PHY2054 Spring 2008 Prof. P. Kumar Prof. P. Avery April 3, 2008 Exam 3 Solutions 1. (Ch. 20) In the circuit shown, L = 56 mH, R = 4.6 Ω and V = 12.0 V. The switch S has been open for a long time then is suddenly closed at t = 0. At what value of t (in msec) will the current in the inductor reach 1.1 A? (1) 6.67 (2) 10.5 (3) 2.88 (4) 19.0 (5) None of these ( ) The current in an RL circuit, starting at 0, is i = imax 1 − e −t / τ , where τ = L / R and imax = V / R and V is the voltage. Here imax = 12.0 / 4.6 = 2.609 and τ = 0.056 / 4.6 = 12.17 msec. The solution is t = −τ ln (1 − i / imax ) = 6.67 msec. 2. Refer to the previous problem. What is the total energy stored in the inductor a long time after the switch is closed? (1) 0.19 J (2) 0.048 J (3) 0.76 J (4) 0.034 J (5) None of these After a long time, the inductor can be ignored and the current in the inductor is i∞ = 12.0/4.6 = 2.61 A. The energy in the inductor is U L = 12 Li∞2 = 0.19 J. 1 PHY2054 Spring 2008 3. (Prob. 20.6) A solenoid 6.00 cm in diameter and 10.0 cm long has 500 turns and carries a current of 15.0 A. Calculate the magnetic flux (in webers) through a circular cross-sectional area in the middle of the solenoid. (1) 2.7 × 10−4 (2) 5.1 × 10−2 (3) 4.7 × 10−4 (4) 7.5 × 10−4 (5) None of these The magnetic flux through the area is φB = π r 2 B = π r 2 μ0 ni , where i is the current, r is the radius of the coil and n is the number of windings/meter. This yields φB = 2.7 × 10−4 webers. 4. (Prob. 21.47) A microwave oven generates electromagnetic waves of frequency 2.4 GHz that are reflected by the walls and form standing waves that heat food as the energy of the waves are absorbed. The standing-wave pattern causes food to be heated unevenly, with burn marks at the wave peaks. If the distance between the burn marks is 4.8 cm, what is the speed of the microwaves inside the food in km/s? (1) 2.3 × 105 (2) 4.7 × 105 (3) 2.9 × 105 (4) 1.2 × 105 (5) None of these The points of highest intensity occur every half wavelength, thus λ = 9.6 cm. The velocity is thus v = λf = 0.096 × 2.4 × 109 = 2.3 × 108 m/s = 2.3 × 105 km/s. 5. (Prob. 21.57) A real inductor has resistance because it is composed of a coil of wire with nonzero resistivity. To measure the inductance of a coil, a student places it across a 12.0-V battery and measures a current of 0.72 A. The student then connects the coil to a 30.0-V (rms) 60 Hz generator and measures an rms current of 0.86 A. What is the inductance of the coil? (1) 0.081 H (2) 0.51 H (3) 34.9 H (4) 30.6 H (5) None of these The resistance is given from the first measurement, so R = 16.67 Ω. The impedance is obtained from the second measurement, Z = R 2 + X L2 = Vrms / irms = 34.88 Ω . This yields XL = 30.64Ω. Using XL = ωL yields L = 0.081 H. 2 PHY2054 Spring 2008 6. (Ch. 21) A series RLC circuit contains a resistor R, a capacitor C = 11 μF and an inductor L = 0.071H all connected to a 120-V (rms) generator of variable frequency. If the power dissipated by the resistor at resonance is 710 W, what is R? (1) 20.3 Ω (2) 10.1 Ω (3) 14 Ω (4) 7.2 Ω (5) None of these The effects of the inductor and capacitor cancel at resonance. Thus P = i2R = V2/R =710 W. This yields R = 20.3Ω. 7. Refer to the previous problem. What is the reactance of C when the circuit is at resonance? (1) 80 Ω (2) 0.012 Ω (3) 32 Ω (4) 11 Ω (5) None of these The resonant angular frequency is ω = 1/ LC = 1130 rad/s . The capacitive reactance can then be calculated to be X C = 1/ ωC = 80Ω . 