Download Exam 4 Solutions L v R

PHY2054 Spring 2008
Prof. P. Kumar
Prof. P. Avery
April 26, 2008
Exam 4 Solutions
1. A doubly charged positive ion of Lithium (3 protons, 4 neutrons) is accelerated through a potential difference of 28 kV. The ion enters a magnetic field perpendicular to the direction of the
field and begins moving in a circular path of radius 15.2 cm. What is the magnitude of the B
field?
(1) 0.30 T
(2) 0.24 T
(3) 0.19 T
(4) 0.42 T
(5) none of these
The radius is found by equating the centripetal force to the magnetic force, or
mv 2 / r = qvB ,where q = 2e and m = 7mp. Solving yields B = mv / qr . From the acceleration
over a potential, we can also write
1 mv 2
2
= qV . Putting these together yields the equation
B = 2mV / q / r , yielding B = 0.30 T.
2. Consider the arrangement shown in the figure. Assume that R = 4.0 Ω and L = 1.35 m, and
that a uniform 2.3 T magnetic field is directed into the page. The bar moves at a constant speed
of 0.56 m/s. What force must be exerted on the bar to cause it to move at this constant speed?
(1) 1.35 N
(2) 0.43 N
(3) 5.4 N
(4) 2.23 N
(5) None of these
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The emf is given by E = BLv, so the current is i = BLv / R. The magnetic force acting on the bar
to slow it down, and therefore the external force required to maintain constant speed, is
F = BLi = B 2 L2v / R or F = 1.35 N.
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PHY2054 Spring 2008
3. A resistor R, a capacitor C and an inductor L are connected in series to a 120-V (rms) generator of variable frequency. If f0 is the resonant frequency, what is the combined reactance (XC and
XL acting together) when the frequency is 2f0?
(1)
(2)
3
2
5
2
L/C
L/C
(3) 2 L / C
(4) 2 L / C
(5) L / 2C
The combined reactance at a given frequency f is XL – XC = ωL – 1/ωC, using the angular frequency ω = 2π f . The resonant angular frequency is ω0 = 1/ LC , so the combined reactance
at twice the resonant frequency is 2 L / LC − LC / 2C =
3
2
L/C .
4. As shown in the figure, a light ray is incident normal to one face of a triangular block of material with index of refraction n2 = 1.78 that is immersed in water. What is the maximum angle α
for which total internal reflection occurs at P?
(1) None of these
(2) 43.8°
(3) 46.2°
(4) 34.2°
(5) 55.8°
From the geometry of the problem, the incident angle at P is 90 – α, thus the condition for total
internal reflection can be written n2 sin ( 90 − α ) = n1 , where n2 = 1.78 and n1 = 1.333. This
yields α = 41.5°.
5. A person stands 1.3 m from a large concave mirror with radius of curvature 3.8 m and uses a
camera to take a picture of her image. For what distance (in m) should the camera lens be focused?
(1) 5.4
(2) 0.53
(3) 4.1
(4) 3.3
(5) None of these
Using the lens equation with p = 1.3 and f = +1.9, we get q = −4.12 m. To photograph this image, we need to set the camera to its distance to the image, or 1.3 + 4.12 = 5.42 m.
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PHY2054 Spring 2008
6. An object placed 9.80 cm from a spherical mirror produces a virtual image 1.5 times larger. If
the object is moved to a new position 20.0 cm from the mirror, what is the position of the image
relative to the mirror?
(1) −63 cm
(2) +8.3 cm
(3) 11.9 cm
(4) −9.9 cm
(5) None of these
The image is virtual, so q = −9.80 × 1.5 = −14.7. The basic way of solving for the final image is
to use the lens equation to solve for f, given the initial p and q, then using f in the lens equation
so solve for the new image position. However, since the lens is the same, we can write directly
1/ 9.8 − 1/14.7 = 1/ f = 1/ 20 + 1/ q′ , and bypass the calculation of f. This yields q = –63 cm.
7. A gray-haired professor sees objects clearly only when they are at distances between 18 cm
and 32 cm from his eyes. If he wears contact lenses to correct his nearsightedness, what is the
minimum distance in cm that he will be able to see an object clearly?
