Download Lecture 17 - Blogs @ Baylor University

Exam 2 is coming !



Tues., March 21, 12:30 to 1:55 pm, in this room.
Covering 5 chapters (27-31)
18 multiple-choice questions (just as before)
 15 conceptual/numerical problems, 1 point each
 3 questions are numerical (like homework problems) - 2 pts.
 I will pass out formula5 sheets at the exam. Please familiarize yourself with it.
Any constants needed will be given. (Only the equations up to “AC Circuits” will
come into play.)
 Personalized exams
 I will enter the grade on your Mastering Physics account (“Exam 2”).

Recovery points
 Set 2 on sPH2435-04 opens March 22 (Weds., noon) and closes March 24 (Fri.,
11:59 pm). Use the “old interface”.
 You need a 4-digit CAPA ID to access it. Will get this from on-class exam.
 You must try to recover everything, not just the ones you missed. You must get a
higher grade on the recovery exam to get any points added to your class score.

What can I bring to the exam?
 Pencil
 eraser
 calculator
 That’s all (no cell phones for example)
Phys 2435: Chap. 27-31, Pg 1
Exam 2 coverage





Chapter 27: Magnetic Field and Forces
 Magnets
 Magnetic Force
Chapter 28: Sources of Magnetic Field
 Biot-Savart Law
 Ampere’s law
Chapter 29: Electromagnetic Induction
 Induced EMF and applications
 General form of Faraday’s Law
Chapter 30: Inductance
 Mutual and Self-inductance
 Energy storage and DC Circuits
Chapter 31: Alternating Current
 AC Circuits
 Resonance and Transformers
7 lectures (including
this one)
6 homework sets
6 quizzes
1 exam
About 1/3 of the work
Phys 2435: Chap. 27-31, Pg 2
Chapter 27: Magnetic Field and
Forces


Magnets and Magnetic Fields
Magnetic Force (Lorentz Force)
 force on a moving charge
 force on a current in a wire
 torque on a current loop
 mass spectrometer
Phys 2435: Chap. 27-31, Pg 3
Magnetism

Natural magnets were observed by Greeks more than 2500
years ago in “Magnesia” (northern Greece)
 certain type of stone (lodestone) exert forces on similar
stones
 Small lodestone suspended with a string aligns itself in
a north-south direction due to Earth’s magnetic field!
Direction of Magnetic Field
tail
out of
in to
page
page
Drawing vectors in
head
Phys 2435: Chap. 27-31, Pg 4
The Magnetic Force

What happens if you put a charged particle in a magnetic field?
 it experiences a magnetic force!
force (Lorentz force)


Magnitude depends on

Charge q

Velocity v

Field B

Angle between v and B
Fmagnetic = q v B sinθ
Direction is “sideways”
Fmag
 force is perpendicular to both v and B!

Vector cross product
!
! !
F = qv ! B
v
B
Phys 2435: Chap. 27-31, Pg 5
Direction of the Magnetic Force
!
! !
F = qv ! B

Use the right-hand rule:
 point your fingers along
the direction of velocity
 curl your fingers towards
the magnetic field vector
 your thumb will then
point in the direction of
the force

F
v
B
Reverse direction if it’s a negative charge!
Phys 2435: Chap. 27-31, Pg 6
Radius of Circular Orbit
magnetic force:
x x x x x x x x x x x x
centripetal force:
x x x x x x x x x x x vx B
x x x x x x x x x x x x
q
v
F
F
R
⇒
⇒
This is an important result:
It relates atomic quantities (m
and q) to quantities we can
measure (R, v, B) in a lab.
Phys 2435: Chap. 27-31, Pg 7
Magnetic Force on a Current-Carrying Wire
! !
!
F = Il ! B
The right-hand-rule
Phys 2435: Chap. 27-31, Pg 8
Torque on a Current Loop
• Define magnetic dipole moment
!
"
µ = NIA

