Universidad de Puerto Rico
Recinto Universitario de Mayagüez
Departamento de Física
Final Exam – Fisi 3162/3172
Name: ______________________________
Section: ____________
December 16, 2008
Prof. ____________________
Read the instructions carefully. Select the best answer and WRITE THE APPROPRIATE
LETTER INTO THE SQUARE BOX. You are required to answer only 20 of the 25 questions. You have to select and identify the 20 questions to be corrected by circulating the
square box of each one. If you choose not to indicate which questions are the chosen
ones, the first 20 questions will be the ones selected. All answers (except question number 20) must be justified by some analytical procedure. If there is no justification for the
answer then it will be considered a wrong answer. In other words: ‘to guess’ the answer
is not valid. The points assigned per question will be: 5 % to a correct answer with a
correct justification, 2.5 % to an incorrect answer with a correct justification, and 0% to a
correct or incorrect answer with either no justification or an incorrect one.
1-The magnitude of the electric force between the proton and the electron of the hydrogen atom,
separated by a distance of 5.29×10-11 m (rounded to two significant figures) is
a) 2.8×10-12 N
Math. operations:
b) 8.2×10-8 N (*)
F = kqq’/r2 = ke2/r2 =8.2×10-8 N
c) 8.2×10-10 N
d) 2.8×10-8 N
e) None of the above.
2-Certain ion with a positive charge of +2e is approaching in vacuum to a little metallic sphere
with total charge of –5.4 nC. When the ion approaches from 100 cm to 60 cm to the center of the
little sphere, its kinetic energy change is:
a) +173 eV
Math. operations:
b) +65 eV (*)
ΔV =k q(1/rf – 1/ri) = (9×109)(-5.4×10-9)(1/0.6 – 1/1) = –32.4 V
c) +32 eV
ΔK = –ΔU = –q’ΔV = –2e×(–32.4) = +65 eV
d) –32 eV
e) –65 eV
3-An electric charge q is at the corner of a cube with side a. The electric flux through any one of
the faces not containing the charge is: (Hint: Analyze all the space around the charge, surrounded
by all the necessary identical cubes, and use symmetry to obtain the answer)
a) q/εo
Math. operations:
b) q/4εo
Φ =1/(3×8) ∫ E · dA = q/24εo
c) q/6εo
q
d) q/8εo
e) q/24εo
(*)
4-An electric charge q is in the x axis at the position x = a, and another one, −q, is at x = −a. The
electric potential at the origin is:
a) +2kq/a
b) −2kq/a
c) 0
Math. operations:
V = Σkq/r = kq/a – kq/a = 0
(*)
d) +2kq/a2
e) None of the above
5-A conducting sphere of radius 0.20 m has an excess of charge q = +1.0 nC. The electric
potential at the center of the sphere is:
a) 0
Math. operations:
b) +22.5 V
V = kq/R = 45 V
c) +45 V (*)
d) +225 V
e) None of the above
6-Two parallel plates, separated 2.0 mm, have no dielectric in between. When the voltage
between the plates is 80 V, the surface density of charge in the plates is:
a) 0.35 µC/m2
(*)
Math. operations:
b) 3.5×102 µC/m2
σ = Eεo = (V/d)εo = 0.35 µC/m2
c) 2.5 pC/m2
(Or: σ = q/A = CV/A =(εo A/d) (V/A) = (V/d)εo )
d) 2.5×103 pC/m2
e) None of the above
7-A 12.0 V battery is applied to a resistance of 50 Ω during 10 minutes. The number of electrons
arriving to the positive terminal of the battery during that time is:
a) 1.5×1019
Math. operations:
b) 24
c) 9.0×1020
n = q/e ; q = I t ;
(*)
I = E /R
; 10 min = 600 s
n = E t/(Re) = 9.0×1020
d) 1.9×1021
e) 2.3×1024
8-In the circuit of the figure, the charge accumulated at each plate of the capacitor a long time
after closing the switch is:
a) 108 µC
Math. operations:
b) 81 µC
q = CV30 ; V30 = IR30 ; I = E /RT ;
c) 40 µC
RT = 90 + 30 = 120 Ω
d) 27 µC (*) So:
e) 5.3 µC
q = CR30 E /RT = 27 µC
(O: q = CV30 y V30/E = R30/RT )
90 Ω
24.0 V
30 Ω
4.5 µF
9-In the circuit of the figure, the currents on the branches are shown. The potential difference
VA − VB between points A and B is:
a) −18 V
Math. operations:
A
c) 0 V
d) +9.5 V
E3=18.0 V
100 Ω
b) −9.5 V (*) VA + 12 - 200×0.0125 = VB
so: VA – VB = −9.5 V
E1=12.0 V
E2=15.0 V
150 Ω
(Same result for any path)
300 Ω
200 Ω
e) +18.0 V
I2 =77.5 mA
I1 =12.5 mA
I3 =65.0 mA
B
10-In the circuit of the question above (9), the total energy delivered by the E1 battery during half
an hour is:
a) 0.225 J
Math. operations:
b) 13.5 J
Q = P1t = E1I1 t = 270 J
con t = 1800 s
c) 56.3 J
(not: 0.1252(100 + 200) or any similar thing)
d) 84.4 J
e) 270 J
(*)
11-A beam of electrons is moving in vacuum from the paper toward you, through a uniform
magnetic field of 0.60 T directed to the right. The electrons have a velocity of 4.8×106 m/s.
