Download lecture 17 - Purdue Physics

LECTURE 17
•  Motional EMF
•  Eddy Currents
•  Self Inductance
Reminder
Magnetic Flux
•  Faraday’s Law
ε =−
dΦB
dt
dA
•  Flux through one loop
ΦB = BAcosθ
•  Flux through N loops
ΦB = NBAcosθ
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Reminder
How to Change Magnetic Flux in a Coil
!
ΔΦB
ΔB
1. B changes:
=N
Acosθ
Δt
Δt
2. A changes:
ΔΦB
ΔA
= NB
cosθ
Δt
Δt
3. θ changes:
Δ ⎡⎣cosθ ⎤⎦
ΔΦB
= NBA
Δt
Δt
4. N changes:
ΔΦB ΔN
=
BAcosθ
Δt
Δt
Unlikely
Example
Ideal Solenoid Within a Coil
120-turn coil of
radius 2.4 cm and
resistance 5.3 Ω
Solenoid with
radius 1.6 cm and
n = 220 turns/cm
Initial current in the solenoid is 1.5 A.
Current is reduced to zero in 25 ms.
What is the current in the coil while the current is
being reduced?
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Example
Ideal Solenoid Within a Coil
•  IDEAL B field inside the
solenoid
–  Constant and parallel to its
axis
B=µ0n I
•  Can easily find flux of
this field through the N
coils
Example
Ideal Solenoid Within a Coil
dϕ m
dt!
ϕ m = NB ⋅ n̂A = NBA
ε =−
No. of turns
in the coil
µ0nI
NOT area
enclosed
by coil
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Example
Ideal Solenoid Within a Coil
At point P1 :
B = µ0nI
At point P2 :
B≅0
Ideal solenoid (closely wound, long):
B = 0 outside the core of the solenoid
∴ relevant area for ϕ m
= area enclosed by solenoid
Example
Ideal Solenoid Within a Coil
2
ϕ m, i = N ⋅ µ0nIi ⋅ π rsolenoid
= 120 ⋅ 4π ×10−7 × 220
2
turns 100 cm
×
×1.5 ⋅ π (0.016)
cm
1m
ϕ m, i = 4.00 ×10−3 T ⋅ m2
ϕ m, f = 0
dϕ m ϕ m, f − ϕ m, i 4.00 ×10−3
T ⋅ m2
=
=
= 160 ×10−3
= −ε
dt
0.025 s
0.025
s
Current is given by Ohm's law
I=
ε 160 ×10−3 V
=
= 30.2 ×10−3 A = 30.2 mA
R
5.3 Ω
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Example
Ideal Solenoid Within a Coil
What is the direction of the current in the coil?
(A)  Same direction as the current in the solenoid
(B)  Opposite the direction of the current in the solenoid
Motional EMF
•  Definition
–  Any emf induced by the
motion of a conductor in a
magnetic field
!
dB
dt
= 0, but
dA
or
dt
d⎡
⎣cosθ ⎤⎦ ≠ 0
dt
!
ϕ m = B ⋅ n̂A = Bn A = Bn ℓx
dϕ m
dx
= Bn ℓ
= Bn ℓv
dt
dt
ε =−
dϕ m
= −Bn ℓv
dt
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Motional EMF
•  Definition
–  Any emf induced by the
motion of a conductor in a
magnetic field
•  Charge Separation
–  + charge experiences
upward force qvB
–  Electric field vB, downward
–  Potential ΔV=vBL
Calculation
A thin conducting rod is
pulled with velocity v along
conducting rails that are
connected by a resistor, R.
A uniform magnetic field B is
directed into the page.
Surface S is increasing
therefore flux is increasing.
What will be the magnitude &
direction of the induced
current?
Lenz Law ⇒ counterclockwise
ε =−
dϕ m
= −Bn ℓv
dt
I=
ε Bℓv
=
R
R
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Energy Conservation
•  What is the rate of work by the applied force?