Then the torque can be written as
a vector cross product
! ! !
" = µ!B

The potential energy is
! !
U = "µ ! B
Phys 2435: Chap. 27-31, Pg 9
Mass Spectrometer

Velocity selector



qE = qvB
 v = E/B
Only those particles with
v can pass through.
Bending in a magnetic field
qvB' = mv /r
2
m = qBB ' r / E
Phys 2435: Chap. 27-31, Pg 10
Chapter 28: Sources of Magnetic
Field

Biot-Savart Law
 moving charge
 a straight wire
 force between parallel
 current loops

wires
Ampere’s Law
 straight wire
 solenoid
 toroidal solenoid
Phys 2435: Chap. 27-31, Pg 11
Biot-Savart Law - Moving Charge
µ 0 qv sin !
B=
2
4" r
!
! µ 0 qv ! rˆ
B=
2
4" r
µ0=4πx10-7 T.m/A is
called permeability of free
space
where rˆ is the unit vector from the source point
to the field point.
Phys 2435: Chap. 27-31, Pg 12
Biot-Savart Law - Curent Segment

Question: how to find B field (both direction and
magnitude) due to a current segment ?
!
! µ 0 qv ! rˆ
B=
2
4" r
!
! µ 0 Idl ! rˆ
dB =
2
4" r
The total magnetic field
is an integral over the
entire wire:
!
! µ 0 Idl " rˆ
B=
2
!
4#
r
Phys 2435: Chap. 27-31, Pg 13
Example: straight wire

Set up the coordinate system
as shown. The field at point P
due to a small segment:
µ0 I dy sin φ
dB =
2
π
4
r

The direction is the same from
any segment, so the total is
µ0 I
B=
4π
+a
xdy
x2 + y 2
∫(
−a
)
3/ 2
µ0 I
2a
=
2π x x 2 + a 2
For a long wire (a ∞):
µ0 I
B=
2!x
Phys 2435: Chap. 27-31, Pg 14
Force Between Two Parallel Current-Carrying Wires



Field at wire 2
due to wire 1:
µ0 I 1
B1 =
2πd
Force on
µ0 I 1 I 2
F2 = I 2 LB1 =
L
wire 2:
2πd
Force per unit length:
F2 = µ 0 I 1 I 2
L
2πd
Phys 2435: Chap. 27-31, Pg 15
Example: a current loop

Set up the coordinate
system as shown. For
a point on the axis
µ0 I dl
dB =
2
π
4 r

By symmetry, total B⊥
is zero, so total B=B||
µ0 I a 2
a µ0 I a
=
B = ∫ dBx = ∫ dB cos θ = ∫ dB =
dl
∫
r 4π r 3
2 r3
Phys 2435: Chap. 27-31, Pg 16
Ampere’s Law

Question: Is there a general relation between a
current in a wire of any shape and the magnetic
field around it?
! !
B
"
d
l
=
µ
I
0
enclosed
!

Ampere’s Law:
The line integral of the magnetic
field around any closed loop is
equal to µ0 times the total
current enclosed by the loop
Phys 2435: Chap. 27-31, Pg 17
Example: a long straight wire

Consider a circular path of radius r around the wire.
 The plane of the path is perpendicular to the wire.

By symmetry, the B field has
the same magnitude at every
point along the path, with a
direction tangential to the
circle by the right-hand-rule.
 
µ 0 I = ∫ B ⋅ dl = B ∫ dl =B (2πr )
Phys 2435: Chap. 27-31, Pg 18
Example: Solenoid


Consider a rectangular path as shown
By symmetry, the only
non-zero contribution
comes from the
segment cd:
  d  
∫ B ⋅ dl = ∫ B ⋅ dl = Bl
c
Bl = µ 0 NI
Where n = N / l is the number loops per unit length
Phys 2435: Chap. 27-31, Pg 19
Example: Toroidal Solenoid
Inside:

consider path 1
B(2πr ) = µ 0 NI

Outside: consider path 2
 The net current passing through is zero
B(2πr ) = 0
Phys 2435: Chap. 27-31, Pg 20
Chapter 29: Electromagnetic
Induction


