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Unformatted text preview: EXAM #13 Fall 2004 // :Melﬁnq : gaff): Moteﬂc ﬁzp—r—gﬁem Instructions: For the appropriate problems circle the letter corresponding to the correct answer. If you are sure that a distance of r = 0.8a—from the center. {1 Lim L :
Answer: (a) 1.9 x10‘7T. (b) 3.1 x 1071“. @341; x10'7T. ( ) 6.9 x10'7T. 7 (e) 9.0 x 10'7 T. (t) . Direction: 0, 0c LOISC
2. A metal conductor that carries a current i is bent in the shape of an L as shown. What is the magnetic k ——4 w 7
ﬁeld at point P, a distance of k meters from the bend in the wire. ? I' (Hint: use the law of Biot and Savart.) ( 2; m2+ K2 A : = \12 2) b =_“:i._’mm
nswer (a) B ln(m+ m +k B 4n<m2+k2)1/JJW?T \
noim _ Hold
® B=____.ﬂ. (d) Bm—kln(k+m) (e) B=—_
4nE(rn2+k2)1/2 ~ . 4mm Direction: [n+0 page. {3); w
LlTr r 1 o 3. Two infinitely long straight wires d and e pierce the plane of the paper as shown. I. /d A I ll m 2at K1 If the magnetic ﬁeld at point G is zero, what must be true about the , { in currents in the wires? (Note: Point G is exactly midway between d and e.) r J ‘ l ' L Answer: (a) The magnitude of the current in wire d must be exactly half of What it is v , ' f G re e.
(b) The currents in wires (1 and e must be in opposite directions. " * .... I 2
The currents in wires d and e must be in the same direction. ' A """" .. ) The currents in wires d and e must be in the opposite directions, and the magnitude of the ﬁ current in wire e must be exactly half what it is in wire d. I 1.,” ' é ‘ © The currents in wires d and e must be in the same direction, and the magnitude of the current in wire e must ~
be the same as in wire d. o x n A: no correct answer is shown, you should write your answer in space (1) . Some questions require ﬁlling
the blanks. Note that some problems/questions havze more than one (1) correct answer. For thosggroblems/questions _
, on should give all the correct answers. , V 9 TV V: [5 1 q (a .,ﬁ; 477x '0 "M//i)(€lv(IP‘) : 438%”) (in.th 36* LL? FVDFQHDVMH f3 L Tram 1115).. .i L I: 1 1. The ﬁgure shows a cross section of a long solid cyf'mdrical conductor whos'e‘ ' ' / ,
radius is a = 5.0 meters that carries a uniform current of 15.0 Amperes. Using [CF I,
' t ‘ th t' fId t . <7 Ampere 3 Law, de ermlne e magne 1c re a ‘30 H L L ' 7; L: T 4. A long straight wire is in the plane of a rectangular conducting loop. The straight wire initially carries a constant current i in the direction shown. Later, while the current in F) x >4
the wire is being reduced, the induced current in the rectangular loop is ind) (Wu/nime I X 3‘ Y X X I
Answer: (a) zero. (b) straight do 11, into the paper. (0) counterclockwise. +9 0 ﬁlm X y
(d) straight up, out of the paper. clockwise. CU/ CHM 599' ‘X “l t ‘_.m0t ¥;§‘_XXX’<
“Q04 _ 5. Two coils are shown in the ﬁgure. Circle the letter corresponding
to the correct statement(s). *1 ,_ ,
Answer: (a) The current in the battery goes through the galvanometer. g ; After switch “S” has been closed for a long time it is suddenly opened. There is a ‘1 I ‘3'“)
current in the galvanometer while switch “S” is being opened. “" There is a steady nonzero reading on the galvanometer as long as the switch “S” remains open. A motional (i.e., induced) emf is generated in the rectangular loop while “S” is being closed.
/(,e)/ Since the two loops are not connected the current in the galvanometer is always zero. 6. Four of the following equations are known as the Maxwell's equations. _ V
(a) [ﬁBods = yoienc + and, (b) [ﬁEodA = i , (c) muodA = o , (d) mEds =  mg?) , (e) id :5. (“‘23) Match the equations above with the information below by writing the letter of the equation in the space by the correct statement(s).
Q This is a ﬁctitious current (iIe., as the current between the plates of a capacitor). 7' t ) Suggests that electric charges
can be of two types having the same magnitude of charge. k 2 Suggests that isolated electric charges can be found in nature.
electric charges may be found in combination such that the net charge of the combination is z ro, hence suggesting
opposite charge polarities. L Ampere—Maxwell's Law. _ Gauss's Law for magnetism: Gauss‘s Law for
elec rostatics.’ Farraday‘s Law. . Predicts that no isolated magnetic monopoles exist in nature,
A changing magnetic ﬁeld produces an electric ﬁeld. 'Q A changing electric‘ﬁeld produces a magnetic ﬁeld. 5 1.
