i( t) L i( t) 56mH 1.1A t = τ ln 1 = ln 1 ln ms

Transcription

1 Exam III PHY 49 Summer C July 16, 8 1. In the circuit shown, L = 56 mh, R = 4.6 Ω an V = 1. V. The switch S has been open for a long time then is suenly close at t =. At what value of t (in msec) will the current in the inuctor reach 1.1 A? (Figure, problem 1 PHY54exam III, Spring 8). (1) 6.67 () 1.5 (3).88 (4) 19. (5) None of these The battery is connecte irectly across the series combination of the inuctor L an resistor R. Therefore the usual LR circuit formulas apply, even though the battery is also connecte irectly across another resistor R. So the current in the inuctor is given by: i( t) = i (1 e ) τ = L / R i = V / R t / τ Solving for time we have, i( t) L i( t) 56mH 1.1A t = τ ln 1 = ln 1 ln ms i R V / R = = 4.6Ω 1 / 4.6Ω. Refer to the previous problem. What is the total energy store in the inuctor a long time after the switch is close? (1).19 J ().48 J (3).76 J (4).34 J (5) None of these After a long time, the current i(t) i, as we can see from the formula above. So therefore the energy is: U = Li = (56 1 ) = 19mJ A real inuctor has resistance because it is compose of a coil of wire with nonzero resistivity. To measure the inuctance of a coil, a stuent places it across a 1.-V battery

2 an measures a current of.7 A. The stuent then connects the coil to a 3.-V (rms) 6 Hz generator an measures an rms current of.86 A. What is the inuctance of the coil? (1).81 H ().51 H (3) 34.9 H (4) 3.6 H (5) None of these The coil has an inuctance an resistance. We can visualize the coil therefore as an inuctor an resistor in series. When we connect the DC battery across the coil, we have therefore an LR circuit. The steay state current is just i = V/R same as above. So V 1 R = = = 16.67Ω i.7 When we connect the AC battery across the coil, then we have: i rms 3 = = = Z (π 6 L).86 The last gives, 1 9 L = 1π.86 = H 4. Sunlight of intensity 134 W/m is incent normally on a 9 km 9 km rectangular sail of negligible mass attache to a spaceship of mass 1, 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 () 18 sec (3) 8.1 sec (4) 4 min (5) 4.8 hr The raiation prouces a pressure on the sail, which is: P = I / c. This gives rise to a force which accelerates the sail (spaceship). Therefore, using simple kinematics equation, an Newton s n Law, we have: 1 1 F m x = at = t m x m x mc x 8 (1)(3 1 )(1) F PA IA 6 (134)(81 1 ) t = = = = = 57s

3 5. A immer for a stage light at a theater consists of a variable inuctor, whose inuctance is ajustable between zero an L, connecte in series with a lightbulb. The electrical supply is 1 V (rms) at 6. Hz; the lightbulb is rate at 1. kw for this voltage. What L is require if you want the rate of energy issipation in the lightbulb to be variable by a factor of 1 from its upper limit of 1. kw? Assume that the resistance of the lightbulb oes not epen on its temperature. (1).11 H ().45 H (3) 1. H (4) 65 mh (5) 1 mh First we ll etermine the resistance of the lightbulb. The problem states that without the inuctor, the average power issipate by the resistor is 1kW. Therefore, P = R 1 R = = = 14.4Ω P 1 If we put the inuctor in, then the current will ecrease, an the power issipate across the resistor will therefore ecrease. So want to know what L will give rate of energy issipation equal to 1.kW/1 = 1W. So then when a inuctor at its imum value, the power issipate across the resistor will be: rms rms rms Z R ω L R + ω L V V V R P = R = R = + R R R R ω L P L = + = P ω L =.11H 6. A small light bulb is place 3 cm inse a sol cube of Lucite (n=1.5) an turne on. What is the raius (in cm) of the illuminate circle that woul be seen by an observer staning above the Lucite block? (1).68 () 3. (3) 3.35 (4) 4.5 (5) 3.4

