Transcript Physics 122B Electromagnetism - Institute for Nuclear Theory

Physics 122B
Electricity and Magnetism
Lecture 23 (Knight: 33.8-33.10)
Inductance and LR Circuits
Martin Savage
Lecture 23 Announcements
 Lecture HW has been posted and is due on
Wednesday at 10 PM.
Midterm examination 3 is this coming Friday--questions from lecture, tutorial AND Lab.
7/21/2015
Physics 122B - Lecture 23
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Generators
The figure shows a coil with
N turns rotating in a magnetic
field, with the coil connected to
an external circuit by slip rings
that transmit current
independent of rotation. The
flux through the coil is:
 m  A  B  AB cos 
 AB cos t
Ecoil  N
d m
d
 ABN  cos t    ABN sin t
dt
dt
Therefore, the device produces emf and current that will vary
sinusoidally, alternately positive and negative. This is called an alternating
current generator, producing what we call AC voltage.
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Example: An AC Generator
A coil with area 2.0 m2 rotates in a 0.10 T magnetic field at a
frequency of 60 Hz. How many turns are needed to generate an AC emf
with a peak voltage of 160 V?
Ecoil   ABN sin t
Emax   ABN
  2 f
Emax
(160 V)
N

 21 turns
2
2 f AB 2 (60 Hz)(2.0 m )(0.10 T)
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Transformers
When a coil wound around an
iron core is driven by an AC voltage
V1cos wt, it produces an oscillating
magnetic field that will induce an
emf V2cos wt in a secondary coil
wound on the same core. This is
called a transformer.
The input emf V1 induces a
current I1 in the primary coil that
is proportional to 1/N1. The flux in
the iron is proportional to this, and
it induces an emf V2 in the secondary
coil that is proportional to N2. Therefore, V2 = V1(N2/N1). From
conservation of energy, assuming no losses in the core, V1I1 = V2I2.
Therefore, the currents in the primary and secondary are related by the
relation I1 = I2(N2/N1).
A transformer with N2>>N1 is called a step-up transformer, which
boosts the secondary voltage. A transformer with N2<<N1 is called a stepdown transformer, and it drops the secondary voltage.
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The Tesla Coil
A special case of a step-up
transformer is the Tesla coil. It
uses no magnetic material, but
has a very high N2/N1 ratio and
uses high-frequency electrical
current to induce very high
voltages and very high
frequencies in the secondary.
There is a phenomenon called
“the skin effect” that causes
high frequency AC currents to
reside mainly on the outer
surfaces of conductors. Because
of the skin effect, one does not
feel (much) the electrical
discharges from a Tesla coil.
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Metal Detectors
Metal detectors like those used at
airports can detect any metal objects,
not just magnetic materials like iron.
They operate by induced currents.
A transmitter coil sends high
frequency alternating currents that
will induce current flow in conductors
in its field. Because of Lenz’s Law,
the induced current opposes the field
from the transmitter, so that net
field is reduced. A receiver coil
detects the reduction in the magnetic
fields from the transmitter and
registers the presence of metal.
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(Self-) Inductance
We define the inductance L of a coil of wire producing flux m as:
m
L
I
The unit of inductance is the henry: 1 henry = 1 H = 1 T m2/A = 1 Wb/A
The circuit diagram symbol used to represent inductance is:
Example: The inductance of a long solenoid with N turns of
cross sectional area A and length l is:
0 NI
per turn  BA
B
l
2
N A
 m 0 N 2 A
 m  N  per turn  NBA  0
I
Lsolenoid 

l
I
l
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Example: Length of an Inductor
An inductor is made by tightly winding 0.30 mm diameter wire around
a 4.0 mm diameter cylinder.
What length cylinder has an inductance of 10 mH?
L
0 N 2 A
l

0 (l / d )2 ( r 2 )
l

0 r 2l
d
2
 1.0 105 H
d 2L
(3.0 104 m)2 (1.0 105 H)
l

 0.057 m  5.7 cm
2
7
3
2
0 r
(4 10 Tm/A) (2.0 10 m)
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Potential Across an Inductor
Ecoil  N
d  per turn
dt
Ecoil  L
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d m

dt
dI
dt
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Potential Across an Inductor (2)
Ecoil  L
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dI
dt
VL   L
Physics 122B - Lecture 23
dI
dt
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Inductors and Sparks
An inductor responds to an interruption in current flow by
developing a very large potential (L dI/dt) across its terminals. This
potential is typically dropped across the object (usually a switch) that
is effecting the change in current. Therefore, breaking a closed circuit
with an inductor often results in a spark across the switch contacts.
Over time, this can erode and destroy the switch.
Spark plugs in automobiles are driven by inductances that are
electronically interrupted.
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Example:
Large Voltage across an Inductor
A 1.0 A current passes through a 10 mH inductor coil.
What potential difference is induced across the coil if the current
drops to zero in 5 s?
dI I
A))
((-10.
1.0 A
-5




