Solution: All electromagnetic waves in vacuum, regardless of their wavelength or frequency, travel at the speed of light, c.

Transcription

1 1. Two electromagnetic waves travel through empty space. Wave A as a wavelength of 700 nm (red light), while Wave B has a wavelength of 400 nm (blue light). Which statement is true? A) Wave A travels faster because it has a longer wavelength. B) Wave B travels faster because it has a higher frequency. C) The speed of the two waves depends on their amplitudes. D) Both waves travel at speed c. E) The speed of the two waves depends on their polarization. Solution: All electromagnetic waves in vacuum, regardless of their wavelength or frequency, travel at the speed of light, c. 2. The following diagram shows a circular loop of wire in a uniform magnetic field that points out of the page. You now perform four separate eperiments: Eperiment I: With the loop held stationary, the strength of the magnetic field is gradually increased. Eperiment II: With the magnetic field held constant, the loop is moved to the right. The loop remains in the plane of the page and never eits the magnetic field. Eperiment III: With the magnetic field held constant, the loop is shrunk to one half its initial radius, always remaining within the plane of the page. Eperiment IV: With the magnetic field held constant, the loop is rotated clockwise about its center, always remaining within the plane of the page. In which of these eperiments will there be an induced current in the loop? Solution: There will be an induced emf if there is a changing magnetic flu through the loop, BA cos. In ept. I, the flu changes because of the increasing field. In ept. III, the flu changes because of the changing area. There is no change in flu in epts. II and IV, since the motion does not change the orientation of the loop with respect to the field. A) I and III B) I, II, and IV C) III only D) I only E) I and IV

2 3. When electricity reaches your house from the power company, it has an rms voltage of 8,500 V and a frequency of 60 Hz. To achieve the desired 120 V (rms) household voltage, a step-down transformer is used, as shown at right. If the outer transformer coil (the one connected to the high-voltage line) contains 25,800 turns of wire, how many turns should the inner transformer coil (the one connected to your house) contain? V rms = 8,500 V N p Solution: Use the transformer equation,, with V V s N p =8500 V, V s =120 V, s and N p =25,800. Solving for N s, you find N s =360 turns. V p A) 550 B) 360 C) D) E) 620

3 4. Two plane mirrors meet at right angles at the origin. An object is placed at the point (3,2), as shown here: y Which diagram best indicates the location of all the images formed by the mirrors? (Images are shown as open circles.) y y A) B) C) y D) y E) y Solution: First of all, you can eliminate any diagrams that contain an image in the same region as the object. Such an image would be a real image, but plane mirrors and form only virtual images. That reasoning eliminates choices A, B, and C. Choice D is correct because it contains the additional image that arises from multiple reflections. You can think of this image as arising from the reflected images of the two mirrors themselves, as shown below. image from mirror 2 mirror 2 Object image of mirror 1 image from image of mirror 2 / image from image of mirror 1 image of mirror 2 image from mirror 1 mirror 1

4 5. A circuit containing a capacitor, an inductor, a generator, and two identical light bulbs is constructed as shown. The generator produces a peak emf of 10 V. The frequency of the generator can be varied. ω C L Bulb A Bulb B Eamine the following statements about this circuit: I. Bulb A shines more brightly than bulb B at large frequencies. II. As approaches zero, bulb A goes out and bulb B continues to shine. III. The two bulbs have equal brightness when 1 LC. Which statement(s) are true? Solution: This is a problem about reactance, the quantity that relates current through an inductor or a capacitor to the applied voltage via the Ohm's Law-like relation V=X L or C I. The main idea here is that, since the bulbs are identical, their brightness is determined entirely by how much current flows through them. Moreover, since the inductor and the capacitor are connected in parallel, they eperience the same driving voltage. So the branch with the most current, and hence the brightest bulb will be the one with the smallest reactance. The reactance of an inductor is X L L and the reactance of a capacitor is X C 1 C. With these ideas in mind, I is true. The capacitative reactance becomes small at high frequency, so it presents little opposition to the flow of current. Conversely, the inductive reactance becomes high, so little current flows through that branch and hence bulb B becomes dim. II is true. As approaches zero, the capacitative reactance becomes infinite. No current flows through that branch and bulb A goes out. Current easily flows through bulb B, since the inductive reactance is approaching zero. III is true. The bulbs will have equal brightness when the two components 1 have equal reactance. This happens when L 1 LC. C A) I and II B) II and III C) I, II, and III D) III only E) I and III

5 6. Lenz's law states that the direction of an induced emf is such that it opposes the change in magnetic flu. If instead the opposite were the case (i.e. the induced emf reinforced the change in flu), what fundamental law of physics would be violated? Solution: In class we saw several eamples of what would happen if Lenz's law went the other way around. Instead of a back emf, generators would spin faster and generate unlimited free power without any work being required to turn them. Conducting bars falling through a magnetic field would gain speed from which we could etract free work. These and other free lunches are forbidden by the law of conservation of energy, which Lenz's law helps preserve. A) No law would be violated; the direction just happens to be this way. B) Newton's second law. C) Conservation of electric charge. D) Gauss's Law. E) Conservation of energy.