8. Refer to the previous problem. Suppose the frequency of the generator is set to 100 Hz. What is the new capacitance (in μF) required to bring the circuit into resonance? (1) 36 (2) 25 (3) 10 (4) 0 (5) None of these We use ω = 1/ LC = 200π rad/s . Solving for C yields C = 36μF. 3 PHY2054 Spring 2008 9. (Example presented in class) Sunlight of intensity 1340 W/m2 is incident normally on a 9 km × 9 km rectangular sail of negligible mass attached to a spaceship of mass 12,000 kg. If the sail is perfectly black, how long will it take the spaceship to move 1 km starting from rest? (1) None of these (2) 180 sec (3) 8.1 sec (4) 24 min (5) 4.8 hr The total force is F = IA/c, where I =1340 is the intensity, A = 90002 is the area and c is the speed of light. The acceleration is a = F/m. Using d = 12 at 2 = 1000 m yields t = 260 sec. 10. (Ch. 21) An inhabitant of a distant planet aims a radar dish at the earth and sends a pulse of microwaves of wavelength 2.6 cm to our planet. The pulse is picked up on earth by a radiotelescope and measured to have wavelength 2.5833 cm. What is the velocity of the planet relative to the earth? (1) 1930 km/s towards the earth (2) 5010 km/s away from the earth (3) 480 km/s towards the earth (4) 1710 km/s away from the earth (5) None of these The change in frequency Δf = f – f0 is given by Δf = f 0 (1 − v / c ) , where f0 = 3 × 108 / 0.026 is the initial frequency and v is the velocity away from the earth. Solving yields v = −1930 km/s, or 1930 km/s towards the earth. 11. (Ch. 22) A block of material is found to have a critical angle with respect to air of 56.0°. What is its critical angle when immersed in water? (1) Does not exist (2) 64.8° (3) 56.0° (4) 48.6° (5) 36.2° The index of refraction of the block is n = 1/ sin 56.0 = 1.21 . Since nwater = 1.333 is greater than n, there is no critical angle going from the block to water. 4 PHY2054 Spring 2008 12. (Prob. 22.40) A small gold ring is placed at the bottom of a swimming pool at a depth of 2.3 m. In an attempt to hide the ring, an opaque square wooden raft with sides L is made to float directly above the ring with the ring just below the center of the raft. What is the minimum value of L for which the ring would not be visible for any observer above the water? (1) 5.2 (2) 4.6 (3) 2.6 (4) 1.7 (5) None of these If the angle from the ring to the sides of the craft is greater than the critical angle, then light rays will be totally internally reflected and the ring will not be visible from the water outside the raft. The minimum value of L is Lmin = 2d tan θc , where d = 2.3 and θc = sin −1 (1./1.333) = 48.6° is the critical angle of water. This yields Lmin = 5.2 m. 13. (Prob. 22.16) A flashlight on the bottom of a 3.70 m deep swimming pool sends a ray upward and at an angle so that the ray strikes the surface of the water 1.45 m from the point directly above the flashlight. What angle (in air) does the emerging ray make with the water's surface? (1) 60.9° (2) 58.5° (3) 31.5° (4) 74.1° (5) None of these The angle the ray makes with the normal in water is given by θ 2 = tan −1 (1.45 / 3.7 ) = 21.4° . The angle in air is obtained by Snell’s law, or sin −1 (1.333sin 21.4° ) = 29.1° , or 60.9° to the surface. 