This preview shows page 1. Sign up to view the full content.
Unformatted text preview: OM  PR
EME D41 1.C P Physics 102 FORM 0  PR
EME D41 1.C OM  PR
EME D41 1.C OM [l] The area of a rectangular loop of wire is 6.4 x 10~3 m 2 . The loop is placed in a magnetic
field t hat changes from 0.20 T to 1.2 T in 1.6 s. The plane of the loop is perpendicular
to the direction of the magnetic field. W hat is the magnitude of the i nduced emf in t hat
loop?
^,
3
[A] 2.1 x 10~ V
f[B]J4.0 x 10~3 V
[C] 6.7 x 10~3 V
[D] 8.4 x 10"3 V
3
[E] 9.8 x I D" V
^  PR
EME D41 1.C OM [2] A circular coil with 600 t urns has a radius of 15 cm. The coil is rotating about an axis
perpendicular to a magnetic field of 0.020 T. If the maximum induced emf in the coil is
1.6 V, at what angular speed is the coil rotating?
[A] 0.30 r ad/s
[B] 0.60 r ad/s
[C] 0.90 r ad/s
[D] 1.4 r ad/s
([E])l.9 rad/s  PR
EME
D4 11. COM [3] A 25mH i nductor is connected in series with a 200 resistor, through a 15V DC power
supply and a switch. If the switch is closed at t = 0 s, w hat is the current after 0.70 s?
[A] 120 mA
[B] 280 mA
[C] 450 mA
[D] 570 mA
(®)750 mA  PR
EME
D4 11. COM [4] According to the little metal plate on a hair dryer, it is r ated at 120 Vrms and an average
power of 1500 W. Assuming the load to be purely resistive, what is the maximum value of
current in the dryer?
_^
[A] 12.5 A
[B] 14.1 A
(fcj)l7.7 A
[D] 25.0 A
[E] 35.3 A COM [5] An electromagnetic wave is propagating towards the west. At a certain moment the
direction of the magnetic field vector associated with the wave points vertically up. What
''ie direction of trhe electricfield v ector?
[B] Vertical and p ointing down
Horizontal and pointing south
[D] Vertical and pointing up
Horizontal and pointing north
[E] Horizontal and pointing east
COM  PR
EM ED4 11. f COM  PR
EM ED4 11. [6] An 8.00mW (average power) laser beam emits a cylindrical beam of light, 0.900 mm in
diameter. What is the rms value of the electric field in t his beam?
_
[A] 412 N/C
[B] 1000 N/C
[C] 1090 N/C
[D] 2000 N/C
([Ej) 2180 N/C  PR
EM ED4 11. [7] A convex mirror has a radius of curvature of R = 1 meter. An object is placed at a distance
of(3.5 meters in front of the mirror. What is the image distance?
(JA])0.25 m
[B] 0.25 m
[C] 0.5 m
[D] 0.5 m
[E] 1 m 11. COM  PR
EM ED4 11. COM [8] An optical device produces an upright virtual image t hat is reduced in size. What could
this optical device be?
[A] convex mirror or convex lens
{LBJjconvex mirror or concave lens
[0] concave mirror or convex lens
[D] concave mirror or concave lens
[E] plane mirror PRE MED 411 .CO MPRE
M ED4 11. COM  PR
EM ED4 [9] A convex lens forms a real, enlarged image. If the object distance is 2.0 meters and the
image distance is 30 meters, what is the focal length and the magnification?
[A] 1.2 meters, 1.5
[B] 1.2 meters, 1.5
 jCj>1.2 meters, 1.5
[D] 1.2 meters, 1.5
[E] none of these OM  PR
EME D41 1.C P Physics 102 FORM 0  PR
EME D41 1.C OM  PR
EME D41 1.C OM [10] A nearsighted person has a far point t hat is 52 cm from her eye. What diopter lens will
correct this person's vision so that she can focus on distant objects? Assume the lens is
placed 2 cm in front of her eye.
[A] 0.02
[B] 0.5
[C] 0.5
(j]DJ)2
[E] 2  PR
EM ED4 11. COM  PR
EM ED4 11. COM  PR
EME
D4 11. COM  PR
EME
D4 11. COM  PR
EME D41 1.C OM [ll] The p icture below shows a zeroresistance rod of mass m = 0.10 kg sliding, due to gravity,
downward on two zeroresistance rails separated by the distance L = 0.20 m. The rails are
connected by a resistor R = 0.10 f i, and the entire system is in a perpendicular uniform
magnetic field with a magnitude B = 0.50 T, pointing out of the plane. Assume t hat
the velocity v of the rod has reached an equilibrium constant value, and find (taking
g — 10 m /s 2 , and ignoring the effect of air resistance)
(a) the magnitude and direction of the magnetic force acting on the rod under such an
equilibrium;
(b) the magnitude and direction (clockwise or counterclockwise) of the induced current in the
circuit;
(c) the magnitude of the induced emf in the circuit;
(d) the equilibrium speed v of the rod.
