oriented at angle θ out of
the plane of the screen.
The emf induced between
the ends of the bar is
x
Quiz of concept 25.1
B
v
φ
If the mobile charge carriers were positive charges, the
motional emf
a) would reverse sign
b) would have the same sign
Quiz of concept 25.1
If the velocity of the conductor in the magnetic field is
reversed, the motional emf
a) would reverse sign
b) would have the same sign
b) LvB cosθ cosφ
d) LvB sinθ cosφ
Physics 1E03 Lecture 25
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Physics 1E03 Lecture 25
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Concept 25.2: The motional emf in a conductor
converts mechanical work to electrical energy.
A conducting bar slides with
constant velocity v along
conducting rails, with a
resistance R between them.
y
z
r r r r dxr r
ε = L•v ×B = L• ×B
dt
d ˆ ˆ
d
ε = Lj • xi × B(− kˆ) = LxB
dt
dt
x
R
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
B
x
v
x
L
x
Express this using the flux of B through the circuit.
r
r r
r
Area A = Akˆ = Lxkˆ and B = − Bkˆ, so LxB = − A • B = −Φ B
ε =−
ε
d
Φ B and I =
flows.
R
dt
d
ΦB
dt
Physics 1E03 Lecture 25
S
N
this is Faraday’s Law. It gives another
source of the electric field – a changing
magnetic field.
S
x
x
x
In these two circuits with
x
and induced emf, the R
force on the bar
x
required to maintain a
constant velocity is
a) zero in both cases
x
b) greater in the resistive
x
circuit
c) greater in the
x
capacitive circuit
d) equal (but not zero) in
C
both circuits.
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
B
N
x
v
x
L
x
x
B
v
x
L
x
Physics 1E03 Lecture 25
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V
v
x
Area enclosed by circuit
Magnetic flux
through the circuit
Concept 25.3: An emf can be induced by any
change in magnetic flux. It can be due to
changes in i) the magnitude of B ii) the direction
of B iii) the area of a loop.
ε =−
Quiz of concept 25.2 x
6
How can a changing magnetic field cause a current to flow
in a loop of wire? Consider a loop moving toward the bar
magnet.
v
S
N
Earlier time, small flux
Later time, large flux
For the magnetic flux to decrease, the number of field lines
inside the loop must decrease. Since magnetic flux lines
cannot end anywhere, the only way is for the field lines to
spread out.
B
End view
Br
x
Br, radial
v
v
component
x
r r
v × B force pushes
Physics 1E03 Lecture 25
charges around loop.
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Physics 1E03 Lecture 25
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2
Summary
Quiz of Concept 25.3
Plot VB – VA = V as a
function of the
magnet’s position as
the magnet moves at
constant speed
through the solenoid.
V
S
N
A
V
V
end
of coil
x
b)
start
of coil
V
c)
start
of coil
B
ε =−
a)
start
of coil
• a conductor moving in ra magnetic field
an emf
r develops
r r
along its length L given by ε = L • v × B
• in a closed circuit, this may written in terms of the flux of
magnetic field through the circuit
end
of coil
x
end
of coil
x
• the sign of the emf is almost impossible to figure out (see
next lecture)
• the induced emf can drive a current in a closed circuit
Practice problems: Chapter 31, #20, 10, 11, 31
d)
start
of coil
d
ΦB
dt
end
of coil
Physics 1E03 Lecture 25
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x
Next lecture: Read sections 31.3
Physics 1E03 Lecture 25
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