PHY2049 Spring 2006
Exam 4 Solutions
Prof. Darin Acosta
Prof. Greg Stewart
May 1, 2006
Exam 4 (Final) Solutions
1. Four charges are arranged into a square with side length a=1 cm as shown in the figure.
The charges (clockwise from top left) are q1 = 10 nC, q2 = -10 nC, q3 = -10 nC, and q4 =
10 nC (1nC = 10-9C). What is the magnitude of the electric field at the center of the
square in N/C?
a. 5.1 × 106 b. 7.2 × 106 c. 1.3 × 106 d. 2.5 × 106 e. 0
Solution: By symmetry the field in the vertical direction will cancel out, and the field will
point in the +x horizontal direction. Magnitude of the field from one charge is:
q
a
E = K 2 where r =
r
2
2q
⇒ Ex = K 2 cos 45D for one charge
a
2q 2
N
all charges
⇒ Ex = 4 K 2
= 5.1× 10−6
C
a 2
2. A charge of 10-6 C is distributed uniformly across a large circular sheet with a radius of
1m that is nonconducting. What is the magnitude of the electric field in N/C at a point 0.5
cm above the center of the circle?
a. 1.8 × 104 b. 3.6 × 104 c. 3.6 × 108 d. 3.2 × 10-7 e. 5.6 × 104
Solution: A 1m sheet compared to a position 0.5 cm above it makes the sheet
essentially infinite. By using Gauss’ Law, we can determine the electric field to be
σ
q 10−6
. The charge density is σ = = 2 = 3.18 × 10−7 C .
constant and given by E =
A πr
2ε 0
4
So the field is 1.8×10 N/C.
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PHY2049 Spring 2006
Exam 4 Solutions
3. Two electrons are fixed 2~cm apart. A third electron is shot from infinity and comes to
a rest midway between the two. What was its initial velocity?
a. 320 m/s b. 80 m/s c. 2.9 × 10-7 m/s d. 1.0 × 105 m/s e. 0
Solution: First find the electric potential at the point between the 2 electrons
−e
−e
assuming V=0 at infinity: V = K
+K
d /2
d /2
The potential energy, then, for an electron (q= -e) to be at rest there, is:
e2
U = qV = 4 K .
d
Thus, an electron must have a kinetic energy equal to U initially:
1
e2
EK = mv 2 = 4 K
2
d
⇒v=
8Ke 2
= 320 m/s
md
4. An air filled parallel plate capacitor has a constant charge q=10-4 C on one of its plates
and the voltage across the capacitor is 100 V. If the plates are pulled apart to 3 times
their original separation, what potential difference in Volts now exists between the
plates?
a. 300 b. 100 c. 33 d. 900 e. 11
A
, so if the plates
d
are pulled apart 3 times further, the capacitance is 3 times smaller. Then since
q = CV , the voltage must increase 3 times to 300 V if the charge is to remain
constant.
Solution: The capacitance of a parallel plate capacitor is C = ε 0
5. Two identical light bulbs (with identical resistances) are connected in series to a
constant voltage source. If the light bulbs are instead connected in parallel to the constant
voltage source, what is the ratio of the brightness of a light bulb in the parallel
configuration to its former brightness when connected in series?
a. 4 b. 0.25 c. 16 d. 2 e. 1
Solution: Let the light bulbs have resistance R and let the voltage be V from the
source. The current through the circuit with the resistors in series is i = V / 2 R . The
brightness of one light is related to the power dissipated by one resistor, which is
V2
P = i2 R =
.
4R
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PHY2049 Spring 2006
Exam 4 Solutions
Now for the resistors in parallel, the current provided by the voltage source is
i = 2V / R , since the effective resistance is R / 2. The current is split exactly in half
between the two lights in parallel, so for one light i1 = V / R . The power dissipated in
one light is thus P = i12 R =
V2
, which is 4 times bigger than before. Each light
R
appears 4 times brighter.
6.
a ______
|
3Ω
⎯ 3V
i2 ->
i1 ->
_____________ ____________
__________
b
i3 ↑ |
|
-
5Ω
|
⎯ 6V
-
5V
|
|
___________________________ ___________
c
d
|
_____________
10Ω
What is the magnitude of the current, in Amperes, across the 10 Ω resistor?
a. 1/15
b. 2/3
c. 3/5
d. 1/3
e. 1/5
Solution.: do Kirchhoff loop equations
currents in and out of point b: choose i1 coming from the left, i2 going out to
the right, and i3 coming in from below: i1 + i3 = i2 as shown in the diagram in red.