4 The raius of the circle seen is the istance between the lightbulb an the point on the interface where light at the critical angle woul hit. Therefore, 1 1 θ c = sin ( n / n1 ) = sin (1/1.5) = 41.8 r = 3cm tan 41.8 =.68cm 7. A horizontal beam of unpolarize light is incent upon a stack of 4 polarizers with axes of polarization, in orer an measure clockwise from the vertical, at 3, 75, 1 an 18. What is the ratio of the intensity of the transmitte beam to that of the incent beam? (1).31 ().63 (3).3 (4).47 (5).63 Since the initial light is unpolarize, half the light woul get through the first polarizer. The fraction that woul get through the secon is cos α, where α is the angle between the incomming light an the polarization axis of the polarizer. Since the light coming out of the 3 eg. polarizer is now polarize at an angle of 3 eg., α woul be 45 for the secon polarizer. An so on so we en up with. 1 I = cos 45cos 45cos 6 I =.31 I An oscillating LC circuit consists of a mh inuctive coil an a 4μF capacitor. The capacitor has a potential ifference of.75 V when the current through the coil is 3 ma. What is the imum possible charge on the capacitor [or current through the inuctor]? (1) 4 x 1-6 C () 3 x 1-6 C (3) x 1-6 C (4) 8 x 1-9 C (5).75 C

5 We can use conservation of energy. The energy store in the circuit is: Q( t) U = LI( t) + CV ( t) = LI( t) + C 1 1 = + = (mh)(3ma) (4μF)(.75).μJ an this is constant for all times. Now conser the particular time when Q is at its imum value, (an I is therefore). Then we can also say 1 1 Q U = L + C 1 Q =.μj (4μF) Q =.μj (4μF) = 4.4μC 9. The average intensity of light from an incanescent light bulb is 3 mw/m on a particular surface. Assuming that the light is in the form of an electromagnetic plane wave, what is the imum magnetic fiel amplitue, Bm? (1) 5 x 1-8 T () 7 x 1-8 T (3) 3.5 x 1-8 T (4) 1 T (5) 15 T 1 1 I = B c = B c µ µ B m rms m 7 3 µ I (4π 1 )(3 1 ) 8 = = = c 3 1 5nT where we use the relationship between B rms an B m, namely B rms = B m /. 1. Light traveling horizontally enters a right prism through the hypotenuse, as shown in the figure. The inex of refraction of the prism is n=1.6. At what angle is the light eflecte from horizontal? (1) 36º () 6º (3) 19º (4) 45º

6 (5) 31º Snell s law for the first interface is: θ n 1 = sin sin θ = sin sin 5 = n Angle with top iagonal surface is = An at corner is 5. Therefore angle of ray with the vertical se of the triangle is = Therefore the angle of ray with respect to the normal to the vertical se is = 1.4. Snell s law then gives, θ 1.6 = = 1 1 sin sin An arrangement for generating a traveling electromagnetic wave in the shortwave raio region of the spectrum works as follows: an LC oscillator prouces a sinusoal current in the antenna, which generates the wave, traveling outwar at the spee of light. What is the wavelength (in meters) of the wave emitte by this system if L =.33μH an C = 45.pF? (1) 7.19 m () 1.14 m (3) 719 m (4) 114 m (5) None of these The LC circuit s current will oscillate at ω = 1/ (LC). An this will be the (angular) frequency of the EM wave as well. From the frequency we can obtain the wavelength. So, 1 1 f = = = π LC π (.33 µ H )(45 pf) 8 c 3 1 λ = = = 7.m 6 f MHz

7 1. A current of 1A is use to charge a parallel plate capacitor with square plates. If the area of each plate is.6m the isplacement current through a.3m area wholly between the capacitor plates an parallel to them is: (1).5A () 1A (3) A (4).7A (5).5A The isplacement current through a loop of area A is efine as: i Φ E t t E A Our loop is inse the capicator an has an area A of.3m. The Electric fiel, E, through the loop is parallel to the area vector, A at all points. It is also constant over the surface of the loop. But E oes change with time since the charge on the plates (which is changing with time) etermines the fiel between the plates. Therefore we may write, EA E A EA A E t ε ε t = = t t To etermine the rate of change of the electric fiel, we will relate the fiel between the plates, to the potential ifference between the plates, an then finally to the charge on the plates. So we have, E = V / = Q / C where Q is the charge on the plates, C the capacitance of the plates, an the istance between them. So now we have, Q ε A Q ε A t C C t C A = = i Now let A C be the area of the plates of the capacitor. An then we ll have, ε A A = i = i ε AC AC So our answer is:.3m (1 ).5 = A = A.6m