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Exam2aQ

Fundamentals of Physics, (Chapters 21- 44) (Volume 2)

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Unformatted text preview: JUL-24-2008 14:24 PHYSICS DEPT/BOSTON UNIV +1 61'? (353 9393 P.001 Please 0,44 P54 'CL‘G‘flc" 30 PY212 Name Discussion Time Exam 2 m (20) 1. Multiple Choice: Please circle the correct answer: (5) a. A wire of current is coming at you between two magnetic poles as shown. In which direction will the beam deflect? includegraphicslscale=0.25]mc1.jpg (5) b. Charged particle beam deflects to the left when entering a region of uniform magnetic field as shown. What is the charge of these particles? at 1. positive ii. negative iii. neutral iv. no way to tell (5) c. 1 Tesla (T) is equivalent to i. 1 N-m/A ii. 1 N-A/m iii. 1N/A-m iv. 1 N/C-m (5) (:1. Which of the following statements is n_ot true? i. A magnetic field exerts a force on electric current. ii. An electric current creates a magnetic field. iii. Any electric charge creates a magnetic field. JUL-24-ZOOB 14:24 PHYSICS DEPT/BOSTON UNI);r +1 61'? (3513 3333 P.002 2 (25) 2. In the figure, R1 = 6.00 Q, 122 2: 18.0 0, and the ideal battery has emf 6’ = 12.0 V. Give the following values: (5) a. The equivalent resistance in the circuit (5) b. What is the potentiai across R1? (5) c. The current 2'1 (5) d. The direction of 2'1 (5) e. How much energy is dissipated in the circuit per minute? JUL-Ztl-ZOOB 14:26 PHYSICS DEPT/BOSTON UNIV +1 61'? (353 3333 P.003 3 (3D) 3. In the figure, a charged particle moves into a region of uniform magnetic field 3, goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends 130 us in that region. (5) a. Is the charge positive or negative? (10) b. What is the magnitude of 5;? (10) c. If the particle is sent back through the magnetic field (along the same initial path) but with 2.00 times its previous kinetic energy, how much time does it spend in the field during the trip? (5) d. What kind of instrument could you build using the ideas learned here? JUL-24-2008 14:26 PHYSICS DEPT/BOSTON UNIV +1 61'? (353 9393 P.004 4 (10) 4. A straight conductor carrying current 2' = 5.0 A splits into identical semicircular arcs as shown in the figure. What is the magnetic field at the center 0 of the resulting circular loop? (15) 5. A ZOO-turn) solenoid having a length of 25 cm and a diameter of 10 cm carries a current of 0.29 A. (1 D) a. Calculate the magnitude of the magnetic field B” inside the solenoid. (5) b. What is the magnitude of the magnetic dipole moment if of the solenoid? JUL-24-2008 14:26 PHYSICS DEPT/BOSTON UNIV +1 61'? (353 9393 P.006 5 (20) 6. In the figure, a 120—turn coil of radius 1.8 cm and resistance 5.3 Q is coaxial with a solenoid of 220 turns/cm and diameter 3.2 cm. The solenoid current drops from 1.5 A to zero in the time interval At : 25 ms. (10) a. What is the total induced emf in the coil? (10) b. What is the curreht induced in the coil during At? 9393 P.006 JUL-24-2008 14:26 PHYSICS DEPT/BOSTON UNIV .- fl 617 853 ." ) Coulomb’s Lawi F = 143$; 60 = 8.85 x 10“12 (Sm/N - m2 Electric Field: E 2 ~51; For a. point charge E == 515%? Diodes: peqd; e = 13x E; U a “13" E‘ Electric Flux: e a f E - (oi “ Gauss’ Law: it = 1;? {or equivalently ea 39 E"- dfl = qmc Electric Potential: AU 2-: —-W; AV = ”lg; .. Potential due to a. point charge: V €353? ‘(V = 0 at infinity) V: ~ v; = — If e - are ' E3 = --%—:— etc. Capacitance: q =-- CV . Parallel plate capacitor C == 5555-; Cylindrical capacitor C = 271mm; Spherical-capacitor C : firefi- Capacitors in parallel (3".3r =2 01 + C; + ...; Capacitors in series 31—; = 31—: —_}- 51; + Potential Energy of a capacitor: U 2—- 53; _ _ Resiétances and Current: Ohm’s Law: VL-ziR ; Resistance R = %’ Current density finefid; E ; pf Power (resistive dissipation) P=iV Resistances inSeriee: R6,, 2R1 +Rg+...;Resistancwinpa.rallel 4—— - -1—+ «3}:+... Junction Rule: The sum of the currents entering any junction must be equal to the sum of the currents leaving that junction. ‘ RC Circuits: time constant 1' = R0 Charging capacitor: .9; = 06 (1 - e"*/RC') ; i = g 2 fired/RC discharging capacitor: q =_q0e“*/RC ; i z 33% = "%e-t/Ro Magnetic Force on a charged particle: F3 = 91'} x E Circulating charged particle in a magnetic field: r z: 3%.“. .. .. .. Magnetic Force on a current carrying wire: F5 = iL x B; dFB :5 x B Torque on a current carrying coit- i‘ = it x f)"; ,u. = NiA Bio-Saran Law: dfi = g ing ; p9 = 4r x 10‘7T - m/A Magnetic field of an infinitely long wire: B = gig; . ‘ Force between parallel wires carrying currents: FM 2 ébLBasinQDo = Ifli—ig‘i Ampere’s law: f 3' - :13 == pole,” Magnetic field of an ideal solenoid: B = min Faraday’s Law of Induction: a =,.—N Lin 3 = f E - d5 " ‘> " x . Self-Induction: L = 13?; For a solenoid: % = nonzA; 6' == ~11 % Series RL Circuits: i = %(1 —— (rt/'1') (rise of current); i = i0 3“!“ (decay of current); TL == L/R u . 2 Energy Dena-mes: 11.3 = 60%“; “B = 5%; LC Oscillations: U3 2 3%; UB = 92*: q = Qcos(wt+ ¢); 3’ = -—wQsin(wt+ (g5) _. 1 w " 7L0” : AC Circuits: 8 = 8m sin wdt ; i:- I sin (wdt— 45) Xo=m1§ ;XL nde ; 2:..— WRZ-i—(XL-w—XHOF, tafl¢= Knfiafg ”03¢: 1% TOTAL P.006 ...
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