Unformatted text preview: PHY2049 Fall 2006 Prof. Yasu Takano Prof. Paul Avery Dec. 11, 2006 Exam 4 Solutions
(First answer is correct)
1. (Exam 1) Consider a cylinder of radius 6 m and height 15 m. At each point on the cylindrical surface, an electric field of 10 V/m points out radially. There is no electric field on the top and bottom surfaces of the cylinder. What is the charge enclosed by the cylinder in coulombs? (1) 5.0 10-8 (2) 2.5 10-8 (3) 1.7 10-8 (4) 8.3 10-9 (5) 3.3 10-8 We use Gauss' law, and note that only the E field perpendicular to the curved surface contributes. Using the area of a cylinder, Gauss' law yields E ( 2 rh ) = qenc / 0 , where r is the cylinder radius and h is its height. Solving for the enclosed charge yields qenc = 5.0 10-8 C. 2. (WebAssign 24.5) Two large, parallel, conducting plates are 1.2 mm apart and have equal and opposite charges on their facing surfaces. An electrostatic force of 4 10-13 N acts on an electron placed anywhere between the two plates. What is the potential difference in volts between the plates? (1) 3000 (2) 360 (3) 1.88 106 (4) 1880 (5) 2250 The force on an electron is F = eE. The potential difference between two plates is Ed. Thus the potential difference can be written V = Fd/e = 3000 V. 1 PHY2049 Fall 2006 3. (WebAssign 25.12) In the figure shown, the battery has a potential difference of 36 V. If C1 = C4 = C5 = 5 F and C2 = C3 = 10 F, what is the charge on C5 in C? (1) 120 (2) 240 (3) 180 (4) 360 (5) 60
C1 + V C2 C3 C5 C4 The total capacitance in the right hand branch is C2345 = 3.33 F. The charge on this equivalent capacitance is the same as the charge on C5, because C5 is in series with the rest of the branch C234 (as discussed in lecture and the textbook). Thus q5 = 3.33 36 = 120 C. 4. (Exam 2) A current of 15 A flows through a copper wire whose radius is 1.25 mm. Assuming that the current is uniform over the wire cross section, calculate the electron drift speed in m/s. Take the free electron density to be 7 1028 electrons/m3. (1) 2.7 10-4 (2) 7.1 10-3 (3) 1.2 10-5 (4) 2.5 10-2 (5) 1.7 10-7 The total current I = JA is given by I = JA = ene vd A , where J is the current density, e is the electronic charge, ne is the free electron density, vd is the drift velocity and A = R2 is the area. With the numbers provided, we obtain vd = 2.7 10-4 m/s. 5. (Exam 2) An electron moves at a speed of v = 4 107 m/s parallel to and a distance d = 15 cm from a conducting wire carrying a current of I = 60 A. If the directions of the electron and current are as shown in the figure, what is the magnitude (in N) and direction of the force experienced by the electron? (1) 5.1 10-16 toward the wire (2) 7.7 10-16 away from the wire (3) 1.8 10-16 away from the wire (4) 6.1 10-17 toward the wire (5) 2.8 10-16 toward the wire I
d e- The B field a distance d from the wire is B = 0 I / 2 d , so the force on the electron is thus F = evB = ev0 I / 2 d . Plugging in the numbers yields F = 5.1 10-16 N, and the right hand rule (accounting for negative charge) shows the force is towards the wire (as expected since we effectively have two currents in the same direction). 2 PHY2049 Fall 2006 6. (WebAssign 29.20) In figures (a) and (b), identical currents are set up in wire loops consisting of a semicicrcle of radius 10 cm, a smaller concentric semicircle of radius 4 cm, and two radial straight lengths, all in the same plane. The magnitude of the magnetic field produced at the center marked by the dot is 10 T in configuration (a). What is the magnitude of the field (in T) at the center in configuration (b)? (1) 23.3 (2) 4.3 (3) 35.0 (4) 6.0 (5) 2.6
(a) (b) The B field from a semicircular current (radius R and current I) is B = 0 I / 4 R . If a is the outer for (b) is Bb = 1 0 I (1/ a + 1/ b ) . The ratio of (b) to (a) is Bb / Ba = (1/ a + 1/ b) ) / (1/ a - 1/ b ) or 4
Bb / Ba = ( a + b ) / ( b - a ) .With a = 10 cm and b = 4 cm, we obtain Bb = 23.3 T. radius and b is the inner radius, then the B field for (a) is Ba = 1 0 I (1/ a - 1/ b ) and the B field 4 7. (Webassign 30.15) As shown in the figure, a stiff wire bent into a semicircle of radius 5 cm is rotated at constant frequency 55 Hz in a uniform magnetic field of 5.5 T. The resistance R is 16. What is the amplitude of the emf (in volts) generated in the circuit? (1) 7.46 (2) 0.52 (3) 2.41 (4) 0.20 (5) 3.26
