Download Parallel Charged Plates - Physics with Calculus - Solved Exam and more Exams Physics in PDF only on Docsity!
PHY2049 Fall 2008 Profs. Y. Takano, P. Avery, S. Hershfield
Final Exam Solution
Note that each problem has three versions, each with different numbers and answers (separated by | ). The numbers for each question and its correct answer are listed in order for all three versions.
- A new particle having twice the charge of an electron and (55 | 65 | 75) times its mass circles a carbon nucleus (6 protons, 6 neutrons) with a velocity of (32, 000 | 37 , 000 | 42 , 000) km/s. What is the radius of its orbit (Note that 1 fm = 10−^15 m.)
Answer: 53.8 fm | 34.1 fm | 22.9 fm
Solution: For circular motion the magnitude of the force is |F | = mv^2 /r. Also, by Coulomb’s law the magnitude of the force is |F | = k(2e)(6e)/r^2. Setting these two forces equal, the radius of the orbit is r = k(2e)(6e)/(mv^2 ).
- An electron moves rightward between two parallel charged plates separated by distance d = 52 cm as shown in the figure. The plate potentials are V 1 = (10. 0 | 12. 0 | 14 .0) V and V 2 = (12. 6 | 15. 6 | 18 .6) V. If the initial speed of the electron at the left plate is (1100 | 1300 | 1500) km/s, what is its speed just as it reaches plate 2?
d
V 1 V 2
Answer: 1460 km/s | 1720 km/s | 1970 km/s
Solution: Using conservation of energy,
mev^2 i − eV 1 =
mev^2 f − eV 2 , solve for the final velocity, vf , which is the only unknown in this equation.
- Two batteries with emfs V = A and V ′^ = (2A | 4 A | 6 A) and three resistors of identical resistance R are connected as shown in the diagram. Find the potential difference Vb − Va. V V'
a b
Answer: 0 | 2 A/ 3 | 4 A/ 3
Solution: Let i 1 be the current going clockwise in the left loop, and let i 2 be the current going counterclockwise in the right loop. Kirchoff’s rules imply that 0 = V 1 − i 1 R − (i 1 + i 2 )R
0 = V 2 − i 2 R − (i 1 + i 2 )R
If we add the two equations we obtain that (i 1 + i 2 )R = (V 1 + V 2 )/3. Consequently, i 1 R = −(2V 1 − V 2 )/3 is the answer for Vb − Va.
- The Earth (radius 6370 km) has a magnetic dipole moment of 8. 0 × 1022 J/T. If (5 | 6 | 7) turns of wire are wound around the geomagnetic equator, what current in the wire (in units of 10^6 A) would be needed to make the magnetic field far from the earth (55% | 60% | 75%) larger?
Answer: 69 | 63 | 67
Solution: Let μ be the magnetic moment of the Earth, and r the radius of the Earth. The magnetic moment of a loop with N turns, radius r, and current I is N πr^2 I. For the first case set this equal to 55% μ to find I.
- As shown in the figure, two radio towers are spaced a distance of (18 | 21 | 24) m apart and radiating at a wavelength of (4.3 | 4.5 | 4.7) m. Tower 2 radiates at the same intensity as tower 1 but (38◦^ | 48 ◦^ | 58 ◦) later in phase. If a radio receiver is placed a long distance away, what is the smallest angle θ at which the intensity of the combined signal is a relative minimum? θ
Answer: 5. 4 ◦^ | 4. 5 ◦^ | 3. 8 ◦
Solution: The net phase difference between the two sources is
φ 2 − φ 1 = 2π
360 ◦^
d sin θ λ
For destructive interference the phase difference is an odd multiple of π. The smallest θ will come from π, which implies that sin θ =
λ a
- A diffraction grating of width W produces a (third | fourth | fifth)-order pattern at angle θ for light of wavelength λ. The total number of slits in the grating must be
Answer: (W/ 3 λ) sin θ | (W/ 4 λ) sin θ | (W/ 5 λ) sin θ
Solution: For a diffraction grating the condition for the m order maximum is d sin θ = mλ, where d is the slit separaration. The total number of slits is W/d = (W/mλ) sin θ.
- A consortium (www.tmt.org) plans to build a 30-meter diameter telescope having unprecedented light collection capability and angular resolution. If the angular resolution is limited by diffraction, what is the approximate maximum distance in light years (1 light year = 9. 5 × 1012 km) that the telescope can distinguish two stars separated by the earth-sun distance of 1. 5 × 108 km? Assume that the observation takes place using (1100 | 1300 | 1500) nm infrared light.