••> s vvvv ' V""1 f/ ^ © © v ©I
,^,^«^, «4*,
«^ m
© v © x, ® ©
\
©©©
©
w wm e COM T 11. COM  PR
EM ED4 11. © 0 0/K. © ©  ioo.lt>V= U >IA  PR
EM ED4 11. COM  PR
EM ED4 tx PRE MED 411 .CO MPRE
M ED4 11. COM 7
r ~ b__
6L ui, OM  PR
EME D41 1.C P FORM 1 OM Physics 102 D41 1.C OM  PR
EME D41 1.C [12] An uupolarized light source with an i ntensity I0 is incident on a polarizer A.
(a) Find the relative orientation of a second polarizer (polarizer B) such that the t ransmitted
intensity through it is Jo/10.
(b) Find the relative orientation of a t hird polarizer (polarizer C) w ith respect to polarizer A
such t hat no light is t ransmitted through polarizer C. Assume t hat polarizer B is between
polarizers A and C. TD (S  PR
EME
D4 11. COM Tit) so To TVV1 JO Co<MLtC\  PR
EME
D4 11. COM  PR
EME D41 1.C OM  PR
EME (c) Now you are challenged to maximize the t ransmitted intensity of all three polarizers by
rotating polarizer B to some angle Q with respect to polarizer A. Using the results from
above, write down an expression for the total transmitted intensity through polarizer C in
terms of this 9. You don't have to solve for the d t hat m aximizes the intensity. TD ED4 11. COM  PR
EM ED4 11. COM A ls, _ PRE MED 411 .CO MPRE
M ED4 11. COM  PR
EM ED4 11. COM  PR
EM ED4 11. COM  PR
EM ED4 11. COM  PR
EM  C— S O =^> OM  PR
EME D41 1.C P
OM Physics 102 PRE MED 411 .CO MPRE
M ED4 11. COM  PR
EM ED4 11. COM  PR
EM ED4 11. COM  PR
EM ED4 11. COM  PR
EM ED4 11. COM  PR
EM ED4 11. COM  PR
EME
D4 11. COM  PR
EME
D4 11. COM  PR
EME D41 1.C OM  PR
EME D41 1.C OM  PR
EME D41 1.C [13] A ray of light travels through air (index of refraction n i) and strikes a plane mirror that
has a glass coating with an index of r efraction n^ > ni and a thickness t. The angle
of incidence is 6 and it strikes the glass at a height h above the ground. The ray is
subsequently refracted, reflected and refracted before it hits the ground a distance d away
from the mirror. Find the distance d in terms of the above parameters.
Steps:
(1) Find the angle of refraction as the ray penetrates the glass.
(2) Find the angle of reflection from, the mirror.
(3) Find the angle of refraction as the ray leaves the glass.
(4) Use trigonometry to find d.
6 b 0\\ OM  PR
EME D41 1.C P Physics 102 F ORM 0  PR
EME D41 1.C OM FORMULAE SHEET me = 9.11 x 1(T31%, k = 8.99 X l09Nm2/C2 1.C OM e = 1.60 x lO"19^ IkWh = 3.6 x 106 J. e0 = — = 8.85 x !Qi:iC*/Nm? D41 leV = 1.60 x 10~19J, OM  PR
EME 4:TTK 1.C €Q Wnc = AJT + AZ7, COM 2 2  PR
EME
D4 11. '
WE 2 C' COM
11.
 PR
EME
D4 Ai i 2 .. q = Q^p(t/r), COM T //0 = 4?r x 10~7 ED4 11. T =~NIAB sinQ, N/J.Q! r>  ~ 2^"'
£ = vBl. (coil) £ = NABu> sino^i, ' . rN  (solenoid)
u — 2nf = — T ~R COM T=  PR
EM ED4 11. =
~ tt J=l A  PR
EM
COM
11. :——, At r '  PR
EM ED4 £= V= . .., J=«p(f/T)J (long)' ?, ^ nlCo ...,  =  TV
K
= "' T*R = 5. ^ =
^ p= ^0 AT^ = EAs,  PR
EME =  g 0 AV, go ^ ^Q D41 UB k = ^, E — cB, c= _ = 3.00 x 10Bm/s, \f = c ED4
 PR
EM f
J ~ 7 ( 1 4.  I,
= f (1 ±
\
c ,, UB UE,
: u= 11. COM /
"•
70  —7E=Ep
27r\/LC 11. COM E 0 sin(ife2:  w f), PRE MED 411 .CO MPRE
M ED4 11. COM  PR
EM ED4 f4R
J — ^0 ) )
0
2 111
j ~ 3~~T5
j
~5
da
di
f hi
di
m = — — — — ,,
—
h0
d0 c
v — — ,,
—
ra 7 = ^" + "T'
J
A
/2 —
:Z? X vac
n
=T
J ...
View
Full
Document
This note was uploaded on 10/11/2011 for the course PHY 101 taught by Professor Ashkenkai during the Fall '08 term at FIU.
 Fall '08
 Ashkenkai
 Physics

Click to edit the document details