2E . g.—  1% :
0 1
Q, A  M l 7. A conducting rod AB slides with speed “v” to the top on africtionless “U” shaped
conductor in a uniform magnetic ﬁeld B, as shown. The magnetic ﬁeld is directed as ,5
shown. As the rod moves, the induced current in it moves
Answer: @ﬁom B to A, in the rod. (b) from A to B, in the rod. . . . in a circle, in the rod. (d) up and down, in the rod. (e) there is no current . . . .
in the rod. "frag 8. Which of the following statements is (are) True , T, or False, F? Indicate by writing T or F in the blank spaces. (a) Ferromagnets are repelled by the poles of a magnet l:
(b) Diamagnets are repelled by the poles of a magnet I ‘
j 8 l (l )5 OD
EH) : 2! (p, 5i % (c) Paramagnets are attracted to the poles of a magnet T
9. The electric ﬁeld in a region of space varies according to E(t) = 250.0 (N/C)[sin (15000], where t is in seconds. The T (d) Ferromagnets are what are usually referred to as permanent magnets maximum displacement current through a 15.0 m2 area perpendicular to E is approximately \: 8.0m (OL— ~ Answer: (a) 10 uA. (b) 30 11A. (c) 50 uA ' (d) 70 uA. (e) 90 uA. (f) Lot: 3 OCEE :30 0‘ EA): b
21: 8.8‘5XID'IZCz/N,Mz / ‘ a? 0L m LT: (16 ~————'1
E(t) = (45.8V)sinq)dt is connected to the box the c For problems 10  13. The ﬁgure shows an AC generator'connected to a “BLACK BOX” through
a pair of terminals. The box is assumed to contains the equivalent of a series RLC electronic i(t)
circuit When an alternating voltage source whose voltage can be represented by
cut that is nieasured is E(t
i(t) = (7.88A)sin(codt — 485°). I: .13— 2 f @ 215. ‘5!
I. ,._i 'L— phase wens‘rcm‘l‘ a I
(d) 0.843. (e) .988. 10. The power factor is
Answer: (a) 0.336. (b) 0.540. © 0.663. (i) (30 : (lo 6 mime
11. Circle the true statement. For the “BLACK BOX”, 3 5 I 3 Answer: @The voltage leads the current. (b) The voltage and the current are in phase.
(d) The circuit is in resonance. (c), The current leads the voltage. 12. Assum'ng that the box must contain the equivalent of at least two components, which two must be present?
Answer: @ Resistance and inductance. (b) Resistance and capacitance. (c) Capacitance and inductance. (d) Capacitance and polarizer. (e) Inductance and polarizer. (f) r g ' 5‘ 1 I 1 ~ *  4%.W’ Answer: (a) 64.0 Watts. 120 Watts. (c) 260 Watts. (d) 340 Watts. (e) 720 Watts. (f) r 7 Paula: emsnmscow: M59
14. An inductor is conn ted across a 32.8 uF capacitor that is fully charged. The circuit oscillates at a frequency of
3.50 x 106 Hz. The value of the inductor is Answer: (a)l.6x10'”H. (b)2.8x10'“H. (c)5.0x10‘3H. @6.3x10‘1.1H. (e)9.5x10‘“H. (r) 13. The average rate at whic§energy is dissipated in the “black box” is v‘ _ 3.5 e}? H% 15. The magnetic ﬂux through the loop shown in the ﬁgure increases X X X X X X X X X ; 3‘2? 6" 6 F
according to the relation ¢B(t) = 5.8t3 + 6.0t,where (D50) is in webers X '
and t is in seconds. The resistance of the loop is 15.9 Ohms. :K, C. X 00 : 2T5: )
What is the magnitude and direction of the current that is induced Lnd; "‘ X , ‘ ‘ (o
in the loop at t = 5.0 seconds? LZERALQ‘: Np" :’ Lind/(f) c' X LOL.?’TF(O?(D 53%
Answer: (a) 14 A, counterlockwise. @ 28 A, aunterclockwise. X ’0' ’ e!
(c) 37 A, counterlockwise.. (d) 47 A, counterclockwise. ,L X ' ‘
(e) 65 A, counterlockwise. (f) HILH' + (p L 2 Z i L LC
16. A 250%(2 resistor and 98.6 pH inductor are connected in series acrossgan ideal J— : LOZQ '5: Z L5‘k
battery that supplies 15.0 ngﬁie circuit is shown at the right. What is the minimum time, L ' alter the switch is closed, that the magnetic ﬁeld in the inductor can be considered to be stable.
Ans fer: (a) 1.0 x10'6s. (b) 2.0 x10'6 5. (c) 5.2 x 1063. (d) 2.0 x 10'6 s. (e) 7.5 x10'6 s. L:,RU*C,'.Q+1.L) _ L/Réé) .
17. For the Clrcuit 1n questlon 16, how much energy 15 stored 1n the inductor when the current in the circuit becomes stable after the
switch is closed ? Answer: (a) 0.07 uJ. 0.18 p]. (c) 0.29 u]. (d) 0.44 11]. (e) 0.72 M. (i) “6“; LL“ acacia—r 1 0"“ (f) _ ...
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This note was uploaded on 03/01/2011 for the course PHYS 222 taught by Professor Henry during the Spring '11 term at SUBR.
 Spring '11
 Henry

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