Download Induction - ADDENDUM

ADDENDUM – CHAPTER 21
Mutual inductance –
Circulation of currents in one
coil can generate a field in the
coil that will extend to a second,
close by device.

Flux Changes
Suppose i1 CHANGES
Current (emf) is
induced in 2nd
coil.
Mutual Inductance
3


i2

i1 creates a field that
(partially) passes through
the second coil.
As i1 changes, the flux
through coil 2 changes
and an emf (and current
i2) are created.
The two coils are
mutually linked by what
we call an “inductance”
Induction
5/23/2017
Watch Out!
4

Exam #2 one week from today.







Chapters 20 & 21
Same format but possibly one set of multiple choice questions that you
hate.
You should already be studying.
QUIZ on Friday – Chapter #21
Today we continue with the chapter. We should finish it on
Friday. Maybe.
No study session on Monday next week
We will have a study session on Tuesday morning like last time.
Details to follow.
Induction
5/23/2017
This schedule is now in effect:
PHY2054 Problem Solving/Office Hours Schedule
Room MAP-318
Bindell
Monday
8:30-9:15AM
Tuesday
Bindell
Dubey
11:00-12:00PM
12:00-1:00PM
Wednesday
8:30-9:15AM
10:30 - 11:15
AM*
Thursday
Friday
8:30-9:15AM
10:30-11:30AM
1:30-2:45PM
These sessions will be used both for office hours and problem solving. Students from any
section of 2054 are invited to stop by for assistance in course materials (problems, etc.)
Note: There will be times when the room may not be available. In that case we will use our individual offices.
* In Office
5
Dr. Dubey's hours are for problem
solving only.
Induction
5/23/2017
Mutual Inductance
 B 2  The magnetic flux in
coil #2 due to the current in
coil #1.
6
 B 2 ~ i1
 B 2  ki1
i2
N 2  B 2  M 21 i1
 21
i1
emf2  N 2
 M 21
t
t
mutual Inductance
Induction
5/23/2017
Note the form:
7
M
N something
UNIT: henry
i
Think of this when we define INDUCTANCE (L) of
a small coil in the next section.
Induction
5/23/2017
The two coils
8
Remember – the magnetic
field outside of the solenoid
is pretty much zero.
Two fluxes (fluxi?) are the same!
Induction
5/23/2017
One solenoid is centered inside another. The outer one has a length of 50.0
cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and
0.120 cm in diameter and contains 15 coils. The current in the outer solenoid
is changing at 37.5 A/s. (a) What is the mutual inductance of these solenoids?
(b) Find the emf induced in the inner solenoid
Length = 0.5 meters
N=6750 coils
n=6750/.5=1.35E04 turn/meter
Magnetic field INSIDE the smaller coil
is the same as in the larger coil and
is given by:
B1  0 ni  4 10 1.35 10 i  1.7i
7
9
4
Induction
5/23/2017
One solenoid is centered inside another. The outer one has a length of 50.0
cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and
0.120 cm in diameter and contains 15 coils. The current in the outer solenoid
is changing at 37.5 A/s. (a) What is the mutual inductance of these solenoids?
(b) Find the emf induced in the inner solenoid
B1  0 ni  1.7i
Areasmaller  r 2   (0.06cm 1m / 100cm) 2
coil
Area  1.13 104 m 2
 B 2  BA  1.92 10 4 i
 B 2
i
 1.92 10 4
 1.92 10  4  37.5V  7.2 10 3V
t
t
 B 2
emf 2  N 2
 15  7.2 10 3V  11mV
t
10
Induction
5/23/2017
One solenoid is centered inside another. The outer one has a length of 50.0
cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and
0.120 cm in diameter and contains 15 coils. The current in the outer solenoid
is changing at 37.5 A/s. (a) What is the mutual inductance of these solenoids?
(b) Find the emf induced in the inner solenoid
B1  0 ni  1.7i
emf2 1110
M

i
37
t
 
11
Induction
3
 0.297mH
5/23/2017
Self-inductance –
Any circuit which carries a varying current self-induced
from it’s own magnetic field is said to have INDUCTANCE
(L).

