### 2014 April Final Exam Solutions

```BROCK UNIVERSITY
Final Exam: April 2014
Course: PHYS 1P22/1P92
Examination date: 17 April 2014
Time of Examination: 9:00–12:00
Number of
Number of
Number of
Instructor:
pages: 11 (+ formula sheet)
students: 134
hours: 3
S. D’Agostino
A formula sheet is attached at the end of the test paper. No other aids are permitted except
for a non-programmable, non-graphing calculator.
Solve all problems in the space provided.
Total number of marks: 50
SOLUTIONS
1. [4 marks] Determine the magnitude and direction of the electric field 1.0 mm away
from (a) a proton, and (b) an electron.
Solution: (a) The magnitude of the electric field due to the proton is
KQ
r2
(8.99 × 109 N.m2 /C2 )(1.6 × 10−19 C)
E=
(1.0 × 10−3 m)2
E = 1.44 × 10−3 N/C
E=
The electric field is directed radially away from the proton.
(b) Because the magnitude of the charge on the electron is the same as the magnitude
of the charge on the proton, the magnitude of the electric field due to the electron is
the same as the magnitude of the electric field due to the proton, at equal distances.
The electric field due to the electron is directed radially towards the electron.
2. [4 marks] The square loop in the figure has a resistance of 0.10 Ω and has an induced
current as shown.
(a) Is the magnetic field strength increasing or decreasing? Explain.
(b) Determine the rate of change of the magnetic field, ∆B/∆t.
Solution: (a) The orientation and size of the loop do not change, so the induced
current must be due to a change in magnitude of the magnetic field. (The unspoken
assumption is that the magnetic field is uniform and does not change direction.) Thus,
there are only two possibilities; either the magnetic field increases in magnitude or
decreases in magnitude. Apply Lenz’s law to each possibilitiy and you will determine
that (given the direction of current flow) the magnetic field must be increasing in
magnitude. (The induced current produces a magnetic field that opposes the change
in flux, and so the induced magnetic field is opposite to the applied magnetic field.)
(b) Given the current and the resistance of the loop, we can determine the induced
emf in the loop using Ohm’s law:
E
E
E
E
= IR
= (150 mA)(0.10 Ω)
= 15 mV
= 0.015 V
∆B ∆Φ = A
E = ∆t ∆t Thus, we can calculate the rate of change of the magnetic field as follows:
∆B E
∆t = A
∆B = 0.015 V
∆t (0.08 m)2
∆B ∆t = 2.3 T/s
3. [6 marks] The allowed energies of a quantum system are 0.0 eV, 1.5 eV, 3.0 eV, and
6.0 eV. Determine the wavelengths that occur in the system’s emission spectrum.
Solution: First label the energy levels in order:
E1 = 0.0 eV,
E2 = 1.5 eV,
E3 = 3.0 eV,
E4 = 6.0 eV
Now determine the transitions, and the energy difference for each transition:
Transition
4 −→ 1
4 −→ 2
4 −→ 3
3 −→ 1
3 −→ 2
2 −→ 1
Energy Difference (eV)
E4 − E1 = 6.0 − 0.0 = 6.0 eV
E4 − E2 = 6.0 − 1.5 = 4.5 eV
E4 − E3 = 6.0 − 3.0 = 3.0 eV
E3 − E1 = 3.0 − 0.0 = 3.0 eV
E3 − E2 = 3.0 − 1.5 = 1.5 eV
E2 − E1 = 1.5 − 0.0 = 1.5 eV
Notice that there are only 4 distinct transition energies. The possible wavelengths of
an emitted photon can be calculated using the usual formula (see the lecture notes or
the textbook):
hc
E=
λ
Solving the previous equation for the wavelength, we obtain
λ=
hc
E
Thus,
hc
(6.63 × 10−34 J.s)(3 × 108 m/s)
=
= 207 nm
E41
(6.0 eV)(1.6 × 10−19 J/eV)
Repeating the same calculation for all the other transition energies, we obtain:
λ41 =
λ42 =
hc
(6.63 × 10−34 J.s)(3 × 108 m/s)
=
= 276 nm
E42
(4.5 eV)(1.6 × 10−19 J/eV)
λ43 = λ31 =
(6.63 × 10−34 J.s)(3 × 108 m/s)
hc
=
= 414 nm
E43
(3.0 eV)(1.6 × 10−19 J/eV)
λ32 = λ21 =
hc
(6.63 × 10−34 J.s)(3 × 108 m/s)
= 829 nm
=
E32
(1.5 eV)(1.6 × 10−19 J/eV)
4. [5 marks] The allowed energies of a quantum system are 0.0 eV, 2.0 eV, and 6.0 eV. An
electron travelling at a speed of 1.6×106 m/s collisionally excites the system. Determine
the minimum and maximum speeds the electron could have after the collision.
