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Document 1798734
```Physics 2020
Spring 2008
Stephan LeBohec
1
EXAM 4
Name:_____________________________________
TA (circle one): Aaron
Akiko
Student ID #:___________________________
Jacque
Matt
A
An ideal transformer with 5000 primary turns and 250 secondary turns is used to power an alarm clock
 =2.75W .
radio from the wall outlet ( V P =110V , f =60Hz ). The average power drawn from the outlet is P
Neglect all internal resistances of the transformer.
RMS
1.
[5 points] The alarm clock radio is powered with V S
of
 : 0.55V
 : 3.89V
 : 5.5V
 : 7.78V
 : 15V
as 110V×250/ 5000=5.5V
 : 55V
2.
[5 points] The current ( I S
) delivered to the alarm clock radio is
 : 0.15A
 : 0.25A
 : 0.5A
 : 0.75A
 : 1.0A
as P=2.75W=V S ×I S
& I S =2.75W / 5.5V=0.5A
 : 1.5A
RMS
RMS
RMS
RMS
RMS
3.
[5 points] The instantaneous current delivered by the outlet can be expressed as
I 0 is
 : 25.0mA
 : 35.3mA  : 250mA
 : 353mA
 : 500mA
 : 707mA
P=V P × I P =V P ×I 0 /  2 & I 0= 2×P /V P 0.0353A =35.3mA
I P =I 0 sin2  f t  where
RMS
B
RMS
RMS
An unpolarized electromagnetic wave with a wavelength
S 0=14Wm −2 is propagating through two polarizers.
RMS
=450nm and an intensity of
1.
[5 points] If the transmission axis of the first polarizer is at an angle of 30o with respect to the
vertical, what is the intensity of the electromagnetic wave after the first polarizer?
 : 3.5Wm-2  : 7.0Wm-2  : 9.0Wm-2  : 10.5Wm-2  : 14.0Wm-2  : 28.0Wm-2
2.
[5 points] When the transmission axis of the second polarizer is at a 60o angle from the first
one, the electric field of the transmitted light gets reduced by a factor of:
 :  3 /2
 : 3/2
: 2
: 4
 : 16
 : 32
o 2
The intensity is reduced by cos 60  =4 and the intensity is proportional the electric field squared
3.
[5 points] What must be the angle  between the polarizers for the transmitted radiation power
to be reduced by a factor 2 at the second polarizer?
 : 15o
 : 30 o
 : 45 o
 : 60o
 : 75o
 : 87.3o
S 3 / S 2=cos2  so =cos−1   S 3 /S 2 =cos−1  1/2=45o
Physics 2020
Spring 2008
Stephan LeBohec
Name:_____________________________________
TA (circle one): Aaron
2
EXAM 4
Akiko
Jacque
Student ID #:___________________________
Matt
C.
Two people (A and B) are standing in front of a pair of
vertical mirrors. The mirrors are making a right angle as shown on
the figure which is a top view of the situation (x=2m and y=1m ).
Person B will see several virtual images of A.
x
y
1.
[7 points] On the figure indicate the positions of these
images. How many images are there?
y
y
There are 3 virtual images forming a 2x×2y rectangle with A. One could
also consider there are 4 images, two of them are overlapping.
x
x
2.
[8 points] Clearly indicate which image of A is the furthest
away from B? How far is that image from B?
The image of A that is the furthest away from Bis the one on the lower left corner of the rectangle. It is at a distance from
B of   2x23y2= 2×223×12 =5m
D.
A person with a spherical head of diameter D=0.15m is standing in front of a concave spherical
mirror with a radius of curvature of 4m. The person is standing 1.5m in front of the mirror.
1.
2.
[4 points] Is the image real or virtual?
The focal length is f=R/2=2m and the person stands at a distance 1.5m<f so the image is virtual.
[9 points] How far away from the person is this image?
1 1 1
1 1 −1 1 1 −1
= 
 =−6m that is how far the image is behind the mirror so the distance
so d i= −  = −
f do di
f do
2 1.5
of the image from the person is 1.5m6m =7.5m
3.
[7 points] What is the diameter of the persons head in the image?
The magnification factor is M =−d i /d o =6 /1.5=4 so the diameter of the head in the image is
4×0.15m =0.60m
Physics 2020
Spring 2008
Stephan LeBohec
EXAM 4
Name:_____________________________________
TA (circle one): Aaron
E.
Akiko
Jacque
3
Student ID #:___________________________
Matt
A rectangular conductive loop of width
w=0.05m and height h=0.1m is kept in motion at a
constant velocity v=10m⋅s−1 as on the figure. The loop has a
resistance R=4  . The loop enters a region which extends
for ever to the right and where there is a constant magnetic field
of magnitude B=2T and direction as specified on the figure.
Outside of this region there is no magnetic field.
1.
[18 points] Graph the magnetic field flux and the
magnitude of the electromotive force (EMF) as a function of
time. Use the instant the loop starts entering the magnetic field
region as the origin of times. Notice time is measured in ms.
It takes w/ v=0.05m / 10m⋅s−1 =0.005s=5ms for the loop to fully penetrate in the magnetic field region. The
magnetic flux changes from =0 to =B⋅w⋅h=2×0.05×0.1=0.01T⋅m 2=10×10−3 T⋅m 2 .
The flux changes with time implying an electromotive force
V =−

0.01Tm 2−0Tm 2
=−
=2V while the loop
t
0.005s
penetrates the magnetic field region. Once the loop has completely entered the magnetic field, the flux does not change
any more and the electromotive force drops to 0.
2.
[4 points] The current flows
clockwise
or
counterclockwise
? (Circle one)
The magnetic field produced by the induced current must oppose the change in flux (Lenz's law)
3.
[8 points] Once the loop has fully entered the magnetic field region, how much energy has been
dissipated in the loop?
When the electromotive force is 2V, the dissipated power is 2V2 /4 =1W . This lasts for 0.005s so the total energy
dissipated is 1W×0.005s=0.005J .
3.
[5 points] What is the total amount of work done by the operator who maintains the loop at constant
velocity as it enters the magnetic field region?
The energy dissipated in the loop is provided by the operator maintaining the loop at constant speed while it enters the
magnetic field. The work W done by the operator must equal the energy dissipated and so W=0.005J.
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