Exam 4 (Final) Solutions

PHY049 Spring 006 Prof. Darin Acosta Prof. Greg Stewart May 1, 006 Exam 4 (Final) Solutions 1. Four charges are arranged into a square with side length a=1 cm as shown in the figure. The charges (clockwise from top left) are q 1 = 10 nc, q = -10 nc, q 3 = -10 nc, and q 4 = 10 nc (1nC = 10-9 C). What is the magnitude of the electric field at the center of the square in N/C? a. 5.1 10 6 b. 7. 10 6 c. 1.3 10 6 d..5 10 6 e. 0 Solution: By symmetry the field in the vertical direction will cancel out, and the field will point in the +x horizontal direction. Magnitude of the field from one charge is: q a E = K where r = r q x cos 45 E = K for one charge a q N Ex = K = a C 6 4 5.1 10 all charges. A charge of 10-6 C is distributed uniformly across a large circular sheet with a radius of 1m that is nonconducting. What is the magnitude of the electric field in N/C at a point 0.5 cm above the center of the circle? a. 1.8 10 4 b. 3.6 10 4 c. 3.6 10 8 d. 3. 10-7 e. 5.6 10 4 Solution: A 1m sheet compared to a position 0.5 cm above it makes the sheet essentially infinite. By using Gauss Law, we can determine the electric field to be 6 σ q 10 7 constant and given by E =. The charge density is σ = = = 3.18 10 C. ε 0 A π r So the field is 1.8 10 4 N/C. 1

PHY049 Spring 006 3. Two electrons are fixed ~cm apart. A third electron is shot from infinity and comes to a rest midway between the two. What was its initial velocity? a. 30 m/s b. 80 m/s c..9 10-7 m/s d. 1.0 10 5 m/s e. 0 Solution: First find the electric potential at the point between the electrons e e assuming V=0 at infinity: V = K + K d/ d / The potential energy, then, for an electron (q= -e) to be at rest there, is: e U = qv = 4K. d Thus, an electron must have a kinetic energy equal to U initially: 1 e EK = mv = 4K d 8Ke v = = 30 m/s md 4. An air filled parallel plate capacitor has a constant charge q=10-4 C on one of its plates and the voltage across the capacitor is 100 V. If the plates are pulled apart to 3 times their original separation, what potential difference in Volts now exists between the plates? a. 300 b. 100 c. 33 d. 900 e. 11 A Solution: The capacitance of a parallel plate capacitor is C = ε 0, so if the plates d are pulled apart 3 times further, the capacitance is 3 times smaller. Then since q= CV, the voltage must increase 3 times to 300 V if the charge is to remain constant. 5. Two identical light bulbs (with identical resistances) are connected in series to a constant voltage source. If the light bulbs are instead connected in parallel to the constant voltage source, what is the ratio of the brightness of a light bulb in the parallel configuration to its former brightness when connected in series? a. 4 b. 0.5 c. 16 d. e. 1 Solution: Let the light bulbs have resistance R and let the voltage be V from the source. The current through the circuit with the resistors in series is i = V /R. The brightness of one light is related to the power dissipated by one resistor, which is V P= i R=. 4R

PHY049 Spring 006 Now for the resistors in parallel, the current provided by the voltage source is i = V / R, since the effective resistance is R /. The current is split exactly in half between the two lights in parallel, so for one light i 1 = V / R. The power dissipated in V one light is thus P= i1 R=, which is 4 times bigger than before. Each light R appears 4 times brighter. 6. i 1 -> i -> a b 3Ω i 3 5Ω 3V 6 V - 5 V - d c 10Ω What is the magnitude of the current, in Amperes, across the 10 Ω resistor? a. 1/15 b. /3 c. 3/5 d. 1/3 e. 1/5 Solution.: do Kirchhoff loop equations currents in and out of point b: choose i 1 coming from the left, i going out to the right, and i 3 coming in from below: i 1 + i 3 = i as shown in the diagram in red. Do the sum of voltages in the left hand loop, starting at a., going clockwise: Eq. -3i 1 5 +3 = 0 Do the sum of the voltages in the right hand loop, starting at b., going clockwise Eq. 3-5i 6 10i + 5 =0 From eq., i 1 = - /3 From eq. 3, i = -1/15 (this is what is asked) For completeness, from the current eq., i 3 =i i 1 = +3/5 7. In a classical picture, an electron (q=1.60 10-19 C) moves in a circle of radius 0.5 10-10 m at a velocity of 6 10 5 m/s. What is the magnetic dipole moment, in units of 10-4 Am, of this orbiting electron? (Hint: current is amount of charge past some point per unit time, Δq/Δt, and the circulating electron completes one orbit every period.) a..4 b. 0.6 c. 1. d. 1.8 e. 3.0 Solution.: d=vt, so the time it takes the electron to go around its orbit is (π r=0.5 10-10 m)/v=6 10 5 m/s = 0.54 10-15 s so current = i = Δq/Δt = 1.60 10-19 C(the charge of one electron)/0.54 10-15 s = 3.06 10-4 A. magnetic dipole moment μ=nia = 1 I πr, where r=0.5 10-10 m 3

