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\begin{center}
{\Large {Solution Derivations for Capa \#12\\[20pt]} }
\end{center}
1) An ideal step-down transformer has a primary coil of 300 turns and a
secondary coil of 25 turns. It is plugged into an outlet with $120\ V$ (AC)
from which it draws a current of $0.79\ A$. What is the voltage in the
secondary coil?
\begin{quote}
$N_{p}$ = Given\\
$N_{s}$ = Given\\
$V_{p}$ = Given (rms), subscript meaning ``primary''\\
$I_{p}$ = Given (rms)\\
The given values for voltage and current are the rms values because you know
it is plugged into an electrical outlet. From the ratio
\[
\frac{N_{s}}{N_{p}}=\frac{V_{s}}{V_{p}}
\]
we can find the voltage in the secondary coil (which is also an rms value).
\[
V_{s}=V_{p}\frac{N_{s}}{N_{p}}
\]
\end{quote}
2) Calculate the rms current in the secondary coil.
\begin{quote}
To find the rms current in the secondary coil, use the ratio
\[
\frac{I_{p}}{I_{s}}=\frac{N_{s}}{N_{p}}=\frac{V_{s}}{V_{p}}
\]
or
\[
I_{s}=I_{p}\frac{N_{p}}{N_{s}}
\]
Since the primary current $I_{p}$ was given as an rms quantity, the
secondary current, $I_{s}$, is also an rms value.
\end{quote}
3) Assuming that the transformer secondary is driving a resistive load,
calculate the average power dissipated in the resistor.
\begin{quote}
Average power is calculated from the equation
\[
\bar{P}=I_{rms}V_{rms}
\]
Where the rms current and rms voltage were calculated above.
\end{quote}
4) The figure below shows the time variation of the current through an
electrical heater when it is plugged into a $120\ V$, $60\ Hz$ outlet.
\begin{quote}
$V_{rms}$ = Given\\
$f$ = Given\\
The voltage is an rms quantity since you are told the device is plugged into
an outlet. The frequency confirms that it is a household outlet. The
relationship to rms voltage and peak voltage is
\[
V_{peak}=\sqrt{2}V_{rms}
\]
\end{quote}
5) A power supply that provides a voltage of $49.0 \sin (388t)\ V$ is
connected across a resistor $R = 200\ \Omega$. Calculate the average power
dissipated in the resistor.
\begin{quote}
$V(t)$ = Given\\
$R$ = Given\\
The power dissipated through a resistor is
\[
P=I_{rms}^{2}R=\frac{V_{rms}^{2}}{R}
\]
You are given an equation for the voltage. The peak voltage is simply the
amplitude of the sinusoidal function. The rms value is then calculated by
\[
V_{rms}=\frac{V_{peak}}{\sqrt{2}}
\]
So,
\[
P=\frac{V_{rms}^{2}}{R}=\frac{1}{R}\left( \frac{V_{peak}}{\sqrt{2}}\right)
^{2}