Transcript iClicker Question

Last time
Ampere's Law
Faraday’s law
1
Faraday’s Law of Induction (More Quantitative)
The magnitude of the induced EMF in conducting loop is
equal to the rate at which the magnetic flux through the
surface spanned by the loop changes with time.
dΦB
ε
dt
where  B 

S
B ndA
N
Minus sign indicates the sense of EMF: Lenz’s Law
• Decide on which way n goes
Fixes sign of B
N
• RHR determines the
positive direction for EMF 
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How 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.
3
Today
Faraday’s law
Inductance and RL (RLC) circuit
4
Conducting Loop in a Changing Magnetic Field
Induced EMF has a direction such that it opposes
the change in magnetic flux that produced it.
approaching
 Magnetic moment 
created by induced currrent
I repels the bar magnet.
Force on ring is repulsive.
moving away
 Magnetic moment 
created by induced currrent
I attracts the bar magnet.
Force on ring is attractive.
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Induced Electric Field from Faraday’s Law
• EMF is work done per unit charge:
ε W /q
• If work is done on charge q, electric field E must be present:
ε
E
nc
W   q Enc ds
ds
Rewrite Faraday’s Law in terms
of induced electric field:
E
nc
dΦB
ds  
dt
This form relates E and B!
B
• Note that  E  ds  0for E fields generated by charges at rest
(electrostatics) since this would correspond to the potential difference
between a point and itself. => Static E is conservative.
• The induced E by magnetic flux changes is non-conservative.
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iClicker Question
The magnetic field is decreasing, what’s the direction of the
induced currents in the closed rectangular loop?
A. Clockwise
B. Counterclockwise
C. No induced currents.
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6D-11 Jumping Ring
Is there any
differences in the
two rings ?
Why one can
jump up, the
other can’t ?
http://www.youtube.com/watch?v=
ZL4kbBIf39s
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iClicker Question
The magnetic field is fixed, what’s the direction of the induced
currents in the closed rectangular loop?
A. Clockwise
B. Counterclockwise
C. No induced currents.
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iClicker Question
A current directed toward the top of the page and a rectangular
loop of wire lie in the plane of the page. Both are held in place
by an external force. If the current I is decreasing, what is the
direction of the magnetic force on the left edge of the loop?
a. Toward the right
I
b. Toward the left
c. Toward top of page
d. Toward bottom of page
e. No force acts on it.
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Inductance
R
emfbat
emf 
emf  
0 N 2
d
d mag
emfcoil
dI
R
dt
r  l  B  0
2
dt
N
I
d
Increasing I  increasing B
emfind  L
emfbat
dI
dt
R
emfind
L – inductance, or self-inductance
0 N 2 2
L
R
d
Unit of inductance L:
Henry = Volt.second/Ampere
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Inductance resists changes in current
Demos: 6C-07 Energy Stored in
an Inductive Circuit
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Circuit Analysis Tips
• Simplify using equivalent resistors
• Label currents with arbitary directions
•If the calculated current is negative, the real direction is opposite to the one
defined by you.
• Apply Junction Rule to all the labeled currents.
•Useful when having multiple loops in a circuit.
• Choose independent loops and define loop direction
•Imagine your following the loop and it’s direction to walk around the circuit.
• Use Loop Rule for each single loop
•If current I direction across a resistor R is the same as the loop direction,
potential drop across R is ∆V = −I×R, otherwise, ∆V = I×R
•For a device, e.g. battery or capacitor, rely on the direction of the electric
field in the device and the loop direction to determine the Potential drop
across the device
• Solve simultaneous linear equations
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
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RL Circuits – Starting Current
1.
2.
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
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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
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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.
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Energy Stored By Inductor
1.
2.
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
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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
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iClicker Question
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
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Two Bulbs Near a Solenoid
Varying B is created by AC current in a solenoid
What is the current in this circuit?
 mag   0 sin t 
d
emf  
dt
emf  emf0 cost 
emf emf 0
I

cost 
2R
2R
Advantage of using AC: Currents and emf ‘s behave as sine and
cosine waves.
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Two Bulbs Near a Solenoid
Add a thick wire:
I1
Loop 1: emf  R1I1  R2 I 2  0
Loop 2: R2 I 2  0
Node:
I1  I 2  I 3
emf
I1 
R1
Loop 1
I2  0
I1  I 3
I2
Loop 2
I3
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Exercise
25
Transformer
emf AC
emfloop  
N prim
emfsec
N sec

emf AC
N prim
Energy conservation:
I secemfsec  I primemf AC
N sec
I prim  
I sec
N prim
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Previously asked question: Why use HV to transport
electricity?
Single home current: 100 A service
Vwires=IRwires
Transformer: emfHV IHV = emfhomeIhome
Single home current in HV: <0.1 A
Power loss in wires ~ I2
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(Ideal) LC Circuit
• From Kirchhoff’s Loop Rule
Q
dI
L 0
C
dt
• From Energy Conservation
2
Q 2 1 2  Q peak
E
 LI  
2C 2
 2C
dQ
I
dt
dE
0
dt
same

  const.

Q dQ
dI
 LI
0
C dt
dt
d 2Q  1 

Q  0
2
dt
 LC 
Q  Q peak cos(0t   )
dQ
I
 0Q peak sin( 0t   )
dt
Q
dI
L 0
C
dt
harmonic oscillator with angular
1
frequency
0 
LC
Natural Frequency
LC Oscillations
Q2
1 2
dQ
UE 
, U B  LI , I 
2C
2
dt
No Resistance =
No dissipation
Backups
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iClicker Question
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
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iClicker Question
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
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