# + - Lecture 18-13

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```Lecture 18-1
Eddy Current
A current induced in a solid conducting
object, due to motion of the object in an
external magnetic field.
• The presence of eddy current in the object
results in dissipation of electric energy
that is derived from mechanical motion
of the object.
• The dissipation of electric energy in turn
causes the loss of mechanical energy of
the object, i.e., the presence of the field
damps motion of the object.
Lecture 18-2
Self-Inductance, L
• As current i through coil increases,
magnetic flux through coil increases.
This in turn induces counter EMF
in the coil itself (Lenz)
• When current i is decreasing, EMF is
induced again in the coil itself in such a way
as to slow the decrease maintain the field.
Self-induction
B
L
i
NB
L
i
 L  T  m2 / A  Wb / A  H (henry)
dB
dI
 
 L
dt
dt
Lecture 18-3
DEMO
SELF INDUCTANCE
6D-10
Lecture 18-4
B
L
I
+
-
 L  T  m2 / A  Wb / A  H
I
dQ
dt
(henry)
d 2Q
 L 2
dt
INERTIA
d 2x
compare with f  m 2
dt
This could have been a battery, instead.
With the simplifying advantage that the
applied ε is then ~constant.
Lecture 18-5
Qualitative discussion of currents in L
Note that if the initial current in an inductor, L, is ZERO, then
difference is applied to it. Current I will “accelerate” from
zero in a continuous way, just as the velocity of a mass at rest
will accelerate in a continuous way under the influence of a
force.
At large times, the transients have died out, and since I is no
longer changing, no voltage drop exists across the inductor L
Lecture 18-6
Which of the following statements is incorrect?
A| The inductance of a coil with N turns is proportional to N2 turns.
B| Like a capacitance (C) an inductor (L) depends on geometric factors
and the nature of the material inside C and L.
C| The potential across an inductor depends only on the magnitude
of the current through the inductor.
D| The magnetic energy stored in an inductor is proportional to the
square of the current through the inductor.
Lecture 18-7
Solenoid: Archetypical Inductor
Current i flows through a long solenoid
of radius r with N turns in length l
r  l  B  0
N
i
l
For each turn
N
A   r   B  BA  0 i r 2
l
For the solenoid
NB
N2 2
N
L
 0
 r  0   l r 2
i
l
 l 
or
2
2
L  0n 2 Al Al is the volume of the solenoid
Inductance, like capacitance, only depends on geometry (if made of
conductor and air). Add a magnetically polarizable material, increase L,
just as adding an electrically polarizable material (dielectric) increases C.
Lecture 18-8
Potential Difference Across Inductor
VL  o I r
+V
ΔV
I
internal resistance must
be symbolically
separated from L !!
• “Analogous” to a battery
• An ideal inductor has r =0
V=0
• All dissipative effects are to be included in
the external + internal resistances (i.e.,
including those of the iron core if any)
dI
  IR  L  0
dt
dI
   IR  L  0
dt
Lecture 18-9
1.
2.
Energy Stored By Inductor
Switch on at t=0
As the current tries to begin flowing,
self-inductance induces back EMF, thus
opposing the increase of I.
Loop Rule:
  IR  L
+
dI
0
dt
-
3. Multiply through by I
dI
 I  I R  LI
dt
2
Rate at which battery is
supplying energy
Rate at which energy is
dissipated by the resistor
Rate at which energy is
stored in inductor L
dU m
dI
 LI
dt
dt
Um 
1 2
LI
2
Lecture 18-10
6C07
ENERGY STORED IN AN INDUCTOR
Lecture 18-11
Where is the Energy Stored?
• Energy must be stored in the magnetic field!
Energy stored by a capacitor is stored in its electric field
• Consider a long solenoid where
2
B  0nI , L  0n Al
2
1 2 1
1
B
U m  LI   0n 2 Al  I 2 
Al
2
2
2 0
• So energy density of
the magnetic field is
2
Um 1 B
um 

