Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Additional Formula Sheet for Final Exam
Reading and thoroughly familiarizing yourself with this formula sheet is an important part
of, but it is not a substitute for, proper exam preparation. The latter requires, among other
things, that you have re-worked all assigned homework problem sets (PS) and the in-class
quizzes, studied the posted PS solutions, and worked and studied the assigned conceptual
practice (CP) problems, as well as (optionally) some practice test (PT) problems, as posted
on the LON-CAPA homework and on the PHYS1112 examples and homework web pages.
You should consult the syllabus, and in particular review the Class Schedule on the last
syllabus page (posted on the PYS1112 course web site), to find out which topics you should
cover in preparing for this exam.
Induction
~ normal to the
(1) Definition of Magnetic Flux: For a flat surface S with area vector A
~ subjected to a uniform magnetic field B,
~ with angle
surface and finite surface area A ≡ |A|,
~ B)
~ enclosed between A
~ and B,
~ the magnetic flux through S is defined as:
θ ≡ 6 (A,
Φm ≡ AB cos(θ) .
~ which are not uniform across S,
For surfaces S which are not flat and/or magnetic fields B
Φm must be calculated by: (1) breaking up S into smaller, (approximately) flat pieces, with
~ being (appproximately) uniform across each such surface piece; and by then: (2) adding
B
up the flux contributions from all these surface pieces.
(2) Faraday’s Law: Induced voltage (EMF) E in a tightly wound conducting coil of N
identical turns, with each turn enclosing the same time-dependent magnetic flux Φm (t):
∆Φm
∆t
Sign Convention. A positive EMF, E > 0, drives an induced current around the coil (or
would do so if the open ends of the coil were connected into a closed circuit) in right-hand
~
(RH) direction around the A-vector
that was used to define the flux Φm (t). A negative
EMF, E < 0, drives an induced current around the coil against RH direction around the
~
~ is the direction of the RH 4 fingers when the RH
A-vector.
(The RH direction around A
~
thumb points in A-direction.)
E = −N
The sign convention is equivalent to Lenz’s Rule: the induced current produces a magnetic
field which opposes the change of the inducing magnetic field.
(3) Motional EMF: Induced EMF E and electric field E in a conducting straight rod or
wire of length ` moving at speed v perpendicular to the rod/wire, in a uniform magnetic
~ with field component B⊥ perpendicular to the rod/wire:
field B
|E| ≡ E` = |B⊥ |v` .
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Current I flowing through, magnetic breaking force F acting on, and mechanical power input
P into (=heating power output from), a straight rod/wire with motionally induced EMF,
connected into closed circuit of resistance R:
I = E/R =
|B⊥ |v`
,
R
2
B⊥
v`2
,
R
F = I`|B⊥ | =
P = RI 2 = F v =
2 2 2
B⊥
v `
.
R
(4) Generator-Induced EMF: The induced EMF of a coil of N turns, all of same of area
~
A, rotating in uniform B-field
at constant angular speed ω is
E(t) = N ABω sin(ωt)
~ and the rotating area vector A(t),
~
~ normal
where θ(t) = ωt is the angle between B
with A(t)
to the planar surface enclosed by the turns of the coil.
(5) Self-Induction: A time-dependent current I(t) in any wire or coil produces a selfinduced EMF E which opposes the change of the current and is proprtional to the rate of
change of the current, E = −L∆I/∆t ∝ ∆I/∆t, where the inductance
L≡
|E|
|∆I/∆t|
depends only on the shape and size of the wire or coil and is independent of E or ∆I/∆t.
Inductance of a Long, Thin Solenoid (LTS). For a LTS of N turns over length `, with
cross-sectional area A:
N 2A
L = µo
`
(6) Magnetic Field Energy: The energy UB required to build up a current I in an
inductance L, stored as magnetic field energy, mostly inside the inductance coil, is given by
1
UB = LI 2
2
Magnetic Field Energy Density. Energy per volume, uB , stored in a magnetic field is
given in terms of the field strength B
1 2
B
2µo
uB =
(7) Transformer: The primary (P ) and secondary (S) coil time-dependent currents (IP ,
IS ), induced voltages (VP , VS ), and number of turns (NP , NS ), are related by:
VS
NS
=
,
VP
NP
2
IS
NP
=
IP
NS
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Electromagnetic (EM) Waves
(1) EM Field Energy Density:
u = uE + uB =
o 2
1 2
E +
B
2
2µo
(2) EM Wave Amplitudes, Intensity, Power and Speed: E- and B-field amplitudes,
Emax and Bmax , and time-averaged (TA) wave intensity Iav , TA EM wave power Pav transported through area A, and TA EM wave energy density uav are related by
Iav ≡
1 2
1 ∆U
1
o 2
≡ Pav = cuav = c Emax
=c
B
,
A ∆t
A
2
2µo max
1
Bmax = Emax .
c
Here, ∆U is the energy transported through the area A, of a surface oriented perpendicular
to EM wave propagation direction, during time ∆t. The speed of light c, o and µo are
related by
1
c= √
o µo
(3) EM Wave Momentum and Radiation Pressure: The momentum p carried by an
EM wave packet of energy U is
1
p= U
c
The radiation pressure Π and average pressure force Fav exerted by an EM wave of TA
intensity Iav , on an object exposed to cross-sectional area A of the EM wave beam, are
related by
1
1 ∆p
Iav
Π ≡ Fav ≡
= α
A
A ∆t
c
Here, α = 1 for a 100% absorbing and α = 2 for a 100% reflecting object. Also, ∆p is the
EM wave momentum transferred through area A onto the object during time internal ∆t.