PHY049 Spring 006 so μ=.4 10-4 Am (this is somewhat different than the value given in Table 8-, but the moment we calculated was due to the electron's motion in its orbit, while the table gives the moment due to quantum mechanics (beyond the classical picture we use in Phy 049). 8. Straight wires along the radial direction carry current into and out of a circular arc of wire which subtends an angle of 30 o. If the arc of wire has a radius of 1 cm and the current is 50 A, what is the B-field at the origin, in units of 10-3 T? a. 0.6 b. 0.04 c. 1.0 d. 1.5 e. 1 Solution: B=μ 0 iφ/4πr where Φ is the angle of the circular arc of wire in radians, in this case Φ=(1/1)π, so B=4π 10-7 50 (1/1)π/(4π 0.01) = 618 10-7 =.6 mt 9.An infinitely long insulated wire carrying a current I=0A is bent into the shape shown (a straight line plus a circle of radius R=4cm with the currents in the direction shown). Find the magnitude of the magnetic field at the center of the circle. a. 4.1 10-4 T b..1 10-4 T c. 1.0 10-4 T d. 3.1 10-4 T e. 8. 10-4 T Solution : We calculate the field from both the long wire and the circle separately. From the long wire, the field at the center of the circle, a distance R away, is: μ0i 4 Bwire = = 10 T into the plane of the paper. π R From the circular wire, the field is: μ0i Barc = Φ where Φ = π 4π R μ0i 4 Barc = = 3.1 10 T into the plane of the paper R Since the magnetic field from the arc is in the same direction of that from the long wire, the total magnitude is the difference: 7 ( + 1) 4 10 ( 0)( + 1) πr π ( ) μ0i π π π 4 B= Barc + Bwire = = = 4.1 10 T 0.04 10. An inductor has L=18 H. Suppose 100 A of dc current is flowing through this inductor through a switch, which is suddenly (Δt = 0.01 s) opened. What magnitude of voltage appears across the inductor, in units of V? a. 180,000 b. 18,000 c. 1800 d. 180 e. 18 4

PHY049 Spring 006 Solution.: V = -L di/dt = (in magnitude) 18 100/0.01 = 180,000 V 11. A source of EMF fully charges a circuit in which are an inductor, L, and a 10 Ω resistor. The source of EMF is removed and is replaced by a short; the current flowing through the LR circuit takes 10 s to decay to ½ its original value. What is the value of L, in H? a. 1.44 10 b. 0.69 10 c. 0.30 10 d. 3.3 10 e. 1.44 Solution: i = i 0 e -t/ (L/R) ; for i/i 0 = 0.5, get ln0.5= - 0.6931 = -t/ (L/R), t=10s, R=10Ω, => L=1.44 *100 1. Suppose a capacitor, C = 10-3 F, is charged to 100 V. Suddenly, all of its energy is released in 0.001 s and converted into emitted isotropic electromagnetic radiation. At a distance of 1 m from this source, what is the magnitude of the electric field, E m, of the emitted electromagnetic radiation when the wave passes by? (in units of V/m) a. 550 b. 690 c. 100 d. 50 e. 69 Solution: energy stored in a capacitor is U=1/ CV = ½ 10-3 (100) = 5 J this energy is emitted as isotropic radiation in 0.001 s, so Power= Energy/time = 5 kw The intensity of the electromagnetic radiation, I, at a distance of 1 m when the pulse of radiation (over the millisec) goes by is Power/4πr where r=1 m, so I=5 10 3 /4π. But we know that I=(1/cμ 0 )E m, so E m = cμ 0 I = 3 10 8 4π 10-7 5 10 3 /4π. The factors of 4π cancel and E m = (30 10 4 ) 1/ = 5.5 10 V/m 13. Which statement about alternating current RLC circuits is completely correct? a. in a capacitor, the current leads the voltage; also, the current is in phase with the voltage in a resistor b. the current leads the voltage in an inductor c. the voltage lags the current in an inductor d. all of a, b, and c are correct. e. none of a, b, and c are correct. Solution: As we did in class in the demo multiple times, the current in a charging capacitor in a dc circuit leads the voltage, which is also true in an ac circuit. Of course, the current is in phase with the voltage in a resistor, so statement a. is correct. Statement b. is incorrect (the current lags the voltage in an inductor) and statement c. is just statement b. stated another way, so both b. and c. are incorrect. Thus d. and e. are incorrect, and only a. is correct. 5