Do the sum of voltages in the left hand loop, starting at a., going clockwise:
Eq. 2 -3i1 – 5 +3 = 0
Do the sum of the voltages in the right hand loop, starting at b., going clockwise
Eq. 3 -5i2 – 6 –10i2 + 5 =0
From eq. 2, i1= - 2/3
From eq. 3, i2= -1/15 (this is what is asked)
For completeness, from the current eq., i3=i2 – i1 = +3/5
7. In a classical picture, an electron (q=1.602 10-19 C) moves in a circle of radius 0.5 10-10
m at a velocity of 6 105 m/s. What is the magnetic dipole moment, in units of 10-24 Am2,
of this orbiting electron? (Hint: current is amount of charge past some point per unit
time, Δq/Δt, and the circulating electron completes one orbit every period.)
a. 2.4
b. 0.6
c. 1.2
d. 1.8
e. 3.0
Solution.: d=vt, so the time it takes the electron to go around its orbit is
(2π r=0.5 10-10 m)/v=6 105 m/s = 0.524 10-15 s
so current = i = Δq/Δt = 1.602 10-19 C(the charge of one electron)/0.524 10-15 s =
3.06 10-4 A. magnetic dipole moment μ=NiA = 1 I πr2, where r=0.5 10-10m
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PHY2049 Spring 2006
Exam 4 Solutions
so μ=2.4 10-24 Am (this is somewhat different than the value given in Table 28-2, but
the moment we calculated was due to the electron's motion in its orbit, while the
table gives the moment due to quantum mechanics (beyond the classical picture we
use in Phy 2049).
8. Straight wires along the radial direction carry current into and out of a circular arc of
wire which subtends an angle of 30o. If the arc of wire has a radius of 1 cm and the
current is 50 A, what is the B-field at the origin, in units of 10-3 T?
a. 0.26
b. 0.042
c. 1.20
d. 1.5
e. 12
Solution: B=μ0iΦ/4πR where Φ is the angle of the circular arc of wire in radians, in
this case Φ=(1/12)2π, so B=4π 10-7 50 (1/12)2π/(4π 0.01) = 2618 10-7 = .262 mT
9.An infinitely long insulated wire carrying a current I=20A is bent into the shape shown
(a straight line plus a circle of radius R=4cm with the currents in the direction shown).
Find the magnitude of the magnetic field at the center of the circle.
a. 4.1 × 10-4 T b. 2.1 × 10-4 T c. 1.0 × 10-4 T d. 3.1 × 10-4 T e. 8.2 × 10-4 T
Solution : We calculate the field from both the long wire and the circle separately.
From the long wire, the field at the center of the circle, a distance R away, is:
μi
Bwire = 0 = 10−4 T into the plane of the paper.
2π R
From the circular wire, the field is:
μi
Barc = 0 Φ where Φ = 2π
4π R
μi
Barc = 0 = 3.1×10−4 T into the plane of the paper
2R
Since the magnetic field from the arc is in the same direction of that from the long
wire, the total magnitude is the difference:
B = Barc + Bwire =
μ0i (π + 1) 4π ×10−7 ( 20 )(π + 1)
=
= 4.1× 10−4 T
2π R
2π ( 0.04 )
10. An inductor has L=18 H. Suppose 100 A of dc current is flowing through this
inductor through a switch, which is suddenly (Δt = 0.01 s) opened. What magnitude of
voltage appears across the inductor, in units of V?
a. 180,000 b. 18,000 c. 1800 d. 180 e. 18
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PHY2049 Spring 2006
Exam 4 Solutions
Solution.: V = -L dI/dt = (in magnitude) 18 100/0.01 = 180,000 V
11. A source of EMF fully charges a circuit in which are an inductor, L, and a 10 Ω
resistor. The source of EMF is removed and is replaced by a short; the current flowing
through the LR circuit takes 10 s to decay to ½ its original value. What is the value of L,
in H?
a. 1.44 102
b. 0.69 102
c. 0.30 102
d. 3.32 102
e. 1.44
Solution: i = i0 e-t/ (L/R) ; for i/i0 = 0.5, get ln0.5= - 0.6931 = -t/ (L/R), t=10s, R=10Ω,
=> L=1.44 *100
12. Suppose a capacitor, C = 10-3 F, is charged to 100 V. Suddenly, all of its energy is
released in 0.001 s and converted into emitted isotropic electromagnetic radiation. At a
distance of 1 m from this source, what is the magnitude of the electric field, Em, of the
emitted electromagnetic radiation when the wave passes by? (in units of V/m)
a. 550
b. 690
c. 100
d. 50
e. 69
Solution: energy stored in a capacitor is U=1/2 CV2 = ½ 10-3 (100)2 = 5 J
this energy is emitted as isotropic radiation in 0.001 s, so Power=
Energy/time = 5 kW
The intensity of the electromagnetic radiation, I, at a distance of 1 m when the pulse
of radiation (over the millisec) goes by is Power/4πr2 where r=1 m, so I=5 103/4π.
But we know that I=(1/2cμ0)Em2, so Em2 = 2cμ0I = 2 3 108 4π 10-7 5 103/4π. The
factors of 4π cancel and Em = (30 104)1/2 = 5.5 102 V/m
13. Which statement about alternating current RLC circuits is completely correct?
a. in a capacitor, the current leads the voltage; also, the current is in phase with the
voltage in a resistor
b. the current leads the voltage in an inductor
c. the voltage lags the current in an inductor
d. all of a, b, and c are correct.
e. none of a, b, and c are correct.