Magnetic Flux
Induced EMF
 Faraday’s law
 Lenz’s law


Motional EMF
Applications of Induction
 generators
 motors



Counter EMF
Faraday Law (general form)
Displacement Current and
Maxwell Equations
Phys 2435: Chap. 27-31, Pg 21
Magnetic Flux

Consider the B field lines
that pass through a surface
 define a quantity called
the magnetic flux Φ
Φ ≡ B A cosθ
where θ is angle between magnetic field B
and the normal to the plane.


units of magnetic flux are T.m2 = Weber (Wb)
Scalar product
Phys 2435: Chap. 27-31, Pg 22
Faraday’s Law of Induction
Lenz’s Law
ε = −N ΔΦ
Δ
B
t
induced
emf

rate of change
of flux with time
minus sign comes from Lenz’s Law:
 The induced emf gives rise to a current whose
magnetic field opposes the original change in flux.
Phys 2435: Chap. 27-31, Pg 23
Motional EMF

Consider a conducting rod moving on metal rails in
an uniform magnetic field:
ε

ΔΦ B Δ (BA ) Δ (BLx )
Δx
=
=
=
= BL
Δt
Δt
Δt
Δt
Current will flow counter-clockwise in this “circuit”
x x x x x x x x x
x x x x x x x x x
x x x x x x x x x
x x x x x x x x x
v
L
ε = BLv
x
Phys 2435: Chap. 27-31, Pg 24
Electric Generators

Flux is changing in a sinusoidal manner:
 Φ = B A cos θ = B A cos (ω t)
 this leads to an alternating emf (AC generator)
Φ
ε = N d
dt

B
= NBA
d cos( ω t )
= NBA ω sin( ω t )
dt
This is how most of our electricity is generated !!
 water or steam turns blades of a turbine which rotates
a loop
Phys 2435: Chap. 27-31, Pg 25
Motors
Generators
AC current + B field → rotation rotation + B field → AC current
electrical ⇒ mechanical energy
mechanical ⇒ electrical energy
Phys 2435: Chap. 27-31, Pg 26
Counter EMF in a motor

The armture windings of dc motor have a resistance of 5.0 Ω. The motor
is connected to a 120-V line, and when the motor reaches full speed
against its normal load, the counter emf is 108 V. Calculate
 the current into the motor when it is just starting up
 the current when it reaches full speed.
Phys 2435: Chap. 27-31, Pg 27
Faraday’s Law (general form)
d! B
# ="
dt
d" B
E
$
dl
=
#
!
dt

# = ! E " dl
The integral is taken around
the loop through which the
magnetic flux is changing.
A changing magnetic flux produces an electric field.
It’s a non-conservative field, because
$ = # E " dl ! 0
Phys 2435: Chap. 27-31, Pg 28
Maxwell’s Displacement Current

Can we understand why we must have a “displacement current”?
• Consider applying Ampere’s Law
to the current shown in the
diagram.
• If the surface is chosen as 1,
2 or 4, the enclosed current = I
circuit
• If the surface is chosen as 3,
the enclosed current = 0! (ie
there is no current between
the plates of the capacitor)
Big Idea: The added term is non-zero in this case, since the current I
causes the charge Q on the capacitor to change in time which causes
the Electric field in the region between the plates to change in time. The
“displacement current” ID = ε0 (dφE/dt) in the region between the plates =
the real current I in the wire.
Phys 2435: Chap. 27-31, Pg 29
Maxwell’s Displacement Current

In order to have
for surface 2 to be equal to
for surface 3, we want
the displacement current in the region between the plates to be equal to the
current in the wire.

The Electric Field E between the plates of the capacitor is determined by the
charge Q on the plate of area A: E = Q/(Aε0)
What we want is a term that relates E to I without involving A. The answer: the
time derivative of the electric flux!!