Solution: The initial kinetic energy of the electron is
1
K = mv 2
2
1
K = (9.11 × 10−31 kg)(1.6 × 106 m/s)2
2
K = 1.166 × 10−18 J
1.166 × 10−18 J
K=
1.6 × 10−19 J/eV
K = 7.288 eV
In exciting the quantum system, the incoming electron will lose at least 2.0 eV and at
most 6.0 eV. Thus, after the collision, the electron will have a minimum kinetic energy
of
Kmin = 7.288 − 6.0 = 1.288 eV
and a maximum kinetic energy of
Kmin = 7.288 − 2.0 = 5.288 eV
The resulting possible speeds of the electron after the collision are
r
2Kmin
vmin =
s m
vmin =
2(1.288 eV)(1.6 × 10−19 J/eV)
9.11 × 10−31 kg
vmin = 6.73 × 105 m/s
and
r
vmax =
vmax =
2Kmax
s m
2(5.288 eV)(1.6 × 10−19 J/eV)
9.11 × 10−31 kg
vmax = 1.36 × 106 m/s
5. [6 marks] In the circuit in the figure, determine V1 − V4 , V2 − V4 , and V3 − V4 .
Solution: First determine the effective resistance of the circuit; after a little work,
you’ll find that the effective resistance is 10 Ω. From this, the current flowing from the
battery can be determined using Ohm’s law:
V
10 V
=
= 1.0 A
R
10Ω
According to our usual convention, we’ll choose the “low” side of the battery to be the
zero level for potential; thus, V4 = 0 V. The “high” side of the battery, which we can
label as position 0, therefore has potential 10 V.
The current through the resistor between the battery and position 1 is 1 A, and so the
voltage drop across this resistor is
V0 − V1 = (1 A)(5 Ω) = 5 V
Thus,
V1 = V0 − 5 V = 10 V − 5 V = 5 V
Using Ohm’s law, the current flowing downward from position 1 can be determined:
V1 − V4
5V−0V
=
= 0.5 A
10 Ω
10 Ω
Applying Kirchhoff’s junction law at position 1, we can conclude that the current
flowing between points 1 and 2 is
1 A − 0.5 A = 0.5 A
Thus, the potential at position 2 is
V2 = V1 − (0.5 A)(5 Ω) = 5 V − 2.5 V = 2.5 V
The current flowing downward from point 2 is
2.5 V − 0 V
V2 − V4
=
= 0.25 A
10 Ω
10 Ω
Applying Kirchhoff’s junction law at position 2, we can conclude that the current
flowing between points 2 and 3 is
0.5 A − 0.25 A = 0.25 A
Thus, the potential at position 3 is
V3 = V2 − (0.25 A)(5 Ω) = 2.5 V − 1.25 V = 1.25 V
Thus,
V1 − V4 = 5 V
V2 − V4 = 2.5 V
V3 − V4 = 1.25 V
6. [5 marks] Determine the change in the electrical potential energy of a 3.0 nC point
charge when it is moved from point A to point B in the figure.
Solution: The change in electrical potential energy in moving the 3.0 nC charge from
point A to point B is
KQ1 Q2 KQ1 Q2
−
rB
rA
1
1
UB − UA = KQ1 Q2
−
rB rA
UB − UA =
9
2
2
−9
UB − UA = (8.99 × 10 N.m /C )(25.0 × 10
−9
C)(3.0 × 10
C)
1
1
−
0.015 m 0.05 m
UB − UA = 3.15 × 10−5 J
7. [6 marks] An electron beam of diameter 0.40 mm and carrying a current of 50 µA
strikes a screen.
(a) Determine the number of electrons that strike the screen each second.
(b) Determine the electric field strength needed to accelerate the electrons from rest
to a final speed of 4.0 × 107 m/s over a distance of 5.0 mm.
(c) Each electron transfers its kinetic energy to the screen upon impact. Determine
the power delivered by the electron beam to the screen.
Solution: (a)
∆Q
∆t
∆Q = I∆t
∆Q = (50 × 10−6 C/s)(1 s)
∆Q = 50 × 10−6 C
I=
Therefore, the number of electrons that strike the screen each second is
50 × 10−6 C
= 3.125 × 1014 electrons
1.6 × 10−19 C/electron)
(b) The potential difference experienced by each electron as it is accelerated is:
∆U = Q∆V
∆U
∆V =
Q
1
mv 2
∆V = 2
Q
1
(9.11 × 10−31 kg)(4 × 107 m/s)2
2
∆V =
1.6 × 10−19 C
∆V = 4555 V
Thus, the magnitude of the electric field is
∆V
d
4555 V
E=
5 × 10−3 m
E = 9.11 × 105 V/m
E=
(c) The power delivered by the electron beam to the screen is
energy/s = energy/electron × electrons/s
1
∆Q
P = mv 2 ×
2
t
1
3.125 × 1014
P = (9.11 × 10−31 kg)(4 × 107 m/s)2 ×
2
1s
P = 0.228 W
8. [4 marks] Electrons in a uniform magnetic field travel in circular orbits with a frequency of 2.4 × 109 Hz. Determine the strength of the magnetic field.