6 Questions 7 and 8 refer to the following figure, which depicts a plane electromagnetic wave. z y E B E B E B 7. Eamine the following statements about this wave: I. The wave is propagating in the + direction. II. This wave could be produced by a charge oscillating back and forth along the y-ais. III. The wave is linearly polarized. Which statement(s) are true? Solution: I is true. To get the direction of an EM wave, use the right-hand rule: curl your fingers from E toward B, while etending your thumb. Your thumb points in the direction of propagation, which in this case is +. II is false. As we demonstrated in class with the antenna, the E-field from a charge distribution oscillating in simple harmonic motion is parallel to the direction of oscillation. So a wave like this could be produced by a charge oscillating back and forth in z. III is true. Linear polarization means that there is a definite, fied direction to the E-field. That's clearly the case here. A) I and III B) III only C) I, II, and III D) I only E) II only

7 8. If you wanted to detect this wave using an antenna, it would be best to use Solution: A linear antenna works by detecting the electric portion of the wave, allowing to drive charge back and forth along the antenna ais and producing a current. So a linear antenna should be oriented parallel to the E-field, i.e. Along the z ais here. You can detect the magnetic portion of this wave using a loop antenna and orienting it perpendicular to the B-field. This orientation will result in the greatest change in magnetic flu through the loop, and hence in the greatest induced current. Here, you want to orient a loop antenna in the -z plane. A) A linear antenna oriented parallel to the y-ais, or a loop antenna lying in the -y plane. B) A linear antenna oriented parallel to the -ais, or a loop antenna lying in the y-z plane. C) A linear antenna oriented parallel to the z-ais, or a loop antenna lying in the -z plane. D) A linear antenna oriented parallel to the z-ais, or a loop antenna lying in the y-z plane. E) A linear antenna oriented parallel to the y-ais, or a loop antenna lying in the -z plane. 9. The Voyager 1 spacecraft, launched in 1977 to eplore the outer planets, is presently about 10 billion kilometers away--beyond the orbit of Pluto. However, its radio transmitter continues to send useful data to scientists on Earth. The rms strength of the electric field 10 from these transmissions is about V/m. If the signal is received by a radio dish with an area of 900 m 2, what is the rms power incident on the dish? Solution: The rms power per unit area in an EM wave is given by the Poynting vector: S P rms c 2 0 E A rms. Solving for the power, we have P c 2 0 E rms A m s C N m V m m This is an incredibly weak signal, about 20 billion times smaller than the power output of a digital watch battery. 18 W. A) W B) W C) W D) W E) W

8 10. A real object is placed 25 cm away from a mirror with a radius of curvature of 30 cm, as shown here: Object C The image formed by this mirror will be: A) real, upright, and magnified. B) virtual, upright, and reduced. C) real, upright, and reduced. D) real, inverted, and magnified. E) virtual, inverted, and magnified. Solution: It's easiest to solve this problem by ray tracing, using the three principal rays described in class and in the tet. Note that the focal point F lies midway between the center of curvature C and the face of the mirror. Object P ray Image C ray C F F ray You can see that the image is real, inverted, and magnified.

9 Questions 11 and 12 refer to the following AC circuit. It consists of an inductor and a resistor in series with a generator of frequency f. Internal resistance of the generator is negligible. f L R 11. The impedance of this circuit is: Solution: Construct the phasor diagram: R V = V 0cos ωt Z 2 π fl The total impedance Z is the vector sum of the current phasors for the inductor and the resistor. These two phasors form a right angle, so the total impedance is the length of the hypotenuse. 2 2 fl D) A) R L B) R 2 L 2 C) R In this circuit: R 2 L 2 E) 2 2 R fl f A) resonance will occur if the circuit is driven at the appropriate frequency. B) the generator performs no work. C) all of the power is dissipated in the resistor. D) the voltage and the current are in phase. E) the current leads the voltage, but if the resistor and the inductor trade positions then the voltage will lead the current. Solution: Only resistors dissipate power. The other statements are incorrect.

10 13.You are standing 3.0 m away from the shore of a calm pond that is 26 m wide. An oak tree grows on the opposite shore. From where you are standing, with your eyes 1.6 m above the surface of the water, you can just see the reflection of the treetop in the water nearest to your feet. How tall is the tree? Solution: A picture of the situation is shown here: H h θ θ d w Because the angle of incidence and the angle of reflection of equal, the two triangles are similar. We have tan wh 26 m 1.6 m H 13.9 m. d 3.0 m H h w d A) 18.7 m B) 13.9 m C) 24.6 m D) 21.8 m E) 15.2 m 14. Vertically polarized light with intensity S 0 passes through two successive polarizers. The first has its transmission ais oriented at 30 from the vertical. The second has its transmission ais oriented at 60 from the vertical. The intensity of the light that emerges from the second polarizer is Solution: This problem uses Malus's Law. The first polarizer transmits an intensity S 1 S 0 cos S 0. The transmitted light is now polarized along the direction of the first polarizer. The second polarizer makes an angle of = 30 with respect to this light, so it transmits S 2 S 1 cos S S. 0 A) 9 16 S 0 B) 1 4 S 0 C) 3 8 S 0 D) 3 16 S 0 E) zero

11 15. In the circuit shown here, a conducting bar is free to move on two parallel metal rails. The two rails are connected by a resistor R. A uniform magnetic field of strength B, directed out of the page, covers the entire region of this diagram. R B out of page You now move the bar as shown in the position vs. time graph at right. Which of the graphs below best depicts the current in the resistor as a function of time? A positive value means that the current is circulating clockwise; a negative value means that the current is circulating counterclockwise. t I A) I B) C) I t t t D) E) I I t t See net page for solution/eplanation.

