2009Q - AP® Physics C: Electricity and Magnetism 2009...

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Unformatted text preview: AP® Physics C: Electricity and Magnetism 2009 Free-Response Questions The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the association is composed of more than 5,600 schools, colleges, universities and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools and 3,800 colleges through major programs and services in college readiness, college admissions, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT® and the Advanced Placement Program® (AP®). The College Board is committed to the principles of excellence and equity, and that commitment is embodied in all of its programs, services, activities and concerns. © 2009 The College Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central, SAT, and the acorn logo are registered trademarks of the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. Permission to use copyrighted College Board materials may be requested online at: www.collegeboard.com/inquiry/cbpermit.html. Visit the College Board on the Web: www.collegeboard.com. AP Central is the official online home for the AP Program: apcentral.collegeboard.com. TABLE OF INFORMATION FOR 2008 and 2009 CONSTANTS AND CONVERSION FACTORS Proton mass, m p = 1.67 ¥ 10 -27 kg Neutron mass, mn = 1.67 ¥ 10 -27 kg Electron mass, me = 9.11 ¥ 10 -31 kg Avogadro’s number, N 0 = 6.02 ¥ 1023 mol-1 Universal gas constant, Electron charge magnitude, e = 1.60 ¥ 10 -19 C 1 electron volt, 1 eV = 1.60 ¥ 10 -19 J Speed of light, Universal gravitational constant, Acceleration due to gravity at Earth’s surface, c = 3.00 ¥ 108 m s G = 6.67 ¥ 10 -11 m 3 kgis2 R = 8.31 J (mol iK) g = 9.8 m s2 Boltzmann’s constant, k B = 1.38 ¥ 10 -23 J K 1 unified atomic mass unit, Planck’s constant, Vacuum permittivity, Coulomb’s law constant, k = 1 4 p Vacuum permeability, 1 u = 1.66 ¥ 10 -27 kg = 931 MeV c 2 h = 6.63 ¥ 10 -34 J is = 4.14 ¥ 10 -15 eV is hc = 1.99 ¥ 10 -25 J im = 1.24 ¥ 103 eV i nm 0 0 = 8.85 ¥ 10 -12 C2 N im 2 = 9.0 ¥ 109 N im 2 C2 m0 = 4 p ¥ 10 -7 (T im) A Magnetic constant, k ¢ = m0 4 p = 10 -7 (T im) A 1 atmosphere pressure, meter, kilogram, second, ampere, kelvin, m kg s A K mole, hertz, newton, pascal, joule, mol Hz N Pa J 1 atm = 1.0 ¥ 105 N m 2 = 1.0 ¥ 105 Pa watt, coulomb, volt, ohm, henry, W C V W H farad, tesla, degree Celsius, electron-volt, F T ∞C eV UNIT SYMBOLS PREFIXES Factor Prefix Symbol 10 9 106 103 10 -2 VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES q 30 0 37 45 53 60 90 giga mega kilo centi milli micro nano pico G M k c m sin q cos q tan q 0 1 0 12 32 33 35 45 34 22 22 1 45 35 43 32 1 0 12 3 • 10 -3 10 -6 10 -9 m n p 10 -12 The following conventions are used in this exam. I. Unless otherwise stated, the frame of reference of any problem is assumed to be inertial. II. The direction of any electric current is the direction of flow of positive charge (conventional current). III. For any isolated electric charge, the electric potential is defined as zero at an infinite distance from the charge. -2- ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2008 and 2009 MECHANICS ELECTRICITY AND MAGNETISM u = u0 + at x = x0 + u0 t + 12 at 2 u 2 = u0 2 + 2 a ( x - x0 )  F = Fnet = ma F= dp dt a F f h I J K k J = Ú F dt = Dp p = mv Ffric £ m N W= K= P= ÚF ∑ dr 1 mu 2 2 dW dt P=Fv DUg = mgh = = = = = = = = = L= m= N= P= p= r= r= T= t= U= u= W= x= m= q= t= w= a= acceleration force frequency height rotational inertia impulse kinetic energy spring constant length angular momentum mass normal force power momentum radius or distance position vector period time potential energy velocity or speed work done on a system position coefficient of friction angle torque angular speed angular acceleration F= E= 1 q1q2 4p 0 r 2 F q ∑ A B C d E ÚE dA = dV dr Q 0 E=V= 1 4p 0  rii i q = = = = = e= F= I= J= L= = n= N= UE = qV = C= C= Q V k 0A 1 4p 0 q1q2 r d Cp =  Ci i 1 1 = Cs i Ci I= dQ dt 1 1 QV = CV 2 2 2 P= Q= q= R= r= t= U= V= u= r= area magnetic field capacitance distance electric field emf force current current density inductance length number of loops of wire per unit length number of charge carriers per unit volume power charge point charge resistance distance time potential or stored energy electric potential velocity or speed resistivity fm = magnetic flux k = dielectric constant ac = u2 = w2r r Fs = - kx t=r¥F  t = t net = I a I = Ú r 2 dm =  mr 2 Uc = ÚB ∑ d = m0 I Us = T= 12 kx 2 2p 1 = w f r R= A E = rJ I = Neud A dB = F= m0 I d ¥ r 4p r3 ¥B ÚI d rcm =  mr  m u = rw Ts = 2 p Tp = 2 p FG = - m k g ˆ r Bs = m0 nI fm = Ú B ∑ dA V = IR Rs = L = r ¥ p = Iw K= 12 Iw 2  Ri i e i =- Gm1m2 r2 1 = Rp ÂR i 1 d fm dt dI dt e = -L w = w0 + at q = q0 + w0 t + 12 at 2 UG = - Gm1m2 r P = IV FM = qv ¥ B UL = 12 LI 2 -3- ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2008 and 2009 GEOMETRY AND TRIGONOMETRY Rectangle A = bh Triangle CALCULUS A= Circle 1 bh 2 A = pr 2 C = 2pr Parallelepiped V = wh Cylinder V = pr 2 A= C= V= S= b= h= = w= r= area circumference volume surface area base height length width radius df d f du = dx du dx dn ( x ) = nx n -1 dx dx (e ) = e x dx d (1n x ) = 1 dx x d (sin x ) = cos x dx