Recall flux:
Therefore, if we want ID = I, we need to identify:
Phys 2435: Chap. 27-31, Pg 30
Maxwell’s Equations
! ! Q
# E " dA = ! 0
Gauss’s law for electric field:
electric charges produce
electric fields.
! !
B
!
d
A
=
0
"
! !
d! B
$ E # dl = " dt
J. C. Maxwell (1831 - 1879)
Gauss’s law for magnetic field:
but there’re no magnetic charges.
Faraday’s law: changing B produces E.
Ampere’s law as modified by Maxwell:
! !
d! E
current or changing E
# B " dl = µ0 I + µ0$ 0 dt electric
produces B.
All of electromagnetism is contained in this set of four equations.
Phys 2435: Chap. 27-31, Pg 31
Chapter 30: Inductance






Mutual Inductance
Self-inductance
Energy Stored in a Magnetic Field
LR Circuits (DC)
LC Circuits (DC)
LRC Circuits (DC)
Phys 2435: Chap. 27-31, Pg 32
Mutual Inductance
The magnetic flux in coil 2 created by coil
1 is proportional to I1. Define
N 2 ! 21
M 21 =
I1
M21 is called the mutual inductance. It depends only on the
geometric factors, NOT on the currents.
Faraday’s law:
The reverse situation is
dI 2
" 1 = ! M 12
dt
dI1
" 2 = ! M 21
dt
M 12 = M 21 = M
The SI unit for M is henry (H). 1 H = 1 V.s/A = 1 Ω.s
Phys 2435: Chap. 27-31, Pg 33
Example: Solenoid and coil
The magnetic field inside the solenoid is
N1
B = µ0
I1
l
The magnetic flux through the coil is
N1
! 21 = BA = µ 0
I1 A
l
Hence the mutual inductance is
N 2 ! 21
M=
I1
N1 N 2 A
M = µ0
l
Phys 2435: Chap. 27-31, Pg 34
Self-Inductance
The total magnetic flux in the coil is
proportional to the current I. Define
N! B
L=
I
L is called self-inductance. It depends only on the geometric
factors, NOT on the current. Such coil is called an inductor.
Use Faraday’s law:
dI
" = !L
dt
The SI unit for L is also henry (H). 1 H =1 Ω.s.
Phys 2435: Chap. 27-31, Pg 35
Example: Solenoid inductance
The magnetic field inside the solenoid is
N
B = µ0 I
l
The total magnetic flux through the coil is
N
! B = BA = µ 0 IA
l
Hence the self-inductance is
N! B
L=
or
I
N2A
L = µ0
l
For N=100, l=5 cm, A=0.3 cm2, L=4π
x10-7x1002x0.3x10-4/0.05=7.5 µH.
If filled with an iron core (µ=4000 µ0 ), L= 30 mH.
Phys 2435: Chap. 27-31, Pg 36
Voltage across an inductor

