Lecture 17 Chapter 30
Friday 17 February 2006
Induction and Oscillations
Ch. 30: Faraday’s Law
Ch. 31: AC Circuits
Induced EMF: Faraday’s Law
“Time-dependent B creates induced E”
In particular: A changing magnetic flux
creates an emf in a circuit:
Ammeter or
voltmeter.
Electromagnetic Induction
Current in secondary circuit can be produced by
a changing current in primary circuit.
Demonstrations
• EMF induced in a coil by moving a bar
magnet
• EMF induced in a secondary coil by
changing current in primary coil
Ammeter or
voltmeter.
Application:
Transformer
EMF induced in a coil by
moving a bar magnet
V
EFM depends on how
strong magnet and
how fast we move
in/out
Sorry, we can’t do it in this packed room
… but here is the essence of it
EMF induced in a secondary coil by
changing current in primary coil
V
A
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Lecture 17 Chapter 30
Friday 17 February 2006
Magnetic Flux
Faraday’s Law
We define magnetic flux Φ exactly as we defined
the flux of the electric field. The idea is the
number of lines of B that pass through an area.
G G
Φ = ∫ B ⋅ dA
Φ = BA
Simple case #1: uniform B, ⊥ surface:
Simple case #2: surface is closed:
Φ=0
Changing Magnetic Flux
G G
Φ = ∫ B ⋅ dA
How can we get a time-changing
flux, so that ε = − dΦ ≠ 0 ?
dt
• Change the field: Φ = B(t) A
• Change the area: Φ = B A(t)
• Change the angle: Φ = B A cos θ(t)
ε = − dΦ
dt
The emf induced in any loop or circuit is equal to
the negative rate of change of the magnetic flux
through that loop.
Voltmeter reading
gives rate of change of
the number of lines
linking the loop.
Example 1
A circle of radius 20 cm in the xy plane is formed
by a wire and a 3-ohm resistor. A uniform
magnetic field is in the z direction; its magnitude
decreases steadily from .08 tesla to 0 in a time of 4
seconds.
What emf is generated?
A = π r 2 = 0.13 m 2
ε = − ddtΦ = − A dB
= −(.13 )( −.02) = 2.6 × 10
dt
Lenz’s Law
ε = − dΦ
dt
The direction of the induced emf is such as to create
a current which will oppose the change in the flux.
dB
.08 T
=−
= −.02 T / s
dt
4s
−3
V
Example 2
I push a rod
along metal rails
through a uniform
magnetic field.
(a) What emf is generated?
Motion as shown produces
clockwise current which
makes B field opposing the
increase.
(b) What current will flow?
(c) What power must I supply?
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Lecture 17 Chapter 30
Friday 17 February 2006
Example 2a
Example 2b
L = 20 cm
V = 3.0 m/s
B = .05 T
Resistance of
bar: R = 15 Ω
i
(b) What current will flow?
(a) What emf is generated?
i=
dA
dx
=L
= Lv = 0.6 m 2 / s
dt
dt
ε
R
=
− 30 × 10 −3 V
= −2 mA
15 Ω
Which direction does current flow?
Forget the minus sign. Use Lenz’s Law!
ε = − ddtΦ = − B dA
= −.05 × 0.6 = −30 mV
dt
Flux is increasing outward. Therefore current will
resist that change by flowing clockwise.
Example 2c
Faraday’s Law: General Form
i
(c) What power must I supply?
G
G
G
Magnetic force: F = iL × B
F = .002 × .2 × .05 = 2 ×10 −5 N
Power: P = Fv = (2 × 10
Check Joule heating:
−5
N )(3 m / s ) = 6 ×10 −5 W
ε
P = i 2 R = 6 ×10 −5 W
Inductance
• For any coil of wire, there is a flux Φ through the
coil, which is proportional to the current.
• If that changes, Faraday’s Law requires an emf
induced in the coil, proportional to the rate of change
of the flux.
• Clearly Φ ∝ i and so
ε = − ddtΦ ∝ − dtdi
• Define the proportionality
constant to be the inductance L:
ε
di
= −L
dt
G G
d G G
⋅
=
−
B ⋅ dA
E
d
s
∫C
∫
dt S
Φ
Inductors
If current is increasing, the
induced emf acts against the
increase, giving a voltage drop.
If current is decreasing, the
induced emf acts against the
decrease, giving a voltage rise.
• SI unit of inductance is the henry (H).
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Lecture 17 Chapter 30
Friday 17 February 2006
Energy in an Inductor
Magnetic Field Energy
The energy stored in an inductor equals the
work required to set up the current.
dq
di
dW = Vdq = V
dt = ( L ) idt = Lidi
dt
dt
The energy stored in an inductor is contained in
the magnetic field. The general formula for the
energy density in any magnetic field is
I
W = ∫ dW = L∫ idi = 12 LI 2
0
So energy stored in an inductor is
B2
u=
2µ0
U = 12 Li 2
RL Circuits
Inductors and Resistors
+ ε − iR − L
Voltage changes
going clockwise
around this loop:
+ε
L
Try same kind of solution:
Inductor gives
voltage drop if
current is increasing.
i=
ε {1 − e −t /τ }
R
τ = L/ R
This works, provided
Example
RL Summary
ε = 30V
R = 5000 Ω
Set switch to position a:
ε
i = {1 − e − t /τ }
Set switch to position b:
di
+ Ri = ε
dt
Same equation as for charging a capacitor!
di
− iR − L = 0
dt
R
di
=0
dt
L = 15 mH
(a) What is the time constant?
ε
i=
In either case time constant is:
R
e
− t /τ
τ = L/ R
τ = L/ R =
15 × 10
5 × 10 3
−3
= 3 × 10 −6 = 3 µs
(b) What is current after 1 second?
i=
ε {1 − e −t /τ }=
R
30
(1 − 0) = 6 mA
5000
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Lecture 17 Chapter 30
Friday 17 February 2006
Example 2:
Problem 30-89
Example 2
continued
(a) What happens
immediately after
switch is closed?
(b) What happens
a long time after
switch is closed?
L prevents sudden change so:
We have reached a steady state so:
i2 = 0
So: VR 2 = 0
∴ i = i1 = ε / R1
∴ VL = ε
and
di2
=ε /L
dt
di2
=0
∴ VL = 0 and VR 2 = ε
dt
So: i = ε / R , i = ε / R , i = i + i
1
1
2
2
1
2
Induction and Oscillations
Ch. 30: Faraday’s Law
Ch. 31: AC Circuits
• Chapter 30 Homework for Monday:
– Questions 1, 3, 7
– Problems 3, 5, 29, 44
• Chapter 31 Homework for Tuesday:
– Questions 3, 4, 7
– Problems 5, 19, 39
• WileyPlus chapters 30, 31 for Tuesday.
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