NYB Exam May 2004 with answers

NYB Exam May 2004 with answers - Elam @oflfle [Pmygfiw...

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Unformatted text preview: Elam @oflfle [Pmygfiw Eflefiey of! an inflow TOTAL MARKS: 100 May 21, 2004 TIME: 3 HOURS 2:00 RM. to 5:00 RM. Instructions: 1. Solve only TEN out of the following twelve problems. Ifyou do solve more ‘ than ten, indicate which one(s) you want omitted. . Formulas and constants are given on the last page of this exam. Expressions not given on this last page should be derived. Calculators are allowed. . Start a new problem on a flesh page. . Write only on the right-hand pages of your booklet. Use the left-hand pages for your rough work. . Show all your work, including diagrams, in the examvbooklet. Give your answers to three figures. . Do not tear any page(s) from the examination booklet. 9. Ifyou use more than one booklet, number your booklet. When you hand in your examination, place booklet #2, 3, etc., inside booklet #1. 10. Write your name and the name of your teacher on each booklet. Solve 10 out of the following 12 problems: 1. The charges and coordinates of two charged particles held fixed in the x—y plane are: q] = +3.00 x 10'6 c; x]: 3.50 cm, y1= 0.500 cm, and q; = —4.00 x 10—6 C; x2= 0.00 cm, y; = 0.00 cm. (a) Find the magnitude and direction of the force on (1;. To do this, first draw a diagram showing the charges and their respective positions, draw the force acting on g; and perform the calculations. (b) Find the coordinates of the point where you would locate a third charge (B = +1200 uC such that the total force on g; is zero. Show the force diagram of 432. 2. A thin glass rod is bent into a semicircle of radius a. A charge of +Q is uniformly distributed along the upper half and a charge of —Q is uniformly distributed along the lower half (see the diagram to the right). Derive an expression for the electric fieldJi‘, at point P, the center of the semicircle. Express your answer in unit-vector notation. 3. Consider two large oppositely charged conducting plates. Each plate has a surface charge density of magnitude 0'. Apply Gauss’s law to find the magnitude and direction of the electric field at points a, b and c. Use the suggested gaussian surfaces 8;, 32 or SI S3. Copy the diagram in your booklet and add to it any other useful markings. Show all the steps of the derivation and justify them. 4. Two point charges, q} = —0.120 x 10—9 C and q; x + 0.200 x 10—9 C, are separated by a distance of 5.00 cm as shown in the figure. [1.00mi A B 1110ch WA. ............. "two... I‘—— 5.00 cm —u| (a) Find the potential at point A, 1.00 cm from ql. (b) An electron is released from rest at A, and moves along the line connecting the two charges. What is its speed when it is at B, 1.00 cm fiom qz‘? Physics NYB Final Exam] May 2004 I 5. Four capacitors are connected as shown in the figure. The equivalent capacitance is 8.00 uF. (a) Find the unknown capacitance C4. Cl=6uF (b) Find the charge and the energy stored in C4. (c) If C; is a parallel plate capacitor with plates of C3= 5 [JP area 250 cm2 and has a dielectric constant K = 4.50, find the separation between the plates. 6. A light bulb is marked 60.0 W and 120V. (a) What is its resistance? (b) When plugged into a 120 V outlet, how long does it take for 2.00 coulombs of charge to pass through the bulb? (c) How long does it take for 1.00 joules to be dissipated by the bulb? (d) What is the cost per-year (365 days) if the bulb is left on continuously for the whole year? Assume the price to be 10.0 cents per kilowatt-hour. (e) Ifthis bulb is connected in series with the 120-V power supply and a IOO-uF capacitor, how long does it take to charge the capacitor fi'om q = O to q = 0.500 me? 7. Consider the circuit shown in the figure. Calculate (a) the total resistance the circuit, (b) the current through each battery, (c) the current through the 6.00-9 resistor, (d) the rate at which energy is supplied (the power output) to the remainder of the circuit by the 22.0—V battery, (e) the rate at which electrical energy is absorbed by the 10.0-V battery, and (t) the rate at which energy is dissipated by the resistances; (g) Check the power balance in the circuit, in other words, check that the energy is conserved. 8. Under steady state conditions (the capacitor has been connected for a very long time): ' (a) Find the unknown currents 11, 12, 13 and I4 in the multi100p circuit shown in the figure. (b) Calculate the charge on the capacitor. 