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Physics 1202: Lecture 14
Today’s Agenda
• Announcements:
– Lectures posted on:
www.phys.uconn.edu/~rcote/
– HW assignments, solutions etc.
• Homework #4:
– Due this Friday
• Midterm 1:
– Answers this Friday.
Faraday's Law
n
B
B
N
q
S
v
S
B
N
v
FB = B· A = BAcosq
DF B
e =Dt
B
Schematic Diagram of an AC Generator
D
d (cos( wt))
dt
= - w sin( wt)
D (cos( wt))
DF B
= - NAB w sin( wt )
NAB
=
e= -N
Dt
Dt
(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 14, 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 14, 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 14, 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
Induction
Self-Inductance, RL Circuits
XXX
XXXX
XX
1
L/R
e1
V
f( x ) 0.5
L
0.0183156
0
0
1
2
3
4
t
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.
I
• The flux, therefore, is also proportional
to the current.
FB = B· A = BAcosq µI
• We define this constant of proportionality
between flux and current to be the
inductance L.
DF B
• We can also define the inductance L, e = Dt
using Faraday's Law, in terms of the
emf induced by a changing current.
L º-
DI
=L
Dt
e
DI / D t
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)
L º-
e
DI / D t
SI UNITS for L :
Henry
• Archetypal inductor is a long solenoid, just as a pair of parallel
plates is the archetypal capacitor.
r << l
Calculation
• Long Solenoid:
N turns total, radius r, Length l
(n: number of
turns per unit
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)
Energy of an Inductor
• How much energy is stored in
an inductor when a current is
flowing through it?
DI
• Start with loop rule: e = IR+ L
Dt
• Multiply this equation by I:
DI
e I = I R+ LI
Dt
2
•
•
b
e
DU
DI
= LI
Dt
Dt

I
eR= -RI
L
eL= -L DI / Dt
or Ptot = PR + PL
PL is the rate at which energy is
being stored in the inductor:
PL =
I
a
LI
Li
LiDI
DI
DU = LI DI
The total U stored in the inductor when the
current = I is the shaded triangle:

i
I
Where is the Energy Stored?
• Claim: (without proof) energy is stored in the Magnetic field
itself (just as in the Capacitor / Electric field case).
• To calculate this energy density, consider the uniform field
generated by a long solenoid:
l
• The inductance L is:
r
N turns
• Energy U:
• We can turn this into an energy density by dividing by the
volume containing the field:
RL Circuits
• At t=0, the switch is closed and
the current I starts to flow.
a
I
I
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
Review: RC Circuits
(Time-varying currents)
I
a
I
• Charge capacitor:
C initially empty with Q=0
Connect switch to a at t=0.
Calculate current and charge
as function of time.
• Discharge capacitor:
C initially charged with Q=Ce
Connect switch to b at t=0.
Calculate current and charge
as function of time.
b
e
R
+ +
C
- -
Charging
Ce
RC
2RC
Ce
q
0
Discharging
RC
2RC
q
0
t
t
0
e/R
I
I
0
- e/R
t
t
Lecture 14, ACT 4
• 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?
(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 Circuit (e on)
Current
Max = e/R
L/R
e/
R
2L/R
I
63% Max at t=L/R
0
Voltage on L
Max = e/R
t
e
VL
37% Max at t=L/R
0
t
RL Circuit (e off)
Current
Max = e/R
e/R
L/R
2L/R
I
37% Max at t=L/R
0
Voltage on L
Max = -e
t
0
VL
37% Max at t=L/R
-e
t
e on
e/R
L/R
e off
2L/R
e/R
I
L/R
2L/R
I
0
t
t
0
e
VL
VL
0
0
-e
t
t