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Unformatted text preview: Elam @oﬂﬂe
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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 righthand pages of your booklet. Use the lefthand 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 unitvector 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 peryear (365 days) if the bulb is left on continuously for the whole
year? Assume the price to be 10.0 cents per kilowatthour. (e) Ifthis bulb is connected in series with the 120V power supply and a IOOuF 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 6.009 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 10.0V 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 20.0mT 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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 MS.SIMPSON
 Electric charge, power supply, electrical energy, 15 J, 5.50 m, $52.6

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