2 Physics NYB Final Exam / May 2004 9. A singly charged, negative ion (mass 2.00x10"25kg) with kinetic energy K = 6.42x10'15 J travels undeflected in the velocity selector of a mass spectrometer. In the deflection chamber it deflects upward with a diameter of 16.0cm. The electric field in the velocity selector has an unknown direction. (a) Ifthe magnetic field is the same on both the velocity selector and the deflection chamber, find its magnitude and direction (in or out of the page, explain). (b) Calculate the magnitude of the electric field in the velocity selector. What is its direction? (Explain whether it is up or down.) (c) If another identical ion with lower velocity is fired into the velocity selector, which way would it deflect, up or down? Explain. Trajectory of the ion q E2" :2 Velocity selector Deflection chamber 10. Two hikers are reading a compass under an overhead transmission line that is 5.50 m above the ground and carries a current of 800 A in a horizontal direction from North to South. (a) Find the magnitude and direction of the magnetic field due to the power line at a point on the ground directly under it. Hint: draw a top view diagram. (b) Considering that the magnetic field of the Earth’s is parallel to the wire and its magnitude is about 0.500 ><10t1 T, will the power line affect significantly the compass ' reading? Explain. (c) Give the magnitude and the direction of the magnetic field resulting from the current in ~ the power line and the magnetic field of the Earth. Physics NY'B Final Exam / May 2004 11. We want to produce a 20.0-mT magnetic field to deflect protons with speed of 2.00 ><106 m/s. The motion of the protons is always perpendicular to the field and along a circular path. To produce this field, we use a solenoid. (a) Use Ampere‘s law to derive an expression for the magnitude of the magnetic field near the center of an "ideal" solenoid as a function of the current through its windings. Assume that the field is uniform inside the solenoid and zero outside of it. (b) Ifthe current through the solenoid is 1000 A, how many turns per meter should the solenoid have? (c) What is the magnitude of the force exerted on the proton? (Ignore gravity.) Hint: draw a diagram with a view along the axis of the solenoid. - 12. A rectangular loop with resistance 50.0 Q has 10.0 turns, each of length 40.0 cm and width 20.0 cm as shown below. The loop moves into a uniform magnetic field of magnitude 2.00 Tesla, directed out of the paper, with velocity 50.0 cm/s. Answer the following questions for the loop positions described in i), ii) and iii) below: (a) Determine the magnitude and direction of the induced current in the loop. (b) What are the magnitude and direction of the resultant magnetic force on the loop? (c) Find the external force required to keep the loop in motion. Correct esses without some sort of roof will receive no marks i) Position I: as the right side of the loop enters the magnetic field. ii) Position II: as the loop moves within the field, after it is completely in the field. iii) Position III: as the right side of the loop leaves the field. 4 Physics NYB Final Exam I May 2004 mama»ng EWM Wm 7751?. 2004 1. (a) F}, = 86.3N,8.13°abovethe+x—axis; (b)x=—7.00 cm,y=~»1.00 cm a 41; Q . 2. E=— e ' NR2 J 3. (a) Ea: 0; (b) E, =£,to then'ght ; (c) E, =£,tothe right 80 50 4. (a) VA = —62.9 V; (b) v3 = 3.71 x 106 m/s 5.'(a) C4 = 2.00 11F; (b) Q4 2 25.0 110, U; = 156 1.1]; d= 4.98 x 10‘"? m 6. (a) R = 240 9; (b) 1‘ = 4.00 s; (c) t= 0.0167 s; (d) cost = $52.6; (6) 16.6 ms 7. (a) R = 6.00 9.; (b) I = 2.00 A; (0) I1 = 1.33 A; (d) P51 = 44.0 W; (e) P82 = 20.0 W; (1) PR = 24.0 W; (g) P51 = PE; + PR 8. (a)I1= 1.38 A, I2 = —0.364 A, 13 = 1.02 A, I4 = 0; (6) Q = 66.0 11C 9. (a) E = 1.25 T, 0111: of the page ; (b) E =1.00><106 N/C, toward the top of the page (c) It would deflect down. - 10. (a) 13,", = 2.91x 10—5 T. toward the East; (b) Yes. (c) 3",, = 5.79 x10‘5 T, 302° E of N 11. (a)B= pgm’; (b) 11: 15.9 turns/In; (c)F= 6.40 x 10—1511 12. (a) i) I = 0.0400 A, in a clockwise direction. ii) There is no induced current. iii) I = 0.0400 A, in a counterclockwise direction. (b) i) Fm=—0.016f ii)F,,,=0 iii) lama—0.016? (c i F, =+0.016f 10F, =0 1191?, =+0.0161° pp PP PP Physics NYE Final Exam 1’ May 2004 ...
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NYB Exam May 2004 with answers - Elam @oflfle [Pmygfiw...

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