There is a uniform electric field in the same zone, superimposed to the magnetic one, in such a
way that the electrons are no deviated across the zone. The electric vector field at the zone is:
a)8.0 MV/m, downward
Math. operations:
b)8.0 MV/m, upward
evB = eE , so
c)2.9 MV/m, downward (*)
E = vB = 2.9 MV/m,
d)2.9 MV/m, upward
downward
B
B
e
B
B
e)None of the above
E
E
E
E E
12- A 2-m long wire is carrying a current of 2 A. The wire is placed at an angle of 60° with
respect to a magnetic field. If the wire experiences a force of 0.2 N, what is the strength of the
magnetic field?
a) 0.02T
Math. operations:
b) 0.03T
F = I l B senθ ; B = F/(I l senθ) = 0.0577 N ≈ 0.06 T
c) 0.04T
d) 0.05T
e) 0.06T (*)
13- At what distance from a long straight wire carrying a current of 5.0 A is the magnitude of the
magnetic field due to the wire equal to the strength of the Earth's magnetic field of about
5.0×10-5 T?
a) 1.0cm
Math. operations:
b) 2.0cm (*)
B = µoI / (2πr) so r = µoI / (2πB) = 0.020 m = 2.0 cm
c) 3.0cm
d) 4.0cm
e) 5.0cm
14- A rectangular coil of N turns, length l = 25 cm, and width w = 15 cm, as shown in the figure,
is rotating between two magnets, in a magnetic field of 1.6 T with a frequency of 75 Hz. If the
coil develops a sinusoidal emf of maximum value 56.9 V, what is the value of N?
a) 2 (*)
b) 4
c) 6
d) 8
e) 10
Math. operations:
Eo = NABω with A = wl and ω =2πf.
ω
w
So:
N = Eo /(wlB2πf) = 2.01 ≈ 2
N
l
S
15-A solenoid 20 cm length and 4.0 cm diameter has 500 compacted turns, with air inside. A
current of 2.0 A goes through the solenoid and drops to zero during a time interval of 0.015 s.
During such interval, an average emf is induced between the ends of the solenoid, of:
a) 0.26 V, opposed to the existing current
Math. operations:
b) 0.26 V, which tries to sustain the existing current (*)
Eavg = -LΔI/Δt
c) 0 V, there is no induced emf
with L = µoN2A/l
d) 0.79 V, opposed to the existing current
A = πr2 = πd2/4.
e) 079 V, which tries to sustain the existing current
.
.
and
So: Eavg = - µoN2(A/l) ΔI/Δt
= 0.263 V ≈ 0.26 V
16- An alternating current (AC) generator of 200 Hz is connected to an LCR series circuit, where
the inductance is 50 mH, the capacitance is 40 µF and the resistance is 40 Ω. An AC voltmeter
connected at the out put of the generator reads a voltage of 90 V. The same voltmeter connected
in parallel with the capacitor reads:
a) (2.9 ± 0.1)×10-6 V
b) (30 ± 1) V (*)
Math. operations:
Z = √(ωL − 1/ωC)2 + R2 =√(2πfL − 1/2πfC)2 + R2 = √(62.83 – 19.89)2 + 402
c) (65 ± 1) V
Z = 58.68 Ω ;
Irms =Erms/Z = 1.53 A
d) (90 ± 1) V
VC,rms = IrmsXC = 19.89×1.53 = 30.4 V ≈ 30 V
e) (110± 1) V
17- An electromagnetic plane wave goes along the z positive axis. Its electric field vector
oscillates parallel to y axis and it gets a peak of +900 j (V/m) at some instant in a fixed point of
the z axis. At such an instant and in such a point, the magnetic field vector of the wave has the
peak value:
a) +2.7×10-5 i (T)
Math. operations:
b) −3.0×10-6 i (T)
(*) B = E/c = 3.0×10-6 (T)
c) −3.0×10-6 j (T)
in direction −i (T)
y
d) +2.7×10-5 j (T)
x
e) +3.0×10-6 k (T)
z
18-A laser ray goes through the air and comes into a rectangular transparent prism (see figure)
which refraction index is1.30. The ray arrives to the first face of the prism perpendicularly to that
face. The ray emerges by the inclined face with a refraction angle of: (Note: Analyze carefully
the angle of incidence over the hypotenuse)
a) The ray does not emerge by that face because
it suffers total internal reflection on that face.