(4) EM Wave Polarization: For an unpolarized EM wave beam of TA intensity Io incident
upon a polarization filter, the TA intensity I transmitted through the filter is
1
I = Io
2
Malus’ Law. For a polarized EM wave beam of TA intensity Io and electric field vector
~ o incident upon a polarization filter of transmission axis T , the TA intensity I
amplitude E
~ transmitted through the filter obey
and electric field vector amplitude E
I = Io cos2 (θ) ,
E = Eo | cos(θ)| .
~ o , T ) = 6 (E
~ o , E)
~ is the angle between the incident E
~ o and T with E
~ k T ; and
Here, θ ≡ 6 (E
~ o | and E ≡ |E|.
~
Eo ≡ |E
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Mechanics Memories: Velocity, Acceleration, Force, Energy, Power
(1) Velocity
~v =
∆~r
∆t
if constant; else ~v = lim
~a =
∆~v
∆t
if constant; else ~a = lim
∆~r
∆t→0 ∆t
(2) Acceleration
∆~v
∆t→0 ∆t
(3) Constant-Acceleration Linear Motion: for ∆~r ≡ ~rf − ~ri and ∆~v ≡ ~vf − ~vi
∆~r =
1
(~vi + ~vf ) t ;
2
∆~r = ~vi t +
1
~a t2 ;
2
∆~v = ~a t .
(4) Constant-Speed Circular Motion: for motion at constant speed v ≡ |~v | around a
circular trajectory of radius r.
The velocity vector ~v is always tangential to trajectory and perpendicular to acceleration
vector ~a: ~v ⊥ ~a.
The acceleration vector ~a always points towards the center of the circular trajectory.
Period T and frequency f of revolution, angular velocity ω, and orbital speed v:
T =
1
2πr
2π
=
=
f
v
ω
2π
v
=
T
r
ω = 2πf =
v = ωr = 2πf r =
2πr
T
Circular centripetal acceleration:
v2
= ω2r
r
Orbital angle ∆φ and arc of circumference ∆s covered during time interval ∆t:
a=
∆φ = ω ∆t =
v ∆t
∆s
=
r
r
∆s = v ∆t = ω r ∆t = ∆φ r
(5) Newton’s 2nd Law:
m~a = F~
(6) Kinetic Knergy (KE), Work, Work-KE-Theorem: K=kinetic energy of object of
mass m moving at speed v; W =work done by force F~ on an object moving/moved with
displacement ∆~r, with ∆~r pointing at an angle θ from F~ and 0o ≤ θ ≤ 180o ; ∆K = Kf −Ki =
change of kinetic energy due to work done by total force F~ :
1
K = m v2 ,
2
W = F ∆r cos θ ,
∆K = W .
(7) Energy Conservation Law for ∆K ≡ Kf − Ki and ∆U ≡ Uf − Ui :
Ki + Ui = Kf + Uf
or
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∆K + ∆U = 0
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
(8) Mechanical Power: P =rate of work done by force F~ on an object moving at speed
~v , with ~v pointing at an angle θ from F~ and 0o ≤ θ ≤ 180o :
P = F v cos θ .
Algebra and Trigonometry
az 2 + bz + c = 0
sin θ =
opp
,
hyp
⇒
cos θ =
z=
adj
,
hyp
−b ±
√
b2 − 4ac
2a
tan θ =
opp
sin θ
=
adj
cos θ
sin2 θ + cos2 θ = 1
For very small angles θ (with |θ| 90o ):
sin θ ∼
= tan θ ∼
= θ (in radians)
Numerical Data
Acceleration of gravity (on Earth):
Speed of light in vacuum:
Biot-Savart’s constant:
Permittivity of vacuum:
Permeability of Vacuum:
Electron mass:
Proton mass:
c = 3.00 × 108 m/s
k = 8.99 × 109 Nm2 /C2
Coulomb’s constant:
Elementary charge:
g = 9.81m/s2
km ≡
µo
4π
= 1 × 10−7 Tm/A (exact)
o ≡ 1/(4πk) = 8.85 × 10−12 C2 /Nm2
µo ≡ 4πkm = 4π × 10−7 Tm/A (exact)
e = 1.60 × 10−19 C
me = 9.11 × 10−31 kg
mp = 1.67 × 10−27 kg
Other numerical inputs will be provided with each problem statement.
SI numerical prefixes:
y = yocto =10−24 , z = zepto =10−21 , a = atto =10−18 , f = femto =10−15 , p = pico =10−12 ,
n = nano =10−9 , µ= micro =10−6 , m = milli =10−3 , c = centi =10−2 , d = deci =10−1 ,
da = deca =10+1 , h = hecto =10+2 , k = kilo =10+3 , M = Mega =10+6 , G = Giga =10+9 ,
T = Tera =10+12 , P = Peta =10+15 , E = Exa =10+18 , Z = Zetta =10+21 , Y = Yotta =10+24 .
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