Al 2 0
1
uE   0 E 2
2
area A
(Energy density of
the electric field)
length l
Lecture 18-12
1.
RL Circuits – Starting Current
Switch to ε at t=0
As the current tries to begin flowing,
self-inductance induces back EMF,
thus opposing the increase of I.
+
I0  0
2.
Loop Rule:
-
dI
  IR  L
0
dt
3. Solve this differential equation
I

R
1  et /( L / R )  , VL  L
τ=L/R is the inductive
time constant
dI
  e  t /( L / R )
dt
T  m2 / A T  m2 / A

s
 L / R 

V/A
Lecture 18-13
Warm-up quiz 2
The circuit is turned on at t=0.
Which of the following statements is correct?
R1
V
R2
L
A| At t = 0, the potential drop across the inductor is V;
When t = ∞, the current through R1 is V/R1 yes
B| At t = 0, the potential drop across the inductor is 0;
When t = ∞, the current through R1 is V.
C| At t = 0, the potential drop across the inductor is V;
t = ∞, the current through R1 is V/(R1+R2)
D| At t = 0, the potential drop across the inductor is V;
When t = ∞, the current through R1 is V/R2
X
No fully correct
offered
See next slide
Lecture 18-14
Warm-up quiz 2
The circuit is turned on at t=0.
Which of the following statements is correct?
R1
V
R2
L
Analysis: At t=0, no current has yet started flowing in L.
BUT: Resistor R2 “short-circuits” L, in the sense that resistors are NONINDUCTIVE and current can “instantly” start flowing in them.
So there IS an instant current path available at t=0, and we have a simple
resistive voltage divider chain. VL = V R2/(R1+R2)
At t= I is no longer changing, so L acts like a wire and completely shortcircuits R2. All the voltage drop is across R1, and the current is V/R1, as
in the SECOND part of answer (a)
Lecture 18-15
GROWTH AND DECAY OF CURRENT
OF AN RL CIRCUIT 6C-05
Lecture 18-16
1.
flowing:
  I 0  R1  R 
2.
-
Switch from e to f at t=0,
causing back EMF to oppose the
change.
dI
IR  L  0
dt
3.
Loop Rule:
4.
Solve this differential equation
I

R
+
e  t /( L / R )
dI
VL  L   e  t /( L / R )
dt
like discharging a capacitor
I cannot instantly
become zero!
Self-induction
Lecture 18-17
Starting Current through Inductor vs Charging Capacitor
q  C (1  e t / RC )
IR   1  e  t /( L / R ) 
R
R
I

R
e  t / RC
dI
VL  L   e  t /( L / R )
dt
Note: when “charging”,
Lecture 18-18
Qualitative discussion of currents in L
Note that if the initial current in an inductor, L, is ZERO, then
difference is applied to it. Current I will “accelerate” from
zero in a continuous way, just as the velocity of a mass at rest
will accelerate in a continuous way under the influence of a
force.
At large times, the transients have died out, and since I is no
longer changing, no voltage drop exists across the inductor L
Lecture 18-19
Behavior of Inductors
• Increasing Current
– Initially, the inductor behaves like a battery connected in reverse.
– After a long time, the inductor behaves like a conducting wire.
• Decreasing Current
– Initially, the inductor behaves like a “reinforcement” battery.
– After a long time, the inductor behaves like a conducting wire.
Lecture 18-20
Physics 241 March 22, 2011 9:30 –Quiz 3
The switch in this circuit is initially open for a
long time, and then closed at t = 0. What is the
magnitude of the voltage across the inductor
just after the switch is closed?
a) zero
b) V
c) R / L
d) V / R
e) 2V
Lecture 18-21
Physics 241 March 22 2011 10:30 Quiz 3
The switch in this circuit is closed at t = 0.
What is the magnitude of the voltage across the
resistor a long time after the switch is closed?
a) zero
b) V
c) R / L
d) V / R
e) 2V
Lecture 18-22
Physics 241 March 22, 2011 11:30–Quiz 3
The switch in this circuit has been open for a
long time. Then the switch is closed at t = 0.
What is the magnitude of the current through the
resistor immediately after the switch is closed?
a) zero
b) V / L
c) R / L
d) V / R
e) 2V / R
```