v 2 = v a(x x 0 ) and v = 0 a = v2 0 2d = K md F net = qe = ma E = ma e = K ed = V/m dq = λ ds de r = de cosθ = 1 λ ds = r dθ E r =

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1 Physics 11 Honors Final Exam Spring 003 Name: Section: Closed book exam. Only one formula sheet (front and back side) can be used. Calculators are allowed. Use the scantron forms (pencil only!) for the multiple choice problems. Circle the answers on the examination sheet as well, and return it together with the scantron form. Use the back of these pages, or attach your own pages with solutions for problems that require calculations. The 14 multiple-choice problems are 3.6 points each. Partial credit (up to points) will be given for multiple-choice problems, if you provide a detailed solution on the examination sheets. The six work-out problems are 10 points each. The nominal number of points is 100. Clearly print your first and last name and indicate your section number on both the scantron form and the examination sheet. Good luck! Multiple-Choice Problems Problem 1: The electrons in a particle beam each have a kinetic energy of 1.5 kev. What is the magnitude of the electric field that stops these electrons in a distance of.8 cm? A) V/m B) V/m C) V/m D) V/m K = 1 mv 0 v 0 = K m v = v 0 + a(x x 0 ) and v = 0 a = v 0 d = K md F net = qe = ma E = ma e = K ed = V/m E) V/m Problem : A thin glass rod is bend into a semicircle of radius r = 14.0 cm. A charge of µc is uniformly distributed along the upper half, and a charge of 7.50 µc is uniformly distributed along the lower half. Find the magnitude and direction of the electric field E at point P, the center of the semicircle! A) V/m (down) B) V/m (down) C) V/m (down) D) V/m (left) E) V/m (up) dq = λ ds de r = de cosθ = 1 λ cosθ ds 4πε 0 r λ ds = r dθ E r = cosθ r dθ 45 4πε 0 r λ [ ] +45 (4q/πr) = sinθ 4πε 0 r = q = 45 4πε 0 r π ε 0 r E = E + + E = E r ĵ = ( V/m) ĵ

2 Problem 3: A point charge q p = +q is placed inside a spherical conducting shell with inner radius a and outer radius b. The total charge of the conductor is Q = 4q. What are the surface charges on the inner surface q a and outer surface q b? A) q a = q and q b = 3q B) q a = 3q and q b = 1q q a = q p = q q b = Q q a = 4q ( q) = 3q C) q a = 0 and q b = 4q D) q a = +q and q b = 4q E) q a = q and q b = 4q Problem 4: Two protons are fixed 8.0 mm apart (m p = kg). Another proton is shot from infinity (V = 0!) and stops midway between the two protons. What is the protons initial speed? A) 6.8 m/s B) 8.87 m/s C) 1.6 m/s D) 7.8 m/s K i = ( U f 1 m pv e ) = ev = 4πε 0 (d/) v = e = 6.8 m/s πε 0 m p d E) 39.4 m/s Problem 5: A parallel-plate capacitor is charged with a battery (E = 1.0 V) to the maximum charge q 0. Suppose that the capacitance in the absence of a dielectric is C 0 = 4.8 pf. If the battery is disconnected and a slab of Pyrex (κ = 4.7) is inserted between the plates, what is the change in potential energy U 0 U stored in the capacitor? A) 0 pj B) 7 pj C) 346 pj D) 544 pj E) 180 pj C = κε 0A d = κc 0 and q 0 = E C 0 U 0 = q 0 and U = q 0 C 0 C = q 0 U 0 U = U 0 U 0 κ = U 0 ( 1 1 κ = U 0 κc 0 κ ) = E C 0 ( 1 1 κ ) = 7 pj

3 Problem 6: Aluminum (ρ Al = Ωm) and copper (ρ Cu = Ωm) wires of equal length are found to have the same resistance. What is the ratio of their radii r Al /r Cu? A) 0.61 B) 0.78 C) 1.00 D) 1.8 E) 1.63 R Al = ρ Al = ρ Al A Al πr Al R Al = R Cu ρ Al r Al r Cu = ρal ρ Cu = 1.8 πr Al and R Cu = ρ Cu = ρ Cu A Cu = ρ Cu πr Cu πr Cu Problem 7: All the sides of a cube are made of 1-Ω resistors. What is the equivalent resistance between the two endpoints of one of its body diagonals? A) 1/3 Ω B) /3 Ω R eq = ( ) (3 ) 1 1 ( ) (6 ) 1 1 ( ) (3 ) = 5 1 Ω 1 Ω 1 Ω 6 Ω C) 5/6 Ω D) 1/5 Ω E) 37/6 Ω Problem 8: All the resistors have a resistance of 5 Ω and all the ideal batteries have an emf of.5 V. At what rate is the energy dissipated in the resistor R? A).00 W B) 4.00 W C) 5.00 W D) 6.00 W 0 = E E E E ir i = E R P = i R = 4E R = 5.0 W E) 11.5 W