Solution: Because the electrons travel in a circular orbit at a constant speed, we
can write Newton’s second law of motion and use the expression for the centripetal
acceleration:
F = ma
F =m·
v2
r
The force is caused by the uniform magnetic field, and so F = qvB. Inserting this
expression into the equation above, we can obtain an expression for the magnetic field
strength:
v2
r
mv 2
B=
qvr
mv
B=
qr
qvB = m
We don’t know the speed of the electrons and we don’t know the radius of their orbit
either. However, there must be a connection between these quantities and the frequency
of the electron motions around the circle. You can obtain this connection by realizing
that for motion at a constant speed, distance = speed × time. Using T for the period
of the motion (i.e., the time needed for an electron to go once around the circle), we
have
2πr = vT
2πr = v ·
2πf =
1
f
v
r
Using the equation in the previous line, we can substitute for v/r in the equation for
the magnetic field strength:
mv
qr
m v
=
·
q r
m
=
· 2πf
q
2π(2.4 × 109 Hz)(9.11 × 10−31 kg)
=
1.6 × 10−19 C
= 86 mT
B=
B
B
B
B
9. [10 marks] Circle the best response in each case.
(a) Two strings of different linear density are joined together and pulled taut. A
sinusoidal wave on these strings is travelling to the right, as shown in the figure.
When the wave crosses the boundary from String 1 to String 2, the frequency
does not change and the speed of the wave
i.
ii.
iii.
iv.
increases.
decreases.
does not change.
[The question can’t be answered because we need to know the values of the
linear densities of the strings.]
v. [The question can’t be answered because we need to know the values of the
tensions in the strings.]
Solution: (ii)
Remember that v = f λ for a wave. Thus, if the frequency stays the same but the
wavelength decreases, the speed also decreases.
(b) Electric equipotential surfaces
i. are close together where the electric field is strong, and far apart where the
electric field is weak.
ii. point away from positive charges and towards negative charges.
iii. are such that electric field vectors are tangent to them.
iv. are such that electric potential vectors are tangent to them.
v. [All of the above are true.]
Solution: (i)
Remember, electric potential is a scalar quantity, not a vector quantity.
(c) When two resistors are connected in series, the equivalent resistance of the combination
i. is certainly greater than each individual resistance.
ii. is certainly less than each individual resistance.
iii. might be greater than, less than, or equal to each individual resistance, depending on the current flowing in the circuit.
iv. might be greater than, less than, or equal to each individual resistance, depending on the specific values of the resistances.
v. [None of the above.]
Solution: (i)
(d) The superposition principle
i. is like the “position” principle, only much better.
ii. forms the basis for wave superkinematics, together with the supervelocity
principle and the superacceleration principle.
iii. explains why constructive interference occurs, but not destructive interference.
iv. explains why destructive interference occurs, but not constructive interference.
v. explains why both constructive and destructive interference occur.
Solution: (v)
(e) When two resistors are connected in parallel, the equivalent resistance of the
combination
i. is certainly greater than each individual resistance.
ii. is certainly less than each individual resistance.
iii. might be greater than, less than, or equal to each individual resistance, depending on the current flowing in the circuit.
iv. might be greater than, less than, or equal to each individual resistance, depending on the specific values of the resistances.
v. [None of the above.]
Solution: (ii)
(f) Neutron diffraction provides strong evidence
i.
ii.
iii.
iv.
v.
that crystals are subject to Heisenberg’s uncertainty principle.
for both the Big Bang theory and LeBron James’s big crunch-time abilities.
that neutrons can interfere constructively with crystals.
for Bohr’s atomic model.
that neutrons have a wave-like nature.
Solution: (v)
(g) The ultimate microscopic origin of all magnetic fields is thought to be
i.
ii.
iii.
iv.
v.
an elementary subatomic particle called the Bohr magneton.
an elementary subatomic particle called the magnetic monopole.
circulating electric charges.
interacting subatomic electric dipoles.
interacting subatomic magnetic dipoles.
Solution: (iii)
(h) Experiments involving the photoelectric effect provide strong support for
i.
ii.
iii.
iv.
the wave model of light.
the particle model of light.
Bohr’s atomic model.
Rutherford’s atomic model.
v. the quark theory of particle physics.
Solution: (ii)
(i) Electric field lines
i. are close together where the electric field is strong, and far apart where the
electric field is weak.
ii. point away from positive charges and towards negative charges.
iii. are such that electric field vectors are tangent to them.
iv. never cross where the electric field is nonzero.
v. [All of the above are true.]
Solution: (v)
(j) The figure shows four particles moving to the right as they enter a region of
uniform magnetic field. All particles move at the same speed and have the same
charge. Which particle has the largest mass?
i.
ii.
iii.
iv.
v.
Particle A.
Particle B.
Particle C.
Particle D.
[It’s impossible to determine from the information given.]
Solution: (iv)
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