•  The induced current gives rise to a net magnetic force
F in the loop which opposes the motion:
⎛ ℓBv ⎞
ℓ 2B 2v
FL = IℓB = ⎜
⎟ ℓB =
R
⎝ R ⎠
Energy Conservation
•  External Agent must exert equal but opposite force FR
to move the loop with velocity v
l 2B 2v 2
–  Agent does work at rate P
P = FRv =
R
•  Energy is dissipated in circuit at rate P’
2
⎛ lBv ⎞
l 2B 2v 2
Pʹ = I R = ⎜
⎟ R=
R
⎝ R ⎠
2
⇒
P =P'
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Eddy Currents
•  Previously we considered currents confined to a
loop or a coil
•  Electrical equipment contains other metal parts
which are often located in changing B fields
–  Currents in such parts are called Eddy currents
•  Swirl around
Induced current
anti-clockwise
F
x x
x x
x x
x x
v
x x
x x
Flux
increasing
x
x
x
x
x
x
Induced current
clockwise
x x x
x x x
x x x
x x xF
x x x
x x x Flux
v
B field into
the page
decreasing
Eddy Currents
Replace the loop with a solid Cu plate
•  The energy is dissipated by heating the metal
–  Conducting material near magnetic fields will heat up,
sometimes a lot
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Induction Heating
Laboratory and
industrial processes
Cooking
Eddy Currents
•  Relative motion between a B field and a conductor
induces a current in the conductor
•  The induced current give rises to a net magnetic
force, FM, which opposes the motion
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Demo
Eddy Currents
!
!
!
•  The force F = i ℓ × B opposes the motion
–  This “braking effect” is used for the brakes of trains
Demo
Reduce Eddy Currents
Metal strips with
insulating glue
Cut slots into
the metal
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Inductance and Inductors
•  Capacitance:
Q ∝V
Q = CV
•  Coil carrying current, I, produces a magnetic field
–  Magnetic flux proportional to current
–  L = self-inductance of coil
φm = LI
•  Inductance, L, depends on geometry of conductor
•  Symbol for Inductor:
•  Units: Henry = Webers/Ampere = Tm2/A
Inductors
φm = LI
d φm
dI
Changing current:
= L = −ε
dt
dt
•  Magnetic flux:
• 
•  Changing the current through an inductor induces
an opposing voltage across the inductor
–  Acts like a voltage source in a circuit
ΔV = −ε
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Self-Inductance
•  An induced emf, εL, appears in any
coil in which the current is
changing
•  Close switch, current starts to flow
–  B field produced within loop
dI/dt
X XX X
X XX XX X
X XX X
a
b
•  Flux through loop increases as
current increases
–  EMF is induced in loop opposing change in
current flow
–  Opposes change in flux
•  Self-Induction
–  Changing current through a loop induces
an opposing voltage in that same loop
Self-Inductance in a Coil
L=
ϕm
I
ϕ m = NBA = N ⋅ µ0nI ⋅ A =
µ0N 2IA
N
= µ0n 2IAℓ since n =
ℓ
ℓ
ϕm
= µ0n 2 Aℓ : geometric factors only, just like capacitance
I
Note that Aℓ = volume
L=
12
Magnetic Energy in an Inductor
•  Upon closing switch, S, apply Kirchoff’s loop rule:
ε − IR − L
dI
=0
dt
•  Multiply through by I:
εI = I 2R + LI
power
delivered
by battery
dI
dt
power
dissipated
by resistor
power
delivered
to inductor
Magnetic Energy in an Inductor
•  Inductor stores magnetic energy
If Um = energy in the inductor,
dUm
dI
= LI ⇒ dUm = LI dI
dt
dt
Um =
∫ dU
Um =
1 2
LI
2 f
m
=
∫
If
0
LI dI
Energy stored in
an inductor
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Magnetic Energy in a Solenoid
B = µ0nI ⇒ I =
B
µ 0n
L = µ0n 2 Al
2
⎛ B ⎞
1
1
B2
Um = LI 2 = µ0n 2 Al ⎜
=
Al
⎟
2
2
µ
n
2
µ
⎝ 0 ⎠
0
Magnetic energy density um =
B2
2µ0
1
Electric energy density ue = ε 0E 2
2
Both of
these are
general
results
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