12 Solution to Problem 15: There will be an emf, and hence an induced current, when there's a changing magnetic flu through the loop. Looking at the position vs. time graph, the bar is moving to the right at constant speed during the first two ticks. This means that the area of the loop is changing at a constant rate, so there is a constant emf during this time. Lenz's law requires that the current be clockwise here so that the induced current generates a flu that opposes the change. During the net two ticks, the bar is sitting still ( remains constant). So there is no emf. Likewise, there's a constant countclockwise current during the net tick while the bar moves to the left. During the last two ticks, the bar is moving to the right again, but at a faster rate than before, so there is a larger clockwise induced current. Diagram A depicts these currents. 16. You are designing a generator that will produce a maimum emf of 170 V when rotated at an angular speed of 377 rad/s in a magnetic field of strength T. You plan to do this by winding wire around a square frame whose sides have length 12 cm. How many turns of wire should you use? V peak Solution: A generator with N turns and area A produces an emf of V NAB sin t when rotated with angular speed in a magnetic field B. Since sin t has a maimum value of 1, the peak emf is NAB N V peak AB 170V m 377rad s 0.050T 626 turns. A) 891 B) 1370 C) 626 D) 1020 E) A shaving/makeup mirror produces an image that is upright and magnified by a factor of 2.5 when your face is 32 cm from the mirror. What is the radius of curvature of the mirror? Solution: This problem is essentially the same as CAP #8, problem 3. We need to combine the mirror equation and the magnification equation. Moreover, we know that the mirror is concave, since a conve mirror cannot produce an enlarged image. So the radius of curvature is related to the focal length via f=r/2. We have: d o d i d o md o R, where I used the magnification equation to replace d i with -md o. Solving this equation for R, we find R 2 d o 1 1 m 2 32cm cm. A) 84 cm B) 128 cm C) 46 cm D) 224 cm E) 107 cm

13 18. A distant galay is receding from us at a speed of m/s. An alien in that galay is beaming you a distress signal whose wavelength he measures to be 662 nanometers (nm). What wavelength do you measure for this signal? Solution: The Doppler shift for a light source moving away from you with speed v is f c o f s 1 v c 1 v. Solving for the observed wavelength c c o, we have o 1 s o v c 662 nm 1 s m s m s nm. A) 610 nm B) 680 nm C) 662 nm D) 730 nm E) 644 nm 19. In class, we performed the jumping ring demonstration, in which a metal ring is placed on top of a vertical solenoid. At t=0, a switch was closed, causing the current in the solenoid to rise very quickly from zero to some large value. The ring jumped off the solenoid and hit the ceiling of the lecture hall. Call this the original version of this eperiment. Here are three possible variations: I. Same as the original version, but the metal ring is replaced by a non-conducting rubber ring. II. Same as the original version, but a notch is removed from the ring, so that it resembles the letter C. III. With a large, steady current running through the solenoid, a metal ring is placed on top of the solenoid. At t=0, a switch is opened, causing the current to fall rapidly to zero. In which variation(s) will the ring jump up from the solenoid? Solution: First, recall why the ring jumps in the original version. When the switch is closed, the flu through the ring rapidly rises from zero to some final value. This induces a big emf in the ring. Because the ring is a conducting loop, it can carry a current. This current flows in a direction to oppose the change in flu, so its direction is opposite to that of the current flowing in the coil. The two opposite currents eert a repulsive force on each other, and the ring jumps up. What about the three variations here? I. The emf is the same as before. But since the ring is non-conducting, no current can flow. The rubber ring does not jump. II. The cut-out notch prevents current from flowing around the ring, so the ring does not jump. III. When the switch is opened, the flu through the ring starts to decrease, so the induced current in the ring flows in a way that opposes the change. This time, the current in the ring flows in the same direction as the current in the coil. So there is a force on the ring, but it's an attractive one: the ring is pulled strongly toward the coil, and does not jump. A) III only B) I only C) II only D) none of these E) I, II, and III

14 20. The current in a certain circuit is increased at a constant rate from 2.0 A to 7.0 A over a period of 2.0 s. During this time, an induced emf of 72 mv is observed in the circuit. What is the self-inductance of this circuit? I Solution: The self-inductance, the emf, and the changing current are related according to: V L. Solving for the self-inductance L, we have t V L I 72 mv t 5.0 A 2.0 s 29 mh. A) 48 mh B) 29 mh C) 68 mh D) 18 mh E) 33 mh