PHY049 Spring 006 14. A uniform electric field points in the z-direction with a time-dependent value given by the expression: Ez(t) = a+bt, where a = 60 V/m and b = 1800 V/(m-s). The electric field is confined to a circular region in the x-y plane with radius r = 3 m. What is the magnitude of the magnetic field in Tesla at a point along the circumference of the circle? a. 3 10-14 b. 10-15 c. 10-13 d. 7300 e. 4.5 10-7 Solution: Use Maxwell s Law of Induction dφe B d s = με 0 0 dt de πrb = μ0ε0a dt με 0 0A de π r b rb 14 B = = = = 3 10 πr dt πrc c 15. What is the frequency, in units of Mega Hz, of an electromagnetic wave with wavelength = 0.1 m? a. 3000 b. 30 c. 3 d. 300 e. 0.3 Solution: c = λf f=c/λ = (3 10 8 m/s) / 0.1 m = 3 10 9 Hz = 3000 MHz 16. Light in air (n=1) is incident on a flat-faced cube of glass (n=1.5) at an angle of 45 o to the normal. The light is bent to angle θ from the normal in the glass. The light exits the back of the cube of glass and enters another flat-faced cube of material whose index of refraction is n 3 (no value given). The light from the glass incident on the n 3 material at angle θ from the normal in the glass is bent to angle θ 3 in this n 3 material. The light then exits this material back into air, n = 1, making an angle θ with respect to the normal in the air. In units of degrees, what is θ? a. 45 b. 30 c. 8.1 d. e. 35 6

PHY049 Spring 006 Solution: Snell's law allows the determination of θ, if we need it. Let's just set up the equations for each interface: eq. 1 n air sin45 = n sinθ eq. n sinθ = n 3 sinθ 3 since ray is straight, its angle from the vertical when it enters region 3 is also θ (rule from geometry for the angles for a parallel bisector) eq. 3 but whatever θ 3 is, we know that n 3 sinθ 3 = n air sinθ from eq., we know that n 3 sinθ 3 = n sinθ from eq. 1, we know that n sinθ just equals n air sin45, so n air sinθ (for the region below) in eq. 3 is just equal to n air sin45 from eq. 1, so θ=45 and the light exits at the same angle it entered. We saw this in a demo (the laser beam projector with 5 beams into and out of the block of Lucite on the blackboard) for one material. 17. 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 0.8m. The glass that makes the lens has an index of refraction of n = 1.5. What is the focal length of the lens? a. -1.6 m b. +1.6 m c. +0.8 m d. -0.8 m e. Solution: Note that a light ray facing a convex surface means that the surface is considered to have a positive radius of curvature. 1 1 1 1 1 1 = ( n 1) = ( 0.5) = f r1 r 0.8 1.6 f = 1.6 m A negative focal length implies a diverging lens. 18. The lens of a movie projector is able to project a 3.0m wide image on a screen 18m away from the lens. The picture on the film that is projected is.5cm wide. What is the focal length of the lens in cm? a. 15 b. 7.5 c. 10 d. 1.5 e. 0 Solution: The magnitude of the magnification is 7

PHY049 Spring 006 3.0 m = = 10s 0.05 i m= m= 10 for a real image p i 18 p = = = 0.15 m 10 1 1 1 1 1 = + = + f p i 0.15 18 f = 0.15 19. A phase difference of how much of a wavelength between two waves of equal amplitude results in fully destructive interference? a. 0.5 b. 0.0 c. 1.0 0.6 e. 0.33 Solution: if two waves of equal amplitude are out of phase by 0.5 of a wavelength, then the crest of one matches the trough of the other, and they add to give zero. 0. A double slit interference experiment finds the maximum adjacent to the center, Θ=0, maximum at an angle (measured from the center of the double slit to the viewing screen) of 0.3 o. If the wavelength of the monochromatic light used is 500 nm, what is the separation of the slits, in units of 10-6 m? a. 95 b..6 c. 0.006 d. 150 e. 0.06 Solution: From eq. 35-14, dsinθ = mλ, where for the case discussed, m=1. Thus, d=500 nm/sin0.3 o = 95,493 nm or about 95 μm 8