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Equation Sheet
Constants
˙
9
Coulomb’s law
k = 8.99·10
1
= 4π
0
Permittivity
0 = 8.85·10−12
Permeability
µ0 = 4π·10−7
Electron charge −e=−1.60·10−19
Speed of light
c = 3.0·108
Fields
point charge
infinitesimal Coulomb
given a potential
E
=
~
dE
=
Ex
=
infinite straight current
B
=
solenoid
B
=
~
dB
=
infinitesimal Biot-Savart
C 2 /N m2
T m/A
C
m/s
.
Q
k 2
r
dQ
k 2 r̂
r
∂V
−
∂x
µ0 I
2π r
KB µ0 In, n = N/l
µ0 I d~l × r̂
4π r2
Maxwell’sI Equations
~ · dA
~ =
(Gauss)
E
I
~ · dA
~ =
B
I
~ · d~l =
(Faraday)
E
I
~ · d~l =
(Ampère)
B
Electric Potential
point charge
infinitesimals
given a field
Forces
q1 q2
Coulomb:
k 2
r
~
Lorentz:
q~v × B
2
mv
Centripetal:
r
N m2 /C 2
Flux
in general
~
uniform X
outflux
influx
q
rR
dV = k dQ
Rb r
~ · d~r
Vab = − a E
~
qE
~
lI~ × B
H
~ · dA
~
= X
~ ·A
~
=X
>0
<0
Qenc /0
0
−
∂ΦB
∂t
(= E)
µ0 Ienc + µ0 0
∂ΦE
∂t
Potential Energy
k
ΦX
ΦX
ΦX
ΦX
˙
point charges
point charge and field
capacitor
inductor
Miscellaneous
Vs
Ns
transformer
=N
Vp
p
~
magnetic moment of current loops µ
~ = N IA
torque
~τ = ~r × F~
~
torque on a magnet
~τ = µ
~ ×B
parallel plate capacitance
C = K0 Ad
RC circuit
Q = Q0 e−t/τ ,
q1 q2
r
qV
1
QV = 21 CV 2 =
2
1
LI 2
2
k
τ = RC
Q2
2C
Circuits
current
capacitance
inductance
(Ohm)
power in resistor
series capapacitors
parallel capacitors
series resistors
parallel resistors
series inductors
parallel inductors
(Kirchhoff) junction
I ≡ ∆Q/∆t
C ≡ Q/V
M21 = N2IΦ1 21
L = N ΦI B
V = IR
P = IVP= I 2 R = V 2 /R P
−1
Ceq
=P i Ci−1
V = i Vi
Q = Qi
P
Ceq = Pi Ci
V = VP
Q = i Qi
i
Req = Pi Ri
V = i Vi
I =PIi
−1
−1
Req =P i Ri
V P
= Vi
I = i Ii
Leq = Pi Li
V = i Vi
I =PIi
−1
=
L
L−1
V
=
V
I
= i Ii
i
eq
i i
The sum of all currents entering the junction must equal the sum
of all currents leaving the junction (charge is conserved).
(Kirchhoff) loop
The sum of the changes in the potential around any closed path of
a circuit must be zero.
AC Circuits
current
I(t) = I0 cos(2πf t)
voltage
V (t) = V0 cos(2πf t + φ)
phase angle
tan φ = (X√
L − XC )/R
√
RMS
Irms =
I
/
2
V
=
V
/
2
0
rms
0
p
2
2
Vrms = Irms Z
impedance
Z = R + (XL − XC )
power factor
cosφ
√
resonance
φ=0
f0 = 1/(2π LC)
2
R
power dissipated P = Irms
R
C
L
V = IR
dV
= CI
dt
V = L dI
dt
displacement current
wave equation
Pointing Vector
Math Review
Vrms = Irms R
Vrms = Irms XC
Vrms = Irms XL
−1
XC = (2πf C)
XL = 2πf L
I, V in phase
V lags I
I lags V
φ=0
φ = −90o
φ = 90o
ID = 0 A dE
dt
v = fλ
~ ×B
~
~= 1E
S
µ0
sin θ = (opp)/(hyp),
cos θ = (adj)/(hyp)
Pythagorean theorem: (hyp)2 = (opp)2 + (adj)2
4
Sphere: V = πr3 ,
A = 4πr2
3
Vectors
~ and components (Ax ,Ay ,...)