(1) 41
(2) 11.5
(3) 18.0
(4) 32
(5) None of these
Recognize that gray-haired professor? In order to correct his nearsightedness, he must wear
contacts with a focal length of −32 cm. The contacts take objects at p = ∞ and create an image
at 32 cm (q = −32). However, the distance of the closest object he can see clearly must have its
image located 18 cm from his eye (q = −18 cm). Using the lens equation, we get 1/p − 1/18 =
−1/32, or p = 41 cm as the distance of the closest object he can see clearly.
8. A camera creates an image by focusing light from a lens onto a flat CCD array behind the
lens. The distance from the lens to the CCD array is adjustable. A camera lens has a focal length
of 250 mm and is initially set to properly focus the image of a distant object. How far must the
lens be moved to focus the image of an object 5.1 m distant?
(1) 12.9 mm
(2) 11.7 mm
(3) 62 mm
(4) 15.1 mm
(5) None of these
The image position for objects at infinity is q = +250 mm. When the distance of the object is 5.1
m, then the image is located at q = +262.9 mm. Thus the lens must be moved 12.9 mm.
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PHY2054 Spring 2008
9. In order to study insects in a swarm, researchers set up a large circular listening dish of diameter 1.6 m that picks up noises of the individual insects. What is the minimum frequency of the
insect sounds that the dish must detect in order to distinguish insects 13.5 cm apart when the
swarm is 50 m away? The velocity of sound in air is 343 m/s.
(1) 97 kHz
(2) 1.2 Hz
(3) 2.4 kHz
(4) 630 Hz
(5) None of these
The angular resolution of a circular aperture is 1.22λ/D, where λ is the wavelength of sound and
D is the diameter of the dish. The angle that needs to be resolved is y/d, where y = 0.135 m is
the distance between the insects and d = 50 m is the distance to the swarm. Using 1.22λ/D = y/d,
with f = cs / λ, and cs = 343 the speed of sound, we obtain f = 1.22csd / Dy = 97 kHz.
10. Refer to the previous question. The researchers deploy two of the previously described dishes
with their outputs connected together so that the signals are added coherently. They place the
dishes side by side with their centers a distance of 11 m apart facing the swarm. What is the
minimum frequency of the insect sounds the dishes must detect in order to resolve insects 1.5 cm
apart?
(1) 105 kHz
(2) 260 kHz
(3) 42 kHz
(4) 270 Hz
(5) None of these
The two receivers act as an interferometer with angular resolution λ/L, where L = 11 m is the
distance between the centers of the dishes. Equating this to the angle y/d between the insects
(where now y = 0.015 m), and using f = cs / λ, we obtain f = csd / Ly = 104 kHz.
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PHY2054 Spring 2008
11. Waves from a radio station have an unknown wavelength. They travel by two paths to a
home receiver 20.0 km from the transmitter. One path is a direct path, and the second is by reflection from a mountain 56 m behind the home receiver. What is the maximum wavelength of
the broadcast signal if it produces a destructive interference at the receiver? (Assume that no
phase change occurs on reflection from the mountain.)
(1) 224 m
(2) 56 m
(3) 168 m
(4) 112 m
(5) None of these
The waves must be out of phase by half a wavelength for destructive interference, so the path difference 112 m is equal to (n + ½)λ, or λ = 224 m.
12. Unpolarized light passes through two polaroid sheets. The axis of the first is vertical, and that
of the second is at 36.0° to the vertical. What fraction of the initial light is transmitted?
(1) 0.33
(2) 0.65
(3) 0.81
(4) 0.41
(5) 1.00
The intensity of unpolarized light passing through the first filter is ½ the initial intensity I0, so the
final intensity I2 can be written I 2 = 12 cos 2 36° = 0.33I 0 .
13. A lens with a refractive index of 1.5 is coated with a material of refractive index 1.2 in order
to minimize reflection. If λ denotes the wavelength of the incident light in air, what is the thinnest possible such coating?
(1) None of these
(2) 0.50λ
(3) 0.42λ
(4) 0.33λ
(5) 0.25λ
(
)
The condition for destructive interference is 2d = m + 12 λ / n , where d is the coating thickness
and n is its index of refraction. This equation applies because the rays reflecting from the top
and bottom of the coating both shift their phases by λ/2, so the difference cancels. Solving for d
gives d = 0.21λ.
