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Unformatted text preview: M l .W. :ngnn“ V
nrxlh a, 2001  RECITATION DIVISI X: g > SHOW YOUR WORK. NUMERICAL ANSWERS SHOULD BE
COMPLETELY PROCESSED. K ' :HSTRUCTIONS : l. The two batteries in the multiloo p circuit have zero internal
resistance. <— 2.00 A "_\° 3.00 A m—a. «<——— 5.00 A a. Calculate the current in the 3.00 O resistor. J::8Ma f:— b. Calculate the value of resistor R. (SHOW YOUR WORK.) “Qafm —Hn_?>n+(o.n_ 68:0 f
“211 “(2V + 30‘, :O , W“ C. Calculate the value of emf, 3 (SHOW YOUR WORK.) . 8+ small)” essay) = C3 d. Calculate the electrical
resistor. P=I£R P=(sg)2.(g.l= ISOW \S‘ Qower dissipated by the 6.00 O u guiuqﬂU...¢L¢i \c: Densitive current. illuludulng QEVlCE, w1tn
resistance 15.0 ohms and full scale current 1.00 milliamp) and
a shunt resistor. ‘ a) Calculate the voltage across the ammeter when it reads
50.0 milliam s.
P 7 {seem ’ /
/ “ma —> El: 1mg
@
\ ,7 .—
\“.\ mp1 4L“30b “3k .OOIRG '5.n_.= do‘ﬁv (L.
b. Calculate the value of the shunt‘resistorff”_ﬁ_—K 3. A proton, m : 1.67 X 10’27 kg, q z 1.60 x 10‘19 C has speed
5.00 x 106 m/s in the +X direction at point A, as shown. The
I }ﬁyéﬁ proton moves in a circular path, in a uniform magnetic field
\{m {A of 0.15 T. {"ﬁ‘d a. The direction of the magnetic
V ' XRJ field is A. the +X direction. B. the —X direction. C. the +y direction. the —y direction.
(ii) out of the page. WA . into the page. J G. none of the abov choices. 0\3 ANSWER: E b. Calculate the radius of the circular path of the proton.
"ﬁvz. __ B :: ,27 ' l gnv 2L
F: E _[.Q><\O N \JJXto #3 (5K0 3)
2e;
F b .
15”: ﬁwxm :ﬁVB :\.bmo'qc.5MD M's ‘1
I Lme'BN ‘EeHéB /’ . UL 4.uu a, counLer ciocKWise. The loop l;es in the xy plane as shown. +y1 B o
c (9%
a. Calculate the magnitude of the magnetic moment of the wire loop/£12164 2 Q‘OOH‘ 6‘wcm.8 00¢vi b. The loop is in a uniform magnetic field that points in the
+y direction. What is the direction of the tor e on the The ~x direction.
. The +y direction.
D. The «y direction,
E. Out of the page.
F. Into the page.
G None of the above choices. AhlSWER: E A Circular conducting loop of resistance 4.00 ohm and area
0.300 m2 is in a uniform magnetic field directed into the plane
of the page, as shown. loop? _
r. The +x direction. 7:\“/ﬁ'*E
, Q: Lin. A: 0.20mi
X X X X a. The magnitude of the magnetic field is given by
B(t) = 1.50 + 0.10 t2 where B(t) is in Tesla and t is in
seconds. Calculate the magnetic flux in the loop at
t z 2.50 S. b. Calculate the induced emf at t = 2.50 s. :O‘EB _~Q\(\.ee+.\tllﬁ d(l.50A*‘Rfﬂ k‘ ‘ﬂp’li’ﬁ,i_i 333. at
‘ xv mM/fm . 4,45 /
: “Zmlt ii J.vv cm anu reSiscance lD.U onm, that maKes cc::acc at points
A and B with a U—shaped conductor of negligible resistance.
The device sits in a constant uniform magnetic field of 0.60 T, directed into the page, as shown. The conducting rod
is pulled to the right at constant speed, 0.30 m/s. 5.00 cm
0.30 m/s
E k I
conducting bar
a. The magnetic flux in the conducting loop is
A. constant. =
increasing . EBA
'. decreasing.
D. 08 '1lating.
E. ro //
S
ANSWER:
b. Calculate the induced voltage (motional emf) across the
sliding rod.
e : m .i
€~VBL .3 Ignogm» (m  .063le
a ‘ /
C. The induced current in the conducting loop is
A. clockwise and constant.
B. clockwise and increasing.
C. clockwise and decreasing.
D. counter clockwise and constant. ///
[E. counter Clockwise and increasing. 5
F. counter clockwise and decreasing.
ANSWER: 1:)
d. Calculate the magnitude of the force required to pull the
J”:];B” sliding rod at the constant speed of 0.30 m/s.
' 2.
1* 130°} g
i'Stl'w— .( / " , x “we 15:1th :(omb‘la'ﬁémgw Lame“ A closed curve encircles three current carrying conductors.
The line integral around this curve —) a
9913 ‘dl
is equal to 3.63 x 10'4 Tm. Calculate the net current in the
conductors.
$8 '011 3/1 1cm ﬂ
. 11 _ I ‘
g'togx‘o Tm ‘ﬂojewc "/ ~.\
\ 1
~ \\
z.\\\l
**** USEFUL STUFF ****
—+ —v o #0]:
F = qv x B B =
Zn
—0 —~
(pa = B.A B = ponI
—+ a —v [LON]:
F z 11 x B B =
2a
u = IA
—v a
_. _. _. §Bod1 = MOIenc
r = u x B
d¢8
)3: = 0 a = — ———
dt
EV = O
6 = VBL
RE=R1+R2+R3+ ¢ o d¢a
ﬁE‘dl =  —
dt
1 1 1 1
__=._.__.+.——+—.—+.., P=eI
R R, R2 R
° 3 p = 12R ...
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 Spring '06
 SMITH
 Physics, Magnetic Field, wA, 2L, EBA, AhlSWER

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