Inductor does not oppose current that flows
through it. It opposes the change in the current.
 It’s a current stabilizer in the circuit.
Phys 2435: Chap. 27-31, Pg 37
Energy Stored in a Magnetic Field
When an inductor is carrying a current
which is changing at a rate dI/dt, the
energy is being supplied to the inductor at
a rate
dI
P = I! = LI
dt
The work needed to increase the current
from 0 to I is
I
1 2
W = ! Pdt = ! LIdI = LI
2
0
By energy conservation, the energy
stored in the inductor is
1 2
U = LI
2
Phys 2435: Chap. 27-31, Pg 38
Energy Stored in a Magnetic Field
Question: Where exactly does the energy
reside?
1 2
U=
2
LI
Answer: It resides in the magnetic field.
N2A
Using L = µ 0
l
N
and B = µ 0 I
l
2
1 & N A #& Bl #
1 B2
!! =
!!$$
U = $$ µ 0
Al
2%
l "% µ 0 N "
2 µ0
2
Or energy density
The conclusion is valid for any region of
space where a magnetic field exists.
Compare with the electric case:
1
U = CV 2
2
1 B2
u=
2 µ0
1
u = !0E2
2
Phys 2435: Chap. 27-31, Pg 39
LR Circuits
dI
loop rule : V0 ! IR ! L = 0
dt
Solve differential equation:
I (t ) =
V0
R
(1 " e )
" t /!
Where τ=L/R is called the
time constant.
Phys 2435: Chap. 27-31, Pg 40
LR Circuits
dI
loop rule : ! IR ! L = 0
dt
Solve differential equation
I (t ) =
V0
R
e
" t /!
time constant τ=L/R
Summary: there is always some reaction
time when a LR circuit is turned on or off.
The situation is similar to RC circuits,
except here the time constant is
proportional to 1/R, not R.
Phys 2435: Chap. 27-31, Pg 41
LC Circuits
The capacitor is charged to Q0.
At t=0, the circuit is closed.
What will happen?
Q
dI
loop rule : ! L = 0
C
dt
Using I=-dQ/dt, one gets
d 2Q Q
+
=0
2
dt
LC
Solve differential equation
where
Q(t ) = Q0 cos("t + ! )
! = 1 / LC
Phys 2435: Chap. 27-31, Pg 42
LC Circuits
Q(t ) = Q0 cos("t + ! )
1 Q 2 Q02
UE =
=
cos 2 ("t + ! )
2 C 2C
1 2 Q02
U B = LI =
sin 2 ("t + ! )
2
2C
dQ
I (t ) =
= #Q0" sin ("t + ! )
dt
Charge oscillates! So does the
current and voltage.
What about energy?
The total energy is conserved.
Phys 2435: Chap. 27-31, Pg 43
LRC Circuits
Damped oscillations! 3 scenarios.
A) Under-damped if R2<4L/C.
B) Critical damping if R2=4L/C.
C) Over-damped if R2>4L/C.
For under-damping:
Q(t ) = Q0 e
T=
$
R
t
2L
+ -
cos(" #t + ! )
2#
2# LC
=
$"
R 2C
1!
4L
Phys 2435: Chap. 27-31, Pg 44
Summary of Various Direct Current Circuits
RC circuit, time constant τ = RC
(transient)
LR circuit, time constant τ = L/R
(transient)
LC circuit, oscillation period
T = 2π LC
(oscillator)
2
R
C
=
π
−
LRC circuit, damped oscillation period T 2 LC / 1
4L
(damped oscillator)
Phys 2435: Chap. 27-31, Pg 45
Chapter 31: Alternating Current

AC Circuits (using phasors)