2.0

10
A/s
-6
dt t (5.0 10 s)
VL   L
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dI
 (0.010 H)(2.0 10-5 A/s)  2000 V
dt
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Question
The potential at a is higher than the potential at b.
Which of the following statements about the inductor
current I is consistent with this observation?
a.
b.
c.
d.
e.
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I is from a to b and steady.
I is from a to b and increasing
I is from a to b and decreasing
I is from b to a and steady
I is from b to a and increasing
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Energy in Inductors and
Magnetic Fields
A magnetic field stores considerable energy. Therefore, an
inductor, which operates by creating a magnetic field, stores energy.
Let’s consider how much energy UL is stored in an inductor L carrying
current I:
dI 
dI

P  I V  I   L    LI
dt 
dt

I
U L  L  IdI  12 LI 2
0
A solenoid of length l , area A, and N turns has L = 0N2A/l, so:
U L  LI 
1
2
2
0 N 2 A
1

AlB 2
2 0
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2l
1
  NI 
I 
Al  0 
2 0  l 
uB 
2
2
UL
1 2

B
Al 20
U L  LI
1
2
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1 2
uB 
B
2 0
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Energy Density in the Fields
This is the magnetic analog of the energy stored in an electric field
UE =
e0
2
1 2
uB 
B
2 0
E
2
Magnetic
Electric
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The LR Circuit
VR  VL  0
dI
R
  dt
I
L
 RI  L
I
dI
0
dt
t
dI
R


dt
I I

L0
0
I 
t
ln    
L/R
 I0 
I (t )  I0et /( L / R)
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Example: Exponential Decay
in an LR Circuit
The switch shown has been
in position a for a long time. It
changes to position b at t=0.
a. What is the current in
the circuit at t = 5 s?
b. At what time has the
current decayed to 1% of its
original value?
I0  V / R  (10 V) /(100 )  100 mA
t / 
I  I 0e
(5 s)/(10 s)
 (100 mA)e
 61 mA
  L /( R  R)
 (2.0 10-3 H) /(200 )  10 s
I 
(1 mA)
t   ln    (10 s) ln
 46 s
(100 mA)
 I0 
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Plumber’s RL Analogy
Valve
Constriction
P1
P2
Pump
The “plumber’s analogy” of an RL circuit is a
pump (=battery) pumping water in a closed
loop of pipe that includes a valve (=switch), a
constriction (=resistor), and rotating
flywheel turned by the flow. When the
valve starts the flow, the flywheel begins to
spin until a steady flow is achieved and the
pressure difference across the flywheel
(P2-P3) goes to zero.
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P3
Flywheel
Pump = Battery
Valve = Switch
Constriction = Resistor
Flywheel = Inductor
Pressure = Potential
Water Flow = Current
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Capacitors: Early and Late*
Initially, when a switch closes there is a potential difference of 0
across an uncharged capacitor. After a long time, the capacitor reaches
its maximum charge and there is no current flow through the capacitor.
Therefore, at t=0 the capacitor behaves like a short circuit (R=0), and at
t=∞ the capacitor behaves like at open circuit (R=∞).
Example:
100 V
100 V
12.5 A
100 V
2.5 A
Circuit
* From Lecture 16
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10 A
0 V
at t=0
at t=∞
Calculate initial currents.
IB = 100 V/8  = 12.5 A
Calculate final potentials.
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Inductors: Early and Late
Initially, when a switch closes an inductor appears to have an infinite
resistance and has a maximum potential drop across it. Ultimately, the
inductor reaches a steady current flow with no potential drop across
it. Therefore, at t=0 the inductor behaves like at open circuit (R=∞),
and at t=∞ the inductor behaves like a short circuit (R=0). This behavior
is opposite that of a conductor.
Example:
Circuit
at t=0
at t=∞
Calculate initial potentials. Calculate final currents.
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Question
Rank the order of the time constants in these circuits.
(a) 123 ;
(d) 1>2>3 ;
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(b) 1<2<3 ;
(e) 2>1>3 ;
(c) 2<1<3 ;
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Lecture 23 Announcements
 Lecture HW has been posted and is due on
Wednesday at 10 PM.
Midterm examination 3 is this coming Friday--questions from lecture, tutorial AND Lab.
7/21/2015
Physics 122B - Lecture 23
23