X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X XRX X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X BX X X X X X X X X X X X X X X X X X X The emf is given by Faraday's law E = d B / dt . The semicircular area is 1 r2 2 and, taking into account the rotation of the semicircle, the total flux is B = 1 r 2 cos t , where = 2 f . 2 Taking the derivative yields the emf E = - 1 r 2 B sin t , so the amplitude of the emf is 2
E m = 1 r 2 B = 7.46 V. 2 3 PHY2049 Fall 2006 8. (Webassign 31.43) Consider an RLC circuit with a driving emf of amplitude Em = 15 V, R = 25, L = 0.08 H and C = 3.0F. What is the amplitude of the voltage (in volts) across the inductor at resonance? (1) 98 (2) 220 (3) 66 (4) 345 (5) 178 At resonance only R contributes to the impedance, so the current amplitude is I m = 15 / 25 = 0.6 A. The voltage amplitude across the inductor can be found from the inductive reactance to be VL = X L I m = LI m . Using = 1/ LC at resonance and substituting values yields 98 V. 9. (Webassign 32.32) An electron is placed in a magnetic field B that is directed along a z axis. The energy difference between parallel and antiparallel alignments of the z component of the electron's spin magnetic moment with B is 2.60 10-25 J. What is the magnitude of B in Tesla? (1) 0.014 (2) 0.11 (3) 0.0062 (4) 1.7 10-4 (5) 3.3 10-6 The difference in potential energies of an electron in a magnetic field B is E = 2 B B , where B is the Bohr magneton. The above values yield B = 0.014 T. 10. (Exam 3) A ray of light from a source placed at the bottom of a pool of water (n = 1.333) impinges on the water surface with an angle of incidence of 52. Which of the following statement is true? (1) The ray is totally reflected at the surface. (2) The angle of refraction is 36.2. (3) The angle of refraction is 90. (4) The angle of reflection is 48.2. (5) The angle of reflection depends on the wavelength of the light. The angle of incidence is larger than the critical angle of 48.2, therefore the ray is totally reflected. 4 PHY2049 Fall 2006 11. (WebAssign 34.52) An observer views an object through a diverging lens, whose focal length is 5 cm. If the object is located 10 cm from the lens, what is the lateral magnification of the object? (1) 0.33 (2) 0.50 (3) 1.0 (4) 2.0 (5) 0 We use the lens equation 1/ p + 1/ q = 1/ f , with f = -5. This yields q = -3.33, and the magnification is given by m = - q / p = 3.33 /10 0.33
12. (Chapter 34, checkpoint 3, in class) A person standing between two vertical parallel mirrors A and B looks into mirror A. The distance between the two mirrors is 3 m, and a lit candle is positioned 1 m from mirror A. How deep behind this mirror does the second image of the candle appear, in meters? (1) 5 (2) 2 (3) 3 (4) 4 (5) 6 The first image in A is 1 m behind A, which makes an image in B 4 m behind B, which in turn makes an image in A 7 m behind A. However, the first image of the candle in B is 2 m behind B, which makes an image in A 5 m behind A. 13. (Chapter 34 new) Let p be the distance between an object and a spherical mirror or a lens, whose focal length is f. Which of the following situation produces an inverted image of the object? (1) The spherical mirror is concave, and p > |f|. (2) The spherical mirror is convex, and p > |f|. (3) The lens is a converging lens, and p < |f|. (4) The lens is a diverging lens, and p < |f|. (5) The lens is a diverging lens, and p > |f|. This can be seen by inspecting the other answers, which are all false. 5 PHY2049 Fall 2006 14. (Spring 2006 exam 4) 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 1.2 m. The glass that makes the lens has an index of refraction of n = 1.5. What is the focal length of the lens in meters, including the sign? (1) -2.4 (2) 2.4 (3) -1.2 (4) 1.2 (5) The lensmaker's equation is 1/ f = ( n - 1)(1/ r1 - 1/ r2 ) , Here r1 = and r2 = +1.2 m. This gives f = -2.4 m. 15. (WebAssign 35.32) In a double-slit experiment, the viewing screen is at a distance of 4 m behind the double slits, whose separation is 4.5 m. The wavelength of the light used is 550 nm. Consider point P, which is on the screen and lies at 20 cm from the center of the interference pattern. What is the ratio of the intensity at point P to the intensity at the center of the pattern? (1) 0.080 (2) 0.013 (3) 0.47 (4) 1.2 (5) 2.5 The angle is approximately = 0.2 / 4.0 = 0.05. The intensity is given by the formula (assuming 0 slit width) I = I max cos 2 ( d sin / ) , which yields an intensity of 0.08 times the maximum. 