b) 31o
Math operations:
41o
49o
c) 59o
90o
d) 67o
e) 79o
41o
(*)
Incidence angle on the inclined face: θ1 = 90 – 41 = 49o (It requires to draw the rays and the
normal in order to identify correctly the angle).
Then: n1 senθ1 = n2 senθ2 , so: senθ2 = (n1/n2)senθ1 = (1.30/1.00) sen 49 = 0.981 and
θ2 = 78.8 = 79o
19-A slide with the picture of a rule is placed in a projector to obtain the image of the rule on a
screen. The rule in the slide has a length of 1.4 cm. The lens has a focal length of 4.0 cm and the
slide is placed 4.1 cm in front the lens. The length of the image of the rule on the screen, with the
appropriate sign according to its orientation, is:
a) +6.1×10-3 cm
Math. operations:
b) −2.1×10-3 cm
hI = MhO = −(dI/dO)hO
where dI is obtained from:
c) +35 cm
1/f = 1/dI + 1/do, so dI = fdo/(do – f ) = 164 cm
d) −35 cm
So: hI = −56 cm
e) −56 cm
(*)
20- Suppose an LCR series circuit, in resonance. ¿Which of the next statements is false?
a) Current and applied voltage oscillate in phase.
b) Frequency of the applied voltage is equal to the natural frequency of the LC circuit.
c) Power consumption in the circuit is the maximum possible consumption under the alternating
applied voltage.
d) Peak voltage at the resistance is the maximum possible voltage under the alternating applied
voltage.
e) Peak voltage at the capacitor is the minimum possible voltage under the alternating applied
voltage. (*)
21- According to Bohr’s model, when the electron of the hydrogen atom changes from its fifth
orbital state to its second orbital state, the emitted radiation has a frequency of:
(a) 9.86×1014 Hz
Math. operations:
(b) 8.74×1014 Hz
En = −13.6/n2 (eV) ; hf = ΔE5,1 ; h = 4.14 × 10-15 eV.s
(c) 7.65×1014 Hz
So: hf = –ΔEn and
f = 13.6 (1/22 – 1/52) / (4.14 × 10-15 ) =
(d) 6.90×1014 Hz (*)
= 6.90×1014 Hz
(e) 5.98×1014 Hz
22-A photoelectric cell receives light with three different wavelengths of 480 nm, 540 nm and
620 nm. The work function of the metal used as photo anode is 1.2 eV. The stopping voltage
required to suppress the photo current produced by this cell is:
a) 0.8 V
Math. operations:
b) 1.1 V
hf = Kmax + Wo ,
c) 1.4 V (*)
Using λmin = 480 nm = 480×10-9 m and h = 4.136 eV.s:
d) 2.5 V
Vstop = (hc/λ - Wo) / e = 1.4 V
Kmax = eVstop and f = c/λ
so
eVstop = hc/λ – Wo
e) 3.3 V
23- Two narrow slits are 1.0 mm apart and light of wavelength 544 nm is used to illuminate
them. If a screen is placed 5.0 m from the slits, the distance between adjacent bright fringes is:
(a) 0.10 mm
Math. operations:
(b) 0.82 mm
d senθ = mλ , tanθ = xm/L For small angles: θ = mλ/d and θ = xm/L
(c) 1.2 mm
so mλ/d = xm/L . Incrementing both sides: (Δm) λ/d = (Δxm)/L
(d) 3.5 mm
For Δm = 1: λ/d = (Δx)/L and Δx = Lλ/d = 0.00272 m = 2.7 mm
(e) None of the above. (*)
None of the above
24-Unpolarized light with intensity Io passes through three successive polarizers each of whose
axes makes 45o degree angle with the previous one. The intensity of the transmitted beam is:
(a) 0
Math. operations:
(b) Io/2
First polarizer reduces intensity to Io/2. Next polarizer reduces the intensity by
(c) Io/4
the law: Iout = Iin cos245 = (Io/2) (√2 / 2)2 = Io/4 . The third one does the same:
Iout = Iin cos245 = (Io/4) (√2 / 2)2 = Io/8
(d) Io/8 (*)
(e) Io/16
.