(A or A)
~ points opposite A
~
~−B
~ means A
~ + (−B)
~
−A
A
Component = (sign)(magnitude)(trig fn)
~ =A
~+B
~ means Cx = Ax + Bx and Cy = Ay + By
C
q
~ = C
~ = Cx2 + Cy2
Magnitude of C
Angle to nearest x-axis: θ = Tan−1
|Cy |
|Cx |
Specify axis!
tan θ = (opp)/(adj)
~ B
~ = AB cos θ = Ax Bx +Ay By
A·
~ × B|
~ = AB sin θ
|A
Physics 122
Final Exam
Name:
2010/12/17
19:15-22:15
Workshop TA, time and location:
P1.
P2.
P3.
.
P4.
.
.
.
Sum,%
/90,
.
This exam contains a mix of multiple-choice questions, and problems. For each multiple choice question, circle the letter of the one best response. For full credit on problems, it must be clear to the
grader how you got your answer. Partial credit is available if your work is clear. Show all essential
steps. Be sure your answer has proper units, sign and direction when appropriate.
1. Three charges of +6µC are placed at three corners of a square of side length
0.10 m, as shown at right.
(6)
(a) Calculate the magnitude of the electric field at the fourth corner (point A).
Solution: |EA1p
| = 2|EA2 | = |EA3 |
|EA | = |EA2 | + |EA1 |2 + |EA3 |2 =
kq
l2
· [ 21 +
√
2] = 1.03 · 107 N/C
(2)
(b) A positive charge released at point A will drift towards point B.
A. True B. False
(4)
(c) Calculate the electric potential at point A.
Solution: VA = VA1 + VA2 + VA3 =
(6)
k·q
l
· [1 +
√1
2
+ 1] = 1.46 · 106 V
(d) Calculate the work done by an external force to move a +0.1µC charge from point A to
point B.
Solution: VB = 3 ·
kq
r
=3·
−6
8.99·109 ·6·10
√
(0.05/ 2)
= 1.53 · 106 V
work= q(VB − VA ) = 10−7 (1.53 − 1.46)106 = 0.7J
(3) 2. 1 Tesla (T) is equivalent to
A. 1 N · m/A
B. 1 N · A/m
C. 1 N/(A · m)
D. 1 N/(C · m)
3. A straight wire 1.7 m long carries a current at an angle of 30o to a 1.7 T magnetic field, as shown.
The force on the wire is 5.0 N.
(4)
(a) How large is the electric current?
Solution: F = IBl sin θ
I = Bl Fsin θ = 1.7·1.75sin 30o =
(3)
5
0.5·1.72
(b) What is the direction of the force on the wire?
A. ← B. ↑ C. → D. ⊗ E. ↓ F. Page 1
= 3.46A
Physics 122
Final Exam
2010/12/17
19:15-22:15
4. Two long, concentric, thin-walled hollow cylinders have radii of 1cm
and 2cm, as shown in cross-section at right. The smaller cylinder
carries a current of 3A out of the page. The larger cylinder carries
a static charge with linear density of +3µC/m along its length.
(2)
(a) Draw the electric (solid) and magnetic (dashed) field lines.
(1)
(b) What path shape allows the useful application of Ampere’s Law to this geometry?
Solution: Circular path; a circle.
(1)
(c) What surface shape allows the useful application of Gauss’s Law to this geometry?
Solution: Closed cylindrical surface; The surface of a cylinder segment.
(2)
(d) Find the magnetic field 0.5cm to the left of the center.
Solution: Zero; (0); Radial symmetry, a circular path r < 1cm encloses no current.
(2)
(e) Find the magnetic field 1.5cm to the left of the center.
Solution:
H
~ · d~l = µ0 Ienc
B
With r = 0.015m, Ienc = 3A,
(2)
µ0 Ienc
2πr
~ is ↓
Right hand rule: direction of B
⇒ radial symmetry, circular path ⇒
|B| = 4 · 10−5 T
|B| =
(f) Find the electric field 1.5cm to the left of the center.
Solution: Zero; (0); Radial symmetry, cylindrical surface r < 2cm encloses no charge.
(2)
(g) Find the electric field 2.5cm to the left of the center.
H
~ · dA
~ = Qenc /0 ⇒ radial symmetry, cylindrical surface ⇒ |E| = Qenc
Solution: E
A0
λ
6 N/C, directed left ← (out)
;
λ
=
3µC/m;
r
=
0.025m
⇒
|E|
=
2.158
·
10
|E| = 2πr
0
(2)
(h) Find the electric potential difference from radius 2.5cm to radius 25.0cm.
Solution: Vab = −
(2)
Rb
a
λ
~
E(r)
· d~r = − 2π
0
Rb
dr
a r
λ
= − 2π
[log r|ba =
0
λ
2π0
log ab = −124kV
(i) In which regions is it possible for an electron beam to propogate in a straight line out of
the page ()? Explain.