Answer: 350 | 300 | 260
Solution: The Rayleigh criterion for resolution of two distant objects is ∆θ = 1. 22 λ/D. The Earth-Sun separation at a distance of x light years away extends an angle ∆θ = (1. 5 × 1011 m)/(x 9. 5 × 1015 m). Solve for x.
- A wire (50 | 70 | 90) cm long carries a (0.4 | 0.6 | 0.7) A current along the +x axis through a magnetic field B~ = (4.0 mT)ˆj+(5.0 mT)ˆk. What is the magnitude of the magnetic force on the wire in newtons?
Answer: 1. 3 × 10 −^3 | 2. 7 × 10 −^3 | 4. 0 × 10 −^3
Solution: The current flow is perpendicular to the magnetic field so the magnitude of the force is just F = iL|B|, where the magnitude of the magnetic field in this case is (4^2 + 5^2 )^1 /^2 mT.
- A thin film of kerosene (n = 1.20) covers the surface of water (n = 1.30). A beam of light (λ = (400 | 550 | 630) nm), perpendicular to it, illuminates the surface. What is the minimum film thickness (in nm) that gives the brightest reflection because of constructive interference?
Answer: 167 | 229 | 263
Solution: There are additional phase shifts of π when going from n = 1 to n = 1.20 and when going from n = 1.20 to n = 1.30; however, the relative phase shift for interference is unchanged. The condition for constructive interference is 2 d = m(λ/ 1 .20). The smallest d occurs when m = 1.
- In an L-C circuit the maximum voltage across the capacitor is 3 V. If L = 50 | 40 | 30 mH and C = 6 | 8 | 9 μF, what is the maximum current in the circuit?
Answer: 33 mA | 42 mA | 52 mA
Solution: The maximum current is 3V /XC , where XC = 1/ωC and ω = 1/
LC.
- A thin converging lens has focal length 8 cm. An object of height (3 | 2 | 4) cm is placed 12 cm away from the lens. What is the height of the resulting image, and is the image inverted or noninverted?
Answer: (6 | 4 | 8) cm inverted
Solution: Using the numbers for the first case, the image distance is (1/ 8 − 1 /12)−^1 = 24 cm. Hence, the magnification is m = −i/p = −2. If the object has height 3 cm, then the image will have height 6 cm and be inverted (m < 0). Since the numerical answers above were not in the choices on the exam, everyone was given credit for this problem.
- Three capacitors, C 1 = 2 μF, C 2 = 4 μF, and C 3 = 6 | 8 | 10 μF, are placed in series. If the charge on capacitor C 1 is 2 μC, what is the voltage across capacitor C 3?
Answer: (0. 33 | 0. 25 | 0 .20) V
Solution: For capacitors in series the charge is the same. Hence, the charge on C 3 is also Q 1 , and the voltage across C 3 is V 3 = Q 1 /C 3. Since the numerical answers above were not in the choices on the exam, everyone was given credit for this problem.
- In the figure the wire carries current (2 | 3 | 4) A, and the loop carries current (3 | 2 | 3) A in the indicated directions. The radius of the loop is 2 mm. What is the magnitude of the magnetic field at the center of the loop?
Answer: 0.74 mT | 0.33 mT | 0.54 mT
Solution: As show in the figure the wire produces a magnetic field going out of the page at the center of the loop, and the loop produces a magnetic field going into the page. The magnitude of the magnetic field is
|B| = |
μoiW 2 πr
μoiL 2 r
where iW is the current in the wire and iL is the current in the loop.
- A slit 1.00 mm wide is illuminated by light of wavelength 450 nm. We see a diffraction pattern on a screen (2.00 | 4.00 | 3.00) m away. What is the distance between the first two diffraction minima on the same side of the central diffraction maximum?
Answer: 0.90 mm | 1.80 mm | 1.35 mm
Solution: The condition for a diffraction minimum is a sin θ = mλ. The two closest minima to the central maximum are m = 1 and m = 2. Because θ is small sin θ ≈ θ in radians. The position on the wall is L tan θ ≈ Lθ, where L is the distance to the screen. Putting these together, Lθ 2 − Lθ 1 = Lλ/a.
- If mirror M 2 in the Michelson interferometer at right is moved through 0.300 mm, a shift of (900 | 800 | 1200) bright fringes occurs. What is the wavelength of the light producing the fringe pattern?
Answer: 667 nm | 750 nm | 500 nm
Solution: One shift of fringes occurs for every change in the path length of λ. Consequently, a shift of 900 fringes corresponds to a change in the path length of 900λ. For an interferometer the path length difference is 2L 2 − 2 L 1 so 2∆L 2 = 900λ.