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PHY2054 Spring 2008
14. A thin layer of oil (n = 1.32) is floating on water. What is the minimum thickness of the oil in
the region that strongly reflects a green light (λ = 527 nm)?
(1) 200 nm
(2) 264 nm
(3) 527 nm
(4) 400 nm
(5) None of these
The condition for constructive interference is 2d = mλ / n , where d is the coating thickness and n
is its index of refraction. This equation applies because the rays reflecting from the top and bottom of the coating both shift their phases by λ/2, so the difference cancels. Solving for d gives d
= 527 / (2 * 1.32) = 200 nm.
15. The lens of a movie projector is able to project a 2.0 m wide image on a screen 18 m away
from the lens. The slide that is projected is 2.5 cm wide. What is the focal length of the lens in
cm?
(1) 22
(2) 7.5
(3) 10
(4) 12.5
(5) 15
The magnification M = −80, so p = q / 80. Thus 1/f = 80 / 1800 + 1 / 1800 = 81 / 1800. This
gives f = 1800 / 81 = 22.2 cm.
16. Charges are arranged on a square of side d as shown in the diagram. In what direction is the
electric field at the center of the square?
(1) Second quadrant
(2) Fourth quadrant
(3) First quadrant
(4) Third quadrant
(5) E = 0
The E field from +Q and –Q points upward and the E field from +Q and –2Q points leftward, so
combined they point in the second quadrant. The fields from the two –5Q corner charges cancel
while the net E field from the other corner charges also points in the second quadrant. The net E
field is thus in the second quadrant.
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PHY2054 Spring 2008
17. Refer to the previous problem. What is the potential at the center of the square (in units of
kQ/d), assuming V = 0 at infinity?
(1) −16.1
(2) +1.3
(3) 0
(4) −7.7
(5) None of these
(
)
(
)
The total potential is −10kQ / d / 2 − kQ / ( d / 2 ) = − 10 2 + 2 kQ / d , where the first term
comes from the corner charges and the second term comes from the left and right charges.
Evaluating yields −16.1kQ/d.
18. In the accompanying figure, C1 = 25 μF, C2 = 5 μF, C3 = 45 μF, and ΔV0 = 21 V. Determine
the charge (in μC) stored by C2.
(1) 35
(2) 70
(3) 350
(4) 105
(5) None of these
First, we determine the equivalent capacitance. C2 and C3 are parallel, so their combination is
50 μF. The total equivalent capacitance is Ceq = 25 × 50 / (25 + 50) = 16.67 μF. The total
charge is then qtot = 16.66 × 21 = 350 μC. This is also the charge on C1, so the voltage on C1 is
V1 = qtot / C1 = 350 / 25 =14 V. The voltage across C2 is V2 = 21 – 14 = 7 V. Thus the charge on
C2 is q2 = C2V2 = 5 × 7 = 35 μC.
19. Consider the circuit shown in the accompanying figure. What is the power dissipated (in
watts) in the 6 Ω resistor?
(1) 0.67
(2) 0.33
(3) 0.28
(4) 1.00
(5) None of these
i2
i1
The two equations from Kirchoff’s rules are 3 − 3i1 − 6 ( i1 + i2 ) = 0 and 2 − i2 − 6 ( i1 + i2 ) = 0 .
Solving yields i1 = 1/3 A, i2 = 0. So the power in the 6 Ω resistor is 1/32 × 6 = 0.67 W.
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PHY2054 Spring 2008
20. In the circuit shown, the switch S has been closed long enough to allow C to fully charge.
After the switch is opened the potential difference across C reaches 2.2 V in 1.8 μs. What is the
value of C?
(1) 0.27μF
(2) 2.2μF
(3) 3.5μF
(4) 0.87μF
(5) None of these
When the capacitor is fully charged, the voltage across it is 12 V. After the switch is opened, it
discharges through a resistor combination equivalent to 4.0 Ω. The equation describing the
voltage on the capacitor is 12 exp ( −t / RC ) = 2.2 ,with R = 4 Ω. Solving for C yields
C = 1.8 / ( 4 ln (12 / 2.2 ) ) = 0.27 μ F .
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