13
An inductor resists CHANGES in the
current going through it.
Induction
5/23/2017
14
An inductor resists CHANGES in the
current going through it.
Induction
5/23/2017
15
An inductor resists CHANGES in the
current going through it.
Induction
5/23/2017
Inductance Defined
16
N B
L
i
If the FLUX changes a bit during a short time
t, then the current will change by a small
amount i.
Li  N B
Faraday says
this is the emf!
 B
i
N
L
t
t
This is actually a
calculus equation
Induction
5/23/2017
So …
17
i
E= emf  L
t
There should be
a (-) sign but we
use Lenz’s Law
instead!
The UNIT of “Inductance – L” of a coil is the henry.
SYMBOL:
Induction
5/23/2017
18
Induction
5/23/2017
Consider “AC” voltage
19
Minimum Change@t
V1
Maximum Change@t
Induction
5/23/2017
The transformer
20

emf1  V1  N1
t
FLUX is the same through
both coils (windings).

emf 2  V2  N 2
t
V1 V2

N1 N 2
Induction
5/23/2017
21
Induction
5/23/2017
Input/Output Impedance (Resistance)
22
V2 N 2

V1 N1
Powerin  Powerout (Lossless)
I1V1  I 2V2
So
V1
R

I1 ( N 2 / N1 ) 2
Induction
 looks like an input 


resistance !


5/23/2017
Remember that a Capacitor stored
ENERGY? U=(1/2)CV2
23
U=Area=(1/2)LI2
Power  Vi
i
V L
t
i
P  Li
t
U  Lii
U
Li
LI
i
Li
DU
 Lii
interval
i
I
Induction
5/23/2017
i
SO …
24
Energy Stored in a capacitor
The energy stored in a capacitor with capacitance C and a voltage V is
U=(1/2)LI2
Induction
5/23/2017
The Energy stored is in the Magnetic Field
25
Consider a solenoid with N turns that is very long. We assume that the field is
uniform throughout its length, ignoring any “end effects”. For a long enough
solenoid, we can get away with it for the following argument. Maybe.
N
B   0 ni   0 i
l
Induction
5/23/2017
Energy Storage in Inductor
26
N B
l
1 2 1 N B 2
U  Li 
i
2
2 l
 B  BA
L
1
NiBA
2
N
B  0 i
l
Bl
Ni 
U
1 Bl
1 B 2lA 1 B 2V
U
BA 

2 0
2 0
2 0
U 1 B2
Energy Density  
V 2 0
1
Capacitor ED   0 E 2
2
0
Induction
5/23/2017
27
Back to Circuits
Induction
5/23/2017
Series LR Circuit
28
Induction
5/23/2017
RL or LR Series Circuit
29


Switch is open .. no
current flows for obvious
reasons.
Switch closed for a long
time:
 Steady
current, voltage
across the inductor is
zero. All voltage (E) is
across the resistor.
 i=E/R
Induction
5/23/2017
RL or LR Series Circuit
30
When the switch opens, current change is high and back emf from L is maximum.
i
E/R
t
As the current increases, more voltage is across R, the rate of change of I decreases
and as the current increases, it increases more slowly.
Induction
5/23/2017
RL Circuit
31


When L=0, the
current rises very
rapidly (almost
instantly)
As L increases, it
takes longer for
the current to get
to its maximum.
Induction
5/23/2017
RL Circuit - Kirchoff Stuff
32
i
emf  iR  L  0
i
Solution
E
i  (1  e ( R / L ) t )
R
E
i  (1  e t / )
R
L
  (time constant)
R
Induction
5/23/2017
The Graphic Result – Current Growth
33
 1
1    0.63
 e
}
63% of
maximum
e= 2.71828…
Induction
5/23/2017
Decay – Short out the battery
34



Magnetic field
begins to collapse,
sending its energy
into driving the
current.
The energy is
dissipated in the
resistor.
i begins at maximum
(E/R) and decays.
Induction
5/23/2017
Solution
35
E t / 
i  (e )
R
Induction
5/23/2017
Up and Down and Up and Down and …..
36
Induction
5/23/2017
NEXT: AC Circuits
37
Induction