d (cos x ) = - sin x dx S = 2pr + 2pr 2 Sphere Úx Úe n dx = 1 x n + 1 , n π -1 n +1 x dx = e x V= 43 pr 3 Ú dx = ln x x S = 4pr 2 Right Triangle Ú cos x dx = sin x Ú sin x dx = - cos x c q b 90° a a 2 + b2 = c2 sin q = cos q = tan q = a c b c a b -4- 2009 AP® PHYSICS C: ELECTRICITY AND MAGNETISM FREE-RESPONSE QUESTIONS PHYSICS C: ELECTRICITY AND MAGNETISM SECTION II Time— 45 minutes 3 Questions Directions: Answer all three questions. The suggested time is about 15 minutes for answering each of the questions, which are worth 15 points each. The parts within a question may not have equal weight. Show all your work in this booklet in the spaces provided after each part, NOT in the green insert. E&M. 1. A spherically symmetric charge distribution has net positive charge Q0 distributed within a radius of R. Its electric potential V as a function of the distance r from the center of the sphere is given by the following. 2 Q0 È r˘ -2 + 3 for r < R 4p 0 R Í R˙ Î ˚ Q0 V (r ) = for r > R 4 p 0r V (r ) = () Express all algebraic answers in terms of the given quantities and fundamental constants. (a) For the following regions, indicate the direction of the electric field E(r) and derive an expression for its magnitude. i. r < R ____ Radially inward ii. r > R ____ Radially inward ____ Radially outward (b) For the following regions, derive an expression for the enclosed charge that generates the electric field in that region, expressed as a function of r. i. r < R ii. r > R (c) Is there any charge on the surface of the sphere (r = R) ? ____ Yes ____ No If there is, determine the charge. In either case, explain your reasoning. ____ Radially outward © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. GO ON TO THE NEXT PAGE. -5- 2009 AP® PHYSICS C: ELECTRICITY AND MAGNETISM FREE-RESPONSE QUESTIONS (d) On the axes below, sketch a graph of the force that would act on a positive test charge in the regions r < R and r > R. Assume that a force directed radially outward is positive. © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. GO ON TO THE NEXT PAGE. -6- 2009 AP® PHYSICS C: ELECTRICITY AND MAGNETISM FREE-RESPONSE QUESTIONS E&M. 2. A 9.0 V battery is connected to a rectangular bar of length 0.080 m, uniform cross-sectional area 5.0 ¥ 10 -6 m 2 , and resistivity 4.5 ¥ 10 -4 Wim , as shown above. Electrons are the sole charge carriers in the bar. The wires have negligible resistance. The switch in the circuit is closed at time t = 0 . (a) Calculate the power delivered to the circuit by the battery. (b) On the diagram below, indicate the direction of the electric field in the bar. Explain your answer. (c) Calculate the strength of the electric field in the bar. A uniform magnetic field of magnitude 0.25 T perpendicular to the bar is added to the region around the bar, as shown below. (d) Calculate the magnetic force on the bar. (e) The electrons moving through the bar are initially deflected by the external magnetic field. On the diagram below, indicate the direction of the additional electric field that is created in the bar by the deflected electrons. (f) The electrons eventually experience no deflection and move through the bar at an average speed of 3.5 ¥ 10 -3 m s . Calculate the strength of the additional electric field indicated in part (e). © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. GO ON TO THE NEXT PAGE. -7- 2009 AP® PHYSICS C: ELECTRICITY AND MAGNETISM FREE-RESPONSE QUESTIONS E&M. 3. A square conducting loop of side L contains two identical lightbulbs, 1 and 2, as shown above. There is a magnetic field directed into the page in the region inside the loop with magnitude as a function of time t given by B (t ) = at + b , where a and b are positive constants. The lightbulbs each have constant resistance R0 . Express all answers in terms of the given quantities and fundamental constants. (a) Derive an expression for the magnitude of the emf generated in the loop. (b) i. Determine an expression for the current through bulb 2. ii. Indicate on the diagram above the direction of the current through bulb 2. (c) Derive an expression for the power dissipated in bulb 1. Another identical bulb 3 is now connected in parallel with bulb 2, but it is entirely outside the magnetic field, as shown below. (d) How does the brightness of bulb 1 compare to what it was in the previous circuit? ____ Brighter ____ Dimmer ____ The same Justify your answer. © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. GO ON TO THE NEXT PAGE. -8- 2009 AP® PHYSICS C: ELECTRICITY AND MAGNETISM FREE-RESPONSE QUESTIONS Now the portion of the circuit containing bulb 3 is removed, and a wire is added to connect the midpoints of the top and bottom of the original loop, as shown below. (e) How does the brightness of bulb 1 compare to what it was in the first circuit? ____ Brighter ____ Dimmer ____ The same Justify your answer. END OF EXAM © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. -9- ...
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This note was uploaded on 02/23/2011 for the course PHYSICS 101 taught by Professor Iqbal during the Spring '08 term at The University of British Columbia.

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