4 Problem 9: What is the required radius of a cyclotron designed to accelerate protons to energies of 4.0 MeV using a magnetic field of 4. T? A) 3.9 mm B) 16.9 cm C) 3.8 cm D) 70.8 cm E).97 m K = 1 mv v = K m F net = q v B = m a F net = qvb = ma = m v r r = mv Km qb = = 16.9 cm qb Problem 10: A length of wire is formed into a closed loop of 1/4-circles and 3/4-circles with radii r a = 0.4 m > r b = 0.3 m > r c = 0. m. What are the magnitude and direction of B at point P, if the current is i = 0. A? A) 0.0 B) 0.50 µt C) 0.50 µt D).0 µt E).0 µt i P B a = µ 0iφ a = µ 0i(3π/) 4πr a 4πr ( ) a µ0 iφ b B b = = µ 0i(π/) 4πr b 4πr b B c = µ 0iφ c = µ 0i(3π/) 4πr c 4πr c B = B a + B b + B c = µ 0i = 0.50 µt 8 = 3µ 0i 8r a = µ 0i 8r b = 3µ 0i 8r ( c 3 3 r b r a r c ) Problem 11: At t = 0, an emf of 475 V is applied to a coil that has an inductance of 0.64 H and a resistance of 36.0 Ω. How long does it take the potential energy of magnetic field to reach 0% of its maximum value? A) 10.6 ms B) 14.3 ms C).0 ms D) 381 ms E) 593 ms U = 0.U 1 i = 0. 1 i i = 1 5 i i = E R and i = E R ( ) 1 e t/τ E 5R = E R ( ) 1 e t/τ t = R ln ( ) = 10.6 ms

5 Problem 1: An AC generator consists of 1 turns of wire, each of area A = m, and the total resistance of the wire is 5 Ω. The loop rotates in a 0.17 T magnetic field at a constant frequency of 60.0 Hz. Find the maximum induced emf! A).81 V B) 9.3 V C) 4.9 V E = N dφ B = NAB d (cosωt) = πnab f sinωt dt dt E max = πnab f = 111 V D) 17.6 V E) 111 V Problem 13: In the circuit E = 10 V, R 1 = 30 Ω, R = 0 Ω, and R 3 = 10 Ω. What is the current i 1 a long time after the switch has been closed? A) 0.0 A B).4 A C) 3.3 A R eq = R 1 + R R 3 R + R 3 = 36 3 Ω i 1 = E R eq = 3.7 A D) 8.0 A E) 18 A Problem 14: A generator supplies 60 V to the primary coil of a transformer of 33 turns. If the secondary coil has 00 turns, what is resistive load in the secondary circuit, if the average rate of energy consumption of the generator is 48 kw? A).0 mω B) 0.36 Ω C) 1.5 Ω D).8 Ω V s = V p N s N p and I s = P avg V s R s = V s = V s = V p Ns I s P avg NpP =.75 Ω avg E) 98 Ω

6 Work-Out Problems Please indicate your final answers and make sure that all the steps of your solution are shown! Problem A: Two parallel plate capacitors differ only in the spacing between their (very thin) plates. One, AB, has a spacing of 5 mm and a capacitance of 0 pf, the other, CD, has a spacing of mm. Plates A and C carry charges of +1 nc, while B and D each carry 1 nc. a) What are the potential differences V AB and V CD after the capacitor CD is slid centrally between and parallel to the plates of AB without touching them? b) Provide a sketch before and after the move including, electric field lines, equipotential surfaces, and the values of the surface charges! c) Would it make a difference if CD were not centrally placed between A and B? C = ε 0A d 5(0 pf) C AC = x C 1 d and C CD = 100 pf = 50 pf and C DB = V CD = nc = 40 V 50 pf induced charges! V AB = (1 nc)x (1 nc)(3 x) V AC +V CD +V DB = + 40 V pf 100 pf = (10 V)x + 40 V + (10 V)(3 x) = 70 V 100 pf 3 x x[mm]

7 Problem B: Two (non-physics) students, A and B, living in neighboring college rooms, decided to economize by connecting their ceiling lights in series. They agree that each installs a 100 W bulb in their rooms and that they would pay equal shares of the electricity bill. However, both decide to try to get better lighting at the other s expense. A installed a 00 W bulb and B installed a 50 W bulb. Who gets the better deal? V R A = and R B = V and R fair = V (00 W) (50 W) (100 W) ( V V i = i A = i B = = V R A + R B (00 W) + V ) 1 ( ) 1 (00 W)(50 W) (40 W) = = (50 W) V (00 W) + (50 W) V ( ) (40 W) P A = i V R A = V (00 W) = 8 W ( ) (40 W) P B = i V R B = V (50 W) = 3 W ( ) (50 W) P fair = i fair R V fair = V (100 W) = 5 W Student B is clearly the winner. While receiving 3 W, student B only pays for (3 W + 8 W)/ = 0 W.

