Transcript Lec18

Lecture 18-1
Ways to Change Magnetic Flux
 B  BA cos 
• Changing the magnitude of the field within a conducting loop (or coil).
• Changing the area of the loop (or coil) that lies within the magnetic field.
• Changing the relative orientation of the field and the loop.
motor
generator
http://www.wvic.com/how-gen-works.htm
Lecture 18-2How
to use Faraday’s law to determine the induced
current direction

1.
define the direction of n ; can be any of the two
normal direction, e.g. n point to right
2.
determine the sign of Φ. Here Φ>0
N
3.
determine the sign of ∆Φ. Here ∆Φ >0
4.
determine the sign of  using faraday’s law. Here  <0
5.
RHR determines the positive direction for EMF 
• If >0, current follow the direction of the curled
fingers.
• If <0, current goes to the opposite direction of
the curled fingers.
Lecture 18-3
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 18-4
Self-Inductance
• As current i through coil increases,
magnetic flux through itself increases.
This in turn induces counter EMF
in the coil itself
• When current i is decreasing, EMF is
induced again in the coil itself in such a way
as to slow the decrease.
Self-induction
B
L
i
NB
L
(if flux linked)
i
 L  T  m2 / A  Wb / A  H (henry)
Faraday’s Law:
dB
dI
 
 L
dt
dt
Lecture 18-5
Lecture 18-6
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
 0   H / m
2
2
L  0n 2 Al
Inductance, like capacitance, only depends on
geometry
Lecture 18-7
Potential Difference Across Inductor
V    I r
+
V
I
internal resistance
• Analogous to a battery
• An ideal inductor has r=0
-
• All dissipative effects are to be
included in the internal resistance (i.e.,
those of the iron core if any)
dI
 0  IR  L  0
dt
dI
  0  IR  L  0
dt
Lecture 18-8
1.
2.
RL Circuits – Starting Current
Switch to e at t=0
As the current tries to begin flowing,
self-inductance induces back EMF,
thus opposing the increase of I.
Loop Rule:
+
-
dI
  IR  L
0
dt
3. Solve this differential equation
I

R
1  et /( L / R )  , VL  L
dI
  e  t /( L / R )
dt
τ=L/R is the inductive time constant
T  m2 / A T  m2 / A

s
 L / R 

V/A
Lecture 18-9
Starting Current through Inductor vs Charging Capacitor
q  C (1  e t / RC )
IR   1  e  t /( L / R ) 
I

R
e  t / RC
dI
VL  L   e  t /( L / R )
dt
Lecture 18-10
Warm-up
Which of the following statement is correct after switch S is closed ?
R1
S
V
R2
L
1. At t = 0, the potential drop across the inductor is V;
When t = ∞, the current through R1 is V/R1
2. At t = 0, the potential drop across the inductor is V;
When t = ∞, the current through R1 is V.
3. At t = 0, the potential drop across the inductor is 0;
When t = ∞, the current through R1 is V/(R1+R2)
4. At t = 0, the potential drop across the inductor is V;
When t = ∞, the current through R1 is V/R2
Lecture 18-11
Remove Battery after Steady I already exists in RL Circuits
1.
Initially steady current Io is
flowing:
  I 0  R1  R 
2.
-
Switch 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 18-12
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 18-13
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
stored in inductor L
dU m
dI
 LI
dt
dt
Rate at which energy is
dissipated by the resistor
Um 
1 2
LI
2
Lecture 18-14
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
(Energy density of the
electric field)
area A
length l
Lecture 18-15
Physics 241 –Quiz A
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 18-16
Physics 241 –Quiz B
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 18-17
Physics 241 –Quiz C
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