Solution: As we did in class in the demo multiple times, the current in a charging
capacitor in a dc circuit leads the voltage, which is also true in an ac circuit. Of
course, the current is in phase with the voltage in a resistor, so statement a. is
correct. Statement b. is incorrect (the current lags the voltage in an inductor) and
statement c. is just statement b. stated another way, so both b. and c. are incorrect.
Thus d. and e. are incorrect, and only a. is correct.
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PHY2049 Spring 2006
Exam 4 Solutions
14. A uniform electric field points in the z-direction with a time-dependent value given
by the expression: Ez(t) = a+bt, where a = 60 V/m and b = 1800 V/(m-s). The electric
field is confined to a circular region in the x-y plane with radius r = 3 m. What is the
magnitude of the magnetic field in Tesla at a point along the circumference of the circle?
a. 3 × 10-14 b. 10-15 c. 10-13 d. 7300 e. 4.5 × 10-7
Solution: Use Maxwell’s Law of Induction
dΦ
v∫ B ⋅ ds = μ0ε 0 dt E
dE
2π rB = μ0ε 0 A
dt
μ ε A dE π r 2b
rb
B= 0 0
=
= 2 = 3 × 10−14
2
2π r dt 2π rc
2c
15. What is the frequency, in units of Mega Hz, of an electromagnetic wave with
wavelength = 0.1 m?
a. 3000
b. 30
c. 3
d. 300
e. 0.3
Solution: c = λf f=c/λ = (3 108 m/s) / 0.1 m = 3 109 Hz = 3000 MHz
16. Light in air (n=1) is incident on a flat-faced cube of glass (n=1.5) at an angle of 45o to
the normal. The light is bent to angle θ2 from the normal in the glass. The light exits the
back of the cube of glass and enters another flat-faced cube of material whose index of
refraction is n3 (no value given). The light from the glass incident on the n3 material at
angle θ2 from the normal in the glass is bent to angle θ3 in this n3 material. The light then
exits this material back into air, n = 1, making an angle θ with respect to the normal in the
air. In units of degrees, what is θ?
a. 45
b. 30
c. 28.1
d. 22
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e. 35
PHY2049 Spring 2006
Exam 4 Solutions
Solution: Snell's law allows the determination of θ2, if we need it. Let's just set up
the equations for each interface:
eq. 1 nairsin45 = n2sinθ2
eq. 2 n2sinθ2 = n3sinθ3 since ray 2 is straight, its angle from the vertical when
it enters region 3 is also θ2 (rule from geometry for the
angles
for a parallel bisector)
eq. 3 but whatever θ3 is, we know that n3sinθ3 = nairsinθ
from eq. 2, we know that n3sinθ3 = n2sinθ2
from eq. 1, we know that n2sinθ2 just equals nairsin45, so nairsinθ (for the region
below) in eq. 3 is just equal to nairsin45 from eq. 1, so θ=45 and the light exits at the
same angle it entered. We saw this in a demo (the laser beam projector with 5
beams into and out of the block of Lucite on the blackboard) for one material.
17. A lens has a shape as shown in the figure. For a light ray traveling from the left, the
front face is flat and the rear face is convex with a radius of curvature of 0.8m. The glass
that makes the lens has an index of refraction of n = 1.5. What is the focal length of the
lens?
a. -1.6 m b. +1.6 m c. +0.8 m d. -0.8 m e. ∞
Solution: Note that a light ray facing a convex surface means that the surface is
considered to have a positive radius of curvature.
⎛1 1⎞
1
1 ⎞
1
⎛1
= ( n − 1) ⎜ − ⎟ = ( 0.5 ) ⎜ −
⎟=−
f
1.6
⎝ ∞ 0.8 ⎠
⎝ r1 r2 ⎠
f = −1.6 m
A negative focal length implies a diverging lens.
18. The lens of a movie projector is able to project a 3.0m wide image on a screen 18m
away from the lens. The picture on the film that is projected is 2.5cm wide. What is the
focal length of the lens in cm?
a. 15 b. 7.5 c. 10 d. 12.5 e. 20
Solution: The magnitude of the magnification is
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PHY2049 Spring 2006
Exam 4 Solutions
3.0
= 120 s
0.025
−i
m = ⇒ m = −120 for a real image
p
−i 18
p= =
= 0.15
m 120
1 1 1
1
1
⇒ = + =
+
f
p i 0.15 18
⇒ f = 0.15
m=
19. A phase difference of how much of a wavelength between two waves of equal
amplitude results in fully destructive interference?
a. 0.5
b. 0.0
c. 1.0
0.6
e. 0.33
Solution: if two waves of equal amplitude are out of phase by 0.5 of a wavelength,
then the crest of one matches the trough of the other, and they add to give zero.
20. A double slit interference experiment finds the maximum adjacent to the center,
Θ=0, maximum at an angle (measured from the center of the double slit to the viewing
screen) of 0.3 o. If the wavelength of the monochromatic light used is 500 nm, what is
the separation of the slits, in units of 10-6 m?
a. 95
b. 2.6
c. 0.0026
d. 150
e. 0.026
Solution: From eq. 35-14, dsinΘ = mλ, where for the case discussed, m=1. Thus,
d=500 nm/sin0.3o = 95,493 nm or about 95 μm
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