Practice Exam 3
PHY 2140
Solutions
1. A 10.0 F capacitor and a 2.00 H inductor are connected in series with a 60.0 Hz source
whose rms output is 150 V. What is the value of the rms current in the circuit?
a.
b.
c.
d.
e.
0.199 A
205 mA
0.307 A
566 mA
147 mA
Z = XL-XC XL = 2f L = 754 , XC = 1/2fC = 265  so Z = 489 
therefore Irms = Vrms/Z = 0.307 A
2. A step-up transformer is designed to have an output voltage of 5500 V (rms) when the
primary is connected across a 110 V (rms) source. If there are 100 turns on the primary
winding, how many turns are required on the secondary winding?
a.
b.
c.
d.
e.
50
1500
3500
5000
5500
Ratio = 5500V/110V = 50 so we require 50 x 100 turns = 5000 turns on
the secondary.
3. An electromagnetic wave in vacuum has a magnetic field amplitude of 7.00 x 10-7 T. What
is the amplitude of the corresponding electric field?
a.
b.
c.
d.
e.
210 V/m
320 V/m
430 V/m
540 V/m
650 V/m
E/B = c so E = cB thus E = 210 V/m
4. An electron has a speed of v = 0.90c. At what speed will a proton have a momentum equal
to that of the electron?
a.
b.
c.
d.
e.
1.1 x 104 m/s
2.4 x 104 m/s
2.1 x 105 m/s
3.4 x 105 m/s
4.8 x 106 m/s
p = mv and electron = 2.29 so for the electron: p = 5.64 x 10-22 Kg m/s
we desire that pproton = pelectron = p = pmpvp = 5.64 x 10-22 Kg m/s
p e2
v 2p
pe c
so pvp = pe/mp then 2 
solve for vp: v p 
= 3.37x105 m/s
2
2
2
2
2
mp 1  v p / c
pe  m p c
5. In a certain x-ray machine electrons are accelerated through a potential difference of 2 MeV
before striking a metal target. With what speed do the electrons strike the x-ray target?
a.
b.
c.
d.
e.
0.272c
0.642c
0.867c
0.979c
2.79c
KE = 2 MeV, mc2 = 0.511 MeV, KE = (-1) mc2 solve for  
KE
1.
mc 2
Then expand :
  1/ 1  v 2 / c 2 
KE
1
 1 solve for v  c 1 
 0.979c
2
2
mc
 KE

 2  1
 mc

6. A muon formed high in Earth’s atmosphere travels at a speed v = 0.99c for a distance of
4.6 km before it decays. How long does the muon live, as measured in its reference frame?
a.
b.
c.
d.
e.
zero
2.2 s
11 s
15 s
110 s
In earth frame:  = 4.6x103 m/0.99c = 15.5 sec,   1
1  0.992
The proper time is then: p = / = 15.5 sec/7.09 = 2.2 sec
 7.09
7. What is the minimum accelerating voltage required to produce an x-ray with a wavelength
of 0.0300 nm?
a.
b.
c.
d.
e.
6.63 V
13.8 V
19.9 kV
41.4 kV
62.5 MV
min = hc/eV using hc=1.989x10-25 Jm
Then V = hc/emin = 4.14x104 V = 41.4 kV
8. Assuming that the tungsten filament of a lightbulb is a blackbody, what is its peak
wavelength if its temperature is 2900 K?
a.
b.
c.
d.
e.
99.9 pm
99.9 m
999 nm
9.99 mm
99.9 m
max T = 0.2989 x 10-2 mK so max = 0.2989 x 10-2 mK/2900 K = 999 nm
9. A radio wave has a wavelength of 6 m. What is the frequency of this radio wave?
a.
b.
c.
d.
e.
5 Hz
10 kHz
50 MHz
100 MHz
500 MHz
c = f so f = c/ = (3.0x108 m/s) / 6 m = 50 x 106 /s = 50 MHz
10. What is the energy in electron volts of a photon having a wavelength of 5.00 cm?
a.
b.
c.
d.
e.
3.98 x 10-7 eV
2.49 x 10-5 eV
7.96 x 10-3 eV
3.17 eV
66.3 eV
E = hc/ = 3.978x10-24 J, convert to eV by dividing by e gives:
E = hc/e = 2.49x10-5 eV
11. What is the value the inductor which must be inserted in a 60 Hz circuit in series with a
generator of 170 V maximum output voltage to produce an rms current output of 0.75 A?
Z = Vrms/Irms = 160.3 
Vmax = 170 V, so Vrms = 170V2 = 120 V
I = V/XL = 0.75 A as stated in the problem, thus
XL = 2f L = V/I, giving L = (1/2f) Vrms/Irms = 0.425 H
12. The proper length of one spaceship is three times that of another. The two spaceships are
traveling in the same direction and, while both pass directly overhead, an observer on Earth
measures the two spaceships to be the same length. If the slower ship is moving with a
speed of 0.350c, determine the speed of the faster spaceship.
Let’s use f as the subscript for the fast ship and s for the slow ship.
Then the lengths of the ships when measured in their rest frames are
related by: Lp,f = 3 Lp,s. We choose the proper length of the fast ship to
be the larger since it will be length contracted more than the slow ship so
that from earth both can appear the same length, Lf = Ls = L.
Then by the length contraction definition, Lp,s = s L and Lp,f = f L
Eliminating L gives: L = Lp,s/s = Lp,f/f
Then solving for f = (Lp,f / Lp,s) s = 3 s Expanding the ’s we find:
1
3
and finally, solving for vf we find:

v 2f
vs2
1
1
c2
c2
vf  c 1
1
vs2
9
c 2 = 0.95c