Transcript Phys132 Lecture 5 - University of Connecticut

Physics 1402: Lecture 20
Today’s Agenda
• Announcements:
– Chap.27 & 28
• Homework 06: Friday
• Induction
Faraday's Law
dS
B
N
S
v
B
S
N
v
B
B
A Loop Moving Through a Magnetic Field
Schematic Diagram of an AC Generator
dF B
d (cos( wt))
= - NAB
e= -N
dt
dt
= - NAB w sin( wt))
Schematic Diagram of an DC Generator
(a) As the conducting plate enters the field (position 1), the eddy currents are counterclockwise. As the plate leaves
the field (position 2), the currents are clockwise. In either case, the force on the plate is opposite to the velocity, and
eventually the plate comes to rest. (b) When slots are cut in the conducting plate, the eddy currents are reduced and
the plate swings more freely through the magnetic field.
Demo
E-M Cannon
• Connect solenoid to a source of
alternating voltage.
• The flux through the area ^ to
axis of solenoid therefore
changes in time.
• A conducting ring placed on top
of the solenoid will have a current
induced in it opposing this
change.
• There will then be a force on the
ring since it contains a current
which is circulating in the
presence of a magnetic field.
v
~
side view
F B

B
F
B
top view
Lecture 20, ACT 1
• Suppose two aluminum rings are used in
the demo; Ring 2 is identical to Ring 1
except that it has a small slit as shown. Let
F1 be the force on Ring 1; F2 be the force on
Ring 2.
Ring 1
Ring 2
(a) F2 < F1
(b) F2 = F1
(c) F2 > F1
Lecture 20, ACT 2
• Suppose one copper and one aluminum
rings are used in the demo; the
resistance of the two rings is similar but
the aluminum ring has less mass. Let a1
be the acceleration of ring 1 and a2 be
the acceleration of Ring 2.
(a) a2 < a1
(b) a2 = a1
(c) a2 > a1
Ring 1
Ring 2
Lecture 20, ACT 3
• Suppose you take the aluminum ring, shoot
it off the cannon, and try to nail your
annoying neighbor. Unfortunately, you just
miss. You think, maybe I can hit him (her) if I
change the temperature of the ring. In order
to hit your neighbor, do you want to heat the
ring, cool the ring, or is it just hopeless?
(a) heat
(b) cool
(c) hopeless
Hot Ring
Cool Ring
Lecture 20, ACT 4
• Suppose the alternating magnetic
field is kept at a level where the ring
just levitates, but doesn’t jump off. If
I keep the ring suspended for about
5 minutes, is it safe to pick it up?
(a) No
(b) Yeah, I’ll do it
~
side view
Induction
Self-Inductance, RL Circuits
I
a
I
R
XXX
XXXX
XX
b
L
e
L/R
e1
1
VL
f( x ) 0.5
0.0183156
0
0
1
2
3
4
t
Recap from the last Chapter:
Faraday's Law of Induction
B
N
S
v
B
S
N
v
Can time varying current in
a conductor induce EMF in
in that same conductor ?
• Time dependent flux is generated by
change in magnetic field strength due
motion of the magnet
• Note: changing magnetic field can
also be produced by time varying
current in a nearby loop
dI/dt
B
Self-Inductance
• Consider the loop at the right.
• switch closed  current starts to flow in
the loop.
X XX X
X XX XX X
X XX X
• \ magnetic field produced in the area
enclosed by the loop.
• \ flux through loop changes
• \ emf induced in loop opposing initial emf
• Self-Induction: the act of a changing current through a
loop inducing an opposing current in that same loop.
Self-Inductance
• The magnetic field produced by the
current in the loop shown is
proportional to that current.
• The flux, therefore, is also proportional
to the current.
• We define this constant of proportionality
between flux and current to be the
inductance L.
• We can also define the inductance L,
using Faraday's Law, in terms of the
emf induced by a changing current.
I
Self-Inductance
• The inductance of an inductor ( a set of coils in some
geometry, e.g., solenoid, toroid) then, like a capacitor, can be
calculated from its geometry alone if the device is constructed
from conductors and air.
• If extra material (e.g. iron core) is added, then we need to add
some knowledge of materials as we did for capacitors
(dielectrics) and resistors (resistivity)
Self-Inductance
• The inductance of an inductor ( a set of coils in some
geometry ..eg solenoid, toroid) then, like a capacitor, can be
calculated from its geometry alone if the device is constructed
from conductors and air.
• If extra material (eg iron core) is added, then we need to add
some knowledge of materials as we did for capacitors
(dielectrics) and resistors (resistivity)
SI UNITS for L :
Henry
• Archetypal inductor is a long solenoid, just as a pair of parallel
plates is the archetypal capacitor.
l
r
N turns
r << l
A
++++
d -----
Calculation
l
• Long Solenoid:
N turns total, radius r, Length l
r
N turns
For a single turn,
The total flux through solenoid is given by:
Inductance of solenoid can then be calculated as:
This (as for R and C) depends only
on geometry (material)
RL Circuits
• At t=0, the switch is closed and
the current I starts to flow.
a
I
I
R
b
• Loop rule:
e
Note that this eqn is identical in form to that for the RC
circuit with the following substitutions:
 RCRL:
RC:
\