5 PHY2054 Spring 2008 14. (Prob. 22.24) As shown in the figure, a narrow beam of ultrasonic waves enters the liver at an angle θ = 50°, reflects off a tumor inside the liver and its outgoing direction and location are measured. If the speed of the wave is 7.0% less in the liver than in the surrounding tissue, determine the depth d of the tumor (in cm). Assume that Δ = 12 cm. (1) 5.9 (2) 5.0 (3) 7.8 (4) 7.3 (5) None of these Tissue d Liver Tumor The displacement Δ is given from geometry by Δ = 2d tan θ 2 , where θ2 is the refracted angle whose size is given by Snell’s law (using nliver / ntissue = 1/ 0.93 = 1.075 ). This give θ 2 = 45.4° . Solving for d yields d = 5.9 cm. 15. (Prob. 22.23) Two identical light pulses are emitted simultaneously from a laser (λ = 632.8 nm ). The pulses immediately enter two different media and take parallel paths to a detector 5.90 m away, one passing through air and the other through a block of ice (n = 1.31. Determine the difference in the pulses' times of arrival (in nsec) at the detector. (1) 6.1 (2) 19.7 (3) 4.7 (4) 15.0 (5) None of these The time difference Δt = t2 – t1 is given by Δt = L / ( c / n2 ) − L / ( c / n1 ) = L ( n2 − n1 ) / c . Using the values L = 5.9, n2 = 1.31, n1 = 1.0, we obtain t = 6.1 nsec. 16. Refer to the previous problem. Over what distance (in μm) will the number of wavelengths traversed by the two beams differ by 7? (1) 14.3 (2) 4.4 (3) 5.8 (4) 1.50 (5) 3.4 Let the common distance traveled by called d. Then the difference in the number of waves traveled in each medium is ΔN = d / ( λ / n2 ) − d / ( λ / n1 ) = ( n2 − n1 ) d / λ . Using λ = 632.8 nm, ΔN = 7 and the values of n1 and n2 from the previous problem, we obtain d = 14.3 μm. 6 PHY2054 Spring 2008 17. (Test Bank 22.64) In Huygens' construction, all points on a wavefront (1) act as point sources for the production of secondary spherical waves (2) act as independent particles (3) demonstrate the dual nature of light (4) must be sources of plane waves (5) None of these This was discussed in the text and in class. 18. (Ch. 23) A person stands 3.5 m facing a large concave mirror of radius 3.2 m and uses a camera to take a picture of her image. For what distance (in m) should the camera lens be focused? (1) 0.55 (2) 2.95 (3) 6.45 (4) 4.60 (5) None of these Using the lens equation with p = 3.5 and f = +1.6, we get q = +2.95 m. To photograph this image, we need to set the camera to its distance to the image, or 3.5 – 3.95 = 0.55 m. 19. (Test Bank 23.15) An upright object is located a distance from a concave mirror that is less than the mirror's focal length. The image formed by the mirror is (1) virtual, upright, and larger than the object (2) real, inverted and smaller than the object (3) virtual, upright and smaller than the object (4) real, inverted and larger than the object (5) None of these The object location is found from the lens equation: 1 q = 1 f − 1p . Since p < f, we see immediately that q < 0, so the image is virtual and therefore upright. You can plug in different values of p to show that the image is larger, but there is a simple way of seeing this. Because of the partial cancellation of the two terms, the magnitude of 1/q is smaller than that of 1/p and |q| > |p|. Thus the magnification M = −q / p > +1 and the image is larger than the object. A simple calculation in fact shows that M = p / ( f − p ) = 1/ (1 − p / f ) which is clearly larger than 1. 7 PHY2054 Spring 2008 20. (Ch. 23.18) It is observed that the size of a real image formed by a concave mirror is two times the size of the object when the object is 29.0 cm in front of the mirror. What is the radius of curvature of this mirror in cm? (1) None of these (2) 116 (3) 58 (4) 19 (5) 9.5 The magnification is M = −q/p = −2 (inverted image), so q = 2p . From the lens equation, we get 1 + 1 = 1 . This yields f = 19.3 or R = 38.6 cm. p 2p f 8