25- Light of wavelength 650 nm illuminates a diffraction grating and the third-order fringe is
observed at an angle of 15.0 degrees. The approximate number of lines per cm in grating is:
(a) 355
Math. operations:
(b) 856
mλ = d senθ so d = mλ/senθ = 7.53×10-6 m = 7.53×10-4 cm
(c) 1330 (*)
So: line density = 1/d = 1327 ≈ 1330
(d) 2750
(e) None of the above.
Constantes:
c = 3.00 × 108 m/s ;
µo = 4π × 10-7 T m / A ; εo = 8.85 × 10-12 C2 /Nm2
k = 1/4πεo = 9.0 × 109 Nm2 /C2 ;
h = 6.63 × 10-34 Js = 4.14 × 10-15 eV.s
hc = 1.99 × 10-25 Jm = 1240 eV-nm
;
Electrón: e = 1.60 × 10-19 C, m = 9.11 × 10-31 kg
;
Conversiones: 1 eV = 1.60 × 10-19 J ; 1 nm = 10-9 m
Ecuaciones:
l = 2πr ; A = πr2
;
A = la
v = vo + at ; v2 = vo2 + 2ad;
F = kqq’/r2 ;
τ=r×F ;
;
F = mv2 /r ;
;
;
V= lah ; V = Ah ;
K = ½ mv2 ;
E = kq/r2 ;
P=W/t;
pe = qd ; E = kp/r3 ;
∫ E · dA = q/εo ;
( )
E=K+U;
ρ = q/V ;
E= λ/(2πεor) ; E = σ/εo;
V = U/q ; V =−∫ E · ds ; V = kq/r ; E = - dV/dl ; V = Ed ;
C = q / V ; E = Wnc/q;
C = κ εo A/d ;
u = ½ εo E2 ; I = dq/dt; I = ∫ J · dA ;
P = IV ;
V = πr2h
F = ma ;
E = F /q ;
σ = q/A ; λ = q/l ; ΦE = ∫ E · dA
E = σ/(2εo) ;
A = ½ bh ;
1/Cs = Σ 1/Ci
;
Cp = Σ Ci ; U = ½ CV2 ;
R = V/I ; R = ρl /A ;
ρ = ρo α ΔT ;
1/Rp = Σ 1/Ri ; Rs = Σ Ri ; Σ Ii = 0 ; Σ Vi + Σ Ei = 0 ; τ = RC ; I = (E/R)e-t/τ ;
q = (EC)[ 1 - e-t/τ] ; F = qvB sen θ ; F = IlB sen θ; µ = NIA ;
dB = µo (I/4π r3)ds×r ;
B = µo nI; B = µo I / 2π r; B = µo Iφ/4π r ; E = Blv ; E = NABω senωt ; E=−L dI/dt ;
L = NΦB /I ; L = N2 µo A / l ; U = ½ LIi2 ; u = B2/(2µo) ; τ = L/R ; I = (E/R)(1 − e-t/τ ) ;
____
ω = 1 / √LC ; ω = 2πf ;
∫ B · dA = 0 ; E = −N dΦB /dt ; ∫ E · ds = - dΦB/dt ;
( )
( )
∫ B · ds = µo ( I + ID ); ID = εo dΦE/dt ; ∫ B · ds = µo εo dΦE/dt ; ω = 2πf ;
______________
XC =1/(ω C); XL =ω L ; Z = √( XL − XC )2 + R2 ; Em = Im Z ; tanφ =(XL − XC )/R ;
__
__
Pavg = Erms Irms cosφ ; Vrms = Vm / √2 ; Irms = Im / √2
; S = (E × B) / µo ; c =λf ;
( )
( )
E = cB ; I = EM2 /(2 µo c) ; I = Ps /(4 π r2 ) ; n1 sen θ1 = n2 sen θ2 ; sen θc = n2 / n1 ;
n = c/v ; I = Io /2 ; I = Io cos2 θ ;
tan θB = n2 / n1 ;
1/f = 1/di + 1/do ;
M = -di /do ;
M = hi /ho ; f = R/2 ; 1/f = (n -1)(1/R1 + 1/R2) ; Δl = mλ (max) ; Δl = (m+ 1/2) λ (min) ;
nλ = d sen θn ; λn = λ/n ; mλ = a sen θm ;
E = hf
; hf = Kmax + Wo ; eVo = Kmax ;
E = pc ; K = p2 /2m ; p =h/λ ; Δλ =(h/mc)(1 – cos φ) ; Δx Δp ≥ h ; hf = Ehigh – Elow ;
En = −[me4 /(8εo2 h2 )] / n2 = − 13.6 (eV)/ n2 ;