16. (WebAssign 35.13) Two waves of light, whose wavelength is 600 nm, are initially in phase and travel in air. One of the waves then travels through a plastic sheet which is 4 m thick with an index of refraction of 1.5, while the other wave travels through another plastic sheet which is 3.5 m thick with an index of refraction of 1.6. What multiple of gives the phase difference of the two waves, after they have emerged from the plastic sheets? (1) 0.67 (2) 0.094 (3) 0.13 (4) 0.40 (5) 0.83 First compute the phase (in number of wavelengths) in the material, which is Ln / , where L is the length of material and n is the index of refraction. The term Ln is called the optical length. For the problem exactly as stated, the difference in optical length as the waves emerge from the 6 PHY2049 Fall 2006 sheets is given by the expression L1n1 - L2 n2 = 4 1.5 - 3.5 1.6 = 0.4 m , which is 0.67 wavelengths. If you report the phase after both waves travel the longer distance of 4 m (which is NOT what was asked in the problem, but the WebAssign problem made the same mistaken assumption), then the shorter wave travels an additional 0.5 m, so the total optical length difference is 0.1 m or 0.167 wavelengths. Because of possible confusion, we did not count this problem, giving everyone credit for it. 17. (Chapter 35, Sample problem 35-6, in class) An anti-reflection coating of a lens consists of a thin layer of MgF2, whose index of refraction is larger than that of air but smaller than that of the lens glass. What is the best thickness of the coating that minimizes the reflection? Assume that the light is perpendicular to the lens surface. (1) 1/4 of the wavelength of the light in MgF2 (2) 1/16 of the wavelength of the light in MgF2 (3) 1/8 of the wavelength of the light in MgF2 (4) 1/2 of the wavelength of the light in MgF2 (5) exactly the same as the wavelength of the light in MgF2 The coating must be 1/4 wave thick to make the two reflected waves a total of 1/2 wavelength out of phase. The extra phase of 1/2 from reflection is common to both waves and cancels in the difference. 18. (Webassign 36.40) Some tropical beetles are colored by optical interference that is due to scales whose alignment forms a diffraction grating. When the incident light rays are perpendicular to the grating, the angle between the first-order maxima (on opposite sides of the zeroth-order maximum) is 26 in light with a wavelength of 550 nm. What is the grating spacing of the beetle in m? (1) 2.4 (2) 1.6 (3) 0.068 (4) 1.4 (5) 0.77 The angle between the first maximum from the center is 13. The spacing d can be calculated from d sin = , which gives = 2.4 m. 7 PHY2049 Fall 2006 19. (Webassign 36.50) A line in the spectrum of a certain element is a doublet with wavelengths 581.6 and 583.0 nm. Calculate the minimum number of lines needed in a grating that will resolve this doublet in the second-order spectrum. (1) 208 (2) 166 (3) 350 (4) 83 (5) 450 The resolving power is R ave / = Nm , where N is the number of slits and m is the order. Using m = 2, we solve for N to get N = 582.3 / 2.8 = 208 slits. 20. (Chapter 36 Sample problem 36.3) A person observes a pointillistic painting (made of colored dots spaced by an average distance of 2 mm). As she steps backwards, at approximately what distance (in m) from the painting will she see the red dots in the painting ( = 650 nm) merge together? Assume that her vision is limited by diffraction and that her pupil diameter is equal to 2.5 mm. (1) 6.3 (2) 8.3 (3) 10.2 (4) 3.1 (5) 4.5 The dots will merge when their angular spacing = d / L (d is spacing and L is distance) as seen by the woman is approximately equal to the minimum angle her eyes are able to resolve, which is given by the Rayleigh angular distance 1.22 / D , where is the wavelength of light and D is the diameter of her pupils. Solving for L yields L = 6.3 m. 8 ...
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This test prep was uploaded on 04/17/2008 for the course PHY 2054 taught by Professor Avery during the Spring '08 term at University of Florida.
- Spring '08