L
Lecture 20, ACT 5
• At t=0 the switch is thrown from position b to
position a in the circuit shown:
1A – What is the value of the current I a long
time after the switch is thrown?
I
a
I
R
b
L
e
R
(a) I = 0
(b) I = e / 2R
(c) I = 2e / R
1B • What is the value of the current I0 immediately after the switch
is thrown?
(a) I0 = 0
(b) I0 = e / 2R
(c) I0 = 2e / R
Lecture 20, ACT 5
• At t=0 the switch is thrown from position b to
position a in the circuit shown:
1A – What is the value of the current I a long
time after the switch is thrown?
I
a
I
R
b
L
e
R
(a) I = 0
(b) I = e / 2R
(c) I = 2e / R
1B • What is the value of the current I0 immediately after the switch
is thrown?
(a) I0 = 0
(b) I0 = e / 2R
(c) I0 = 2e / R
RL Circuits
• To find the current I as a fct of
time t, we need to choose an
exponential solution which
satisfies the boundary
condition:
a
I
I
R
b
e
RL

• We therefore write:
• The voltage drop across the inductor is given by:
L
=R
L
RL Circuit (e on)
e/1
R
Max = e/R
Q
Current
L/R
2L/R
1
2
f( x ) I
0.5
63% Max at t=L/R
00
0
Voltage on L
Max = e/R
t
3
4
t
3
4
x
t/RC
1 e1
f( xV
) 0.5
L
37% Max at t=L/R
0.0183156 0
0
1
2
RL Circuits
• After the switch has been in
position a for a long time,
redefined to be t=0, it is moved to
position b.
• Loop rule:
• The appropriate initial condition is:
• The solution then must
have the form:
a
I
I
R
b
e
L
RL Circuit (e off)
e/R
1 1
Current
L/R
2L/R
1
2
f( x ) 0.5
I
Max = e/R
37% Max at t=L/R
0.0183156
0
0
01 0
Q
Voltage on L
Max = -e
t
3
x
4
4
f( x ) V
0.5L
37% Max at t=L/R
-e
0
0
1
2
t
3
4
e on
L/R
e/1
R
e off
2L/R
e/R
1 1
I
f( x ) 0.5
00
2L/R
1
2
f( x ) 0.5
I
0
1
1 e1
2
t
0.0183156
3
4
01 0
Q
0
0
1
2
0
0
x
t/RC
f( xV
) 0.5
L
3156
L/R
t
3
4
t
3
x
4
4
f( x )V
0.5
L
-e0
0
1
2
t
3
4