Solution: In two regions.
(1) In the region radius < 1cm, where there are no fields, an electron beam can propogate
out of the page at any speed.
(2) In the region radius > 2cm, where the electric and magnetic fields are perpendicular,
their forces on the beam cancel at appropriate values of speed and radius.
Page 2
Physics 122
Final Exam
2010/12/17
19:15-22:15
(6) 5. A circular loop of wire (lying in the page) of radius 10 cm carries a clockwise current of 6 A.
Find the magnitude (4 points) and direction (2 points) of the magnetic field at the center of the
loop. (Hint: Use the Biot-Savart Law.)
Solution: Each infinitesimal segment of the wire contributes to the magnetic field at the
center of the ring in the same direction, “into the page”, (⊗), so the final direction is also
⊗, and the magnitude of the field there can be found by scalar addition of the infinitesimal
contributions. Since the current is everywhere perpendicular to the displacement, d~l× r̂ = dl.
r is constant. We integrate around the circumference to account for all infinitesimal segments.
~ =
|B|
Z
0
2πr
µ0 I dl
µ0 I
µ0 I
=
· 2πr =
=
4π r2
4πr2
2r
3.77 · 10−5 T
⊗
6. Draw a configuration of three 1Ω resistors with equivalent resistance of
(2)
(a) 3Ω
Solution:
(4)
(b) 1.5Ω
Solution:
(2)
(c) Which of the above circuits will draw more power from a given battery? A.
B.
(4) 7. Which of the following equations does not describe the circuit to the right?
A. 0 = V1 − I1 R1 − I2 R2
B. 0 = V2 − I3 R3 − I2 R2
C. 0 = V1 − V2 − I1 R1 − I3 R3
D. 0 = I1 + I3 − I2
E. All are true for the circuit at right.
(3) 8. If a plane wave of electromagnetic radiation has electric field direction ← and magnetic field
direction , then it is propogating in the following direction:
A. ← B. ↑ C. → D. ⊗ E. ↓ F. (3) 9. A wire loop lies in the page and a magnetic field points out of the page . When the magnetic
field is turned off, the current generated in the loop is
A. clockwise
B. counter-clockwise
C. zero
D. undecidable.
Page 3
Physics 122
Final Exam
2010/12/17
19:15-22:15
10. The circuit to the right is connected to a 60.0V (rms) AC
generator with frequency 15kHz. The resistor is 85 Ω; the
capacitor is 60µF; the inductor is 3.5 mH.
(6)
(a) Find the impedance of the circuit.
Solution: XC = (2πf C)−1 ;
XL = 2πf L
p
Z = R2 + (XL − XC )2 = 340.8Ω
(4)
(b) Find the phase angle.
Solution: cos φ = R/Z
(3)
⇒
φ=
Cos−1 R
Z
=
1.32
(c) At the moment when the current is at its maximum, what is the voltage drop over the
capacitor?
Solution: The voltage drop over the capacitor is out of phase with the current by −90o ,
so that when the current is at maximum (sin 90o = 1), the voltage across the capacitor
is zero (sin(90o − 90o ) = 0).
(3)
(d) At the moment when the current is at its maximum, what is the voltage drop over the
inductor?
Solution: The voltage drop over the inductor is out of phase with the current by +90o ,
so that when the current is at maximum (sin 90o = 1), the voltage across the inductor
is zero (sin(90o + 90o ) = 0).
(6) 11. An amateur radio operator wishes to receive radio signals with wavelength of 5.8m. What value
of inductance should she connect in series with a 250 pF capacitor to tune in these signals?
Solution: The radio operator would want to make a circuit which resonates at the same
frequency as the radio wave. A radio wave with wavelength 5.8 m has a frequency given by
f = c/λ = 3.0·108 (m/s)/5.8m = 51.724 MHz. The condition for resonance is that XL = XC .
Given the capacitance 250pF, she needs an inductor satisfying
XL = XC
2πf L = 2πf1 C
1
−8 H
L = (2πf1)2 C = 4π2 (51.724)
2 ·250 = 3.79 · 10
Extra Credit (up to 4 points): Write (not more than 3) sentences about any of the following
people: A. Ampère; J. Biot; J. Clerk Maxwell; C. de Coulomb; M. Faraday; C. Gauss; H. Hertz;
G. Kirchhoff; H. Lorentz; H. Lenz; J. Poynting; F. Savart; N. Tesla; A. Volta;
More points for more truth and more entertainment.
Page 4