8 Problem C: In the following circuit, the resistance of the resistor R is variable. a) For what value of R will the ideal battery transfer energy to all resistors at a rate of 60.0 W? b) Find the corresponding value of the potential difference across the 1.0 Ω resistor! c) For what value of R will the ideal battery transfer energy to all resistors at the maximum rate? d) For what value of R will the ideal battery transfer energy to all resistors at the minimum rate? e) What are those rates? P = E (1 Ω)(4 Ω)R and R eq = 7 Ω + R eq (1 Ω)(4 Ω) + (1 Ω)R + (4 Ω)R = R R P = 60 W = E R eq (solve for R) R = 19.5 Ω R eq = 9.6 Ω V 1 Ω = E V 7 Ω = E E R eq (7 Ω) = 6.5 V P max P 1 R eq minimize R eq R = 0 P min P 1 R eq maximize R eq R = R eq,max = 7 Ω Ω = 10 Ω P min = E R eq,max = 57.6 W R eq,min = 7 Ω + 0 Ω = 7 Ω P max = E R eq,min = 8.3 W in [Ω]

9 Problem D: One end of a horizontal track of gauge l and negligible resistance, is connected to a capacitor of capacitance C charged to a potential difference V 0. The inductance of the assembly is negligible. The system is placed in a homogeneous magnetic field B as shown in the figure. A frictionless conducting rod of mass m and resistance R is placed perpendicular on the track. a) Use enz law to derive the polarity of the capacitor such that the rod is repelled from the capacitor when the switch is turned over! b) What is the initial acceleration of the rod? c) What is the maximum velocity of the rod? (Hint: What does this mean for the potential difference across the capacitor V C and across the rod V R? Integrate the equation of motion (Newton s nd law)!) d) What is the minimum charge of the capacitor? e) Under what condition is the efficiency η = K/U C of this electromagnetic rail gun maximal? f) What happened to the energy initially stored in the capacitor? S C R B l V m F net = i B to the right i (down) top plate of capacitor is positively charged! F net = i l B = m a a = ilb m = V 0lB mr E = dφ B dt F net = ma = m dv dt v max = lb ( CV 0 q ) min m lbc = lb dx dt = lbv max = V C = q min C dq = ilb = Bl dt v max = q min lbc mv max = lb(q 0 q min ) V max = BCV 0 m + l B C q min = lbv max C = l B C V 0 m + l B v max V 0 K V0 C ( 1 m η = mv max 1 CV 0 = lb C + lb ) C 0.5 (geometric/arithmetric mean) m η max m = Cl B

10 Problem E: A series RC circuit has R = 515 Ω, =.34 H, C = 6.5 µf, ω = 30 s 1, and E m = 10 V. a) Determine the inductive reactance! b) Determine the capacitive reactance! c) Determine the impedance! d) What is the maximum current in the circuit? e) Find the phase angle between the current and the voltage! f) What is the average power in the circuit? g) Find the maximum voltage across each element! h) Provide a sketch that shows the instantaneous voltages across each element and the instantaneous current! X = ω d = 749 Ω X C = 1 ω d C = 500 Ω Z = R + (X X C ) = 57 Ω I = E m Z = 0.1 A (red) φ = tan 1 X X C = 5.8 R P avg = E rms I rms cosφ = 11.4 W V R = IR = 108 V (green) V C = IX C = 105 V (blue) V = IX = 157 V (orange) E m, v R, v, and v C [V] ω[rad] i [A]

11 Problem F (Extra Credit): Some relations between electrons and magnetism can be explained in terms of the loop model for electron orbits. a) What is the spin magnetic dipole moment? b) What is the orbital magnetic dipole moment? c) Is it possible to directly measure the spin magnetic dipole moment or the orbital magnetic dipole moment? d) Describe in your own words the loop model for electron orbits! e) Use the loop model to explain what happens if a diamagnetic material is placed in a non-uniform magnetic field! f) Provide a sketch of a magnetization curve for a ferromagnetic material and explain the term hysteresis! µ s = e m S and S z = m s h π, m s = ± 1 µ s,z = e m S z = ± eh 4πm = ±µ B µ orb = e m orb and orb,z = m l h π, µ orb,z = m l eh 4πm = m lµ B i = dq dt = πr/v m l = 0,±1,±,...,±(limit) and l = m( r v) l = mrv µ orb = ia = πr/v πr = evr = e m orb oop model for electron orbits Chapter 3-4! Diamagnetism: A diamagnetic material placed in an external magnetic field B ext develops a magnetic dipole moment directed opposite B ext. If the field is non-uniform, the diamagnetic material is repelled from a region of greater magnetic field toward a region of lesser field. Paramagnetism: A paramagnetic material placed in an external magnetic field B ext develops a magnetic dipole moment in the direction of B ext. If the field is non-uniform, the paramagnetic material is attracted toward a region of greater magnetic field from a region of lesser field. Ferromagnetism: A ferromagnetic material placed in an external magnetic field B ext develops a strong magnetic dipole moment in the direction of B ext. If the field is non-uniform, the ferromagnetic material is attracted toward a region of greater magnetic field from a region of lesser field. The lack of retraceability of a ferromagnetic magnetization curve is called hysteresis. Before you turn in your quiz, please double-check you multiple-choice answers. Are they marked correctly on the scantron form? Don t forget to use the correct units in the work-out problems!

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