AC Circuit, R only
AC Circuit, L only
AC Circuit, C only
AC Circuit with LRC
Resonance
Transformers
Phys 2435: Chap. 27-31, Pg 46
RMS current and voltage
i = I cos !t
T
root-mean-square:
2 T
I rms = i
2
2
1 2
I
I
i = ! i dt =
(1 + cos 2"t )dt =
!
T 0
2T 0
2
2
I rms
I
=
2
Vrms
V
=
2
Phys 2435: Chap. 27-31, Pg 47
AC Circuit Containing only Resistance
The AC source is given as
So the voltage across
the resistor is
i = I cos !t
vR = iR = IR cos !t = VR cos !t
The resistor voltage is in phase with the current.
The power dissipated in the resistor
is p=iv, or at an average rate
2
V
2
P = I rms
R = rms
R
Phys 2435: Chap. 27-31, Pg 48
AC Circuit Containing only Inductor
The AC source is given
i = I cos !t
di
So the voltage across
vL = L = ! IL" sin "t = VL cos("t + 900 )
the inductor is
dt
The inductor voltage leads the current by 90 degrees.
Define reactance of inductor XL=ωL, then VL = I XL.
Its unit is Ohm.
Phys 2435: Chap. 27-31, Pg 49
AC Circuit Containing only Capacitor
i = I cos !t
The voltage across the capacitor is
t
q = " idt =
0
I
I
sin #t = cos(#t ! 900 )
#
#
v = q/C
I
0
vC =
cos("t ! 90 )
"C
The capacitor voltage lags the current by 90 degrees.
Define reactance of capacitor XC=1/ωC, then VC = I XC. Its unit
is Ohm.
Phys 2435: Chap. 27-31, Pg 50
AC-driven LRC Circuit in Series
(Graphical Method)
The goal is to cast the total
voltage at any time into the form i = I cos !t
v = vR + vL + vC = V cos("t + ! )
where φ is the phase angle with
which v leads i.
From the
triangle:
V = VR2 + (VL ! VC ) 2
= ( IR) 2 + ( IX L ! IX C ) 2
= I R 2 + ( X L ! X C )2
The impedance
The phase difference
V = IZ with Z = R 2 + ( X L ! X C ) 2
X L ! XC
R
tan " =
or cos" =
R
Z
Phys 2435: Chap. 27-31, Pg 51
The Impedance Triangle
Z = R 2 + ( X L ! X C )2
X L = #L
1
XC =
#C
X L ! XC
R
tan " =
or cos" =
R
Z
General results, valid for any
combination of L, R, C in series.
For example, if LR circuit, set XC=0.
Check special cases:
R only : X L = 0, X C = 0, so Z = R, ! = 0
L only : R = 0, X C = 0, so Z = X L , ! = +90
0
C only : R = 0, X L = 0, so Z = X C , " = !90
0
Phys 2435: Chap. 27-31, Pg 52
Frequency Dependence
i = I cos #t
v = V cos(#t + $ )
V = IZ
Z = R 2 + ( X L ! X C )2
X L = #L
1
#C
X L ! XC
R
tan " =
or cos" =
R
Z
XC =
Phys 2435: Chap. 27-31, Pg 53
The Power Factor
The power is only dissipated in the resistor, so
the average power is
2
P = I rms R
But
P=I
2
rms
R = Z cos !
1
Z cos ! = I rmsVrms cos ! = IV cos !
2
The factor cosφ is called the power factor.
For example:
For a pure resistor (φ=0) , cosφ=1.
For a pure inductor (φ=900) or capacitor (φ=-900), cosφ=0.
Phys 2435: Chap. 27-31, Pg 54
Resonance in AC Circuit
Recall:

V
V
I= =
2
2
Z
(
)
+
−
R
X L XC
For fixed R,C,L the current I will be a maximum at the resonant frequency ω0
which makes the impedance Z purely resistive.
XL = XC
Z is minimum when:
So the frequency at which this condition is satisfied is given
from:
ωoL =
1
ω oC
⇒
ωo =
1
LC
•
Note that this resonant frequency is identical to the natural
frequency of the LC circuit by itself!
•
At this frequency, the current and the driving voltage are in
phase!
R
φ
=
= 1, so φ = 0
cos
Phys 2435: Chap. 27-31, Pg 55
Z
Resonance in AC Circuit
At resonance, I and V are in phase. V, I
and P are at their maximum.
(V=100 V, L=2.0 H, C=0.5 µF)
Phys 2435: Chap. 27-31, Pg 56
Transformers
Transformers change alternating (AC) voltage to a bigger or
smaller value
Changing flux
in secondary
induces emf Vs
Input AC voltage Vp
in the primary
produces a flux
Same ΔΦ /Δt !!
Transformer equation:
Ns
Vs = Vp
Np
Phys 2435: Chap. 27-31, Pg 57
Transformers

Nothing comes for free, however!
 voltage increase comes at the cost of current
 output power cannot exceed input power
 power in = power out (assume no heat loss)
I pVp = I sVs
If voltage increases, then current decreases
If voltage decreases, then current increases
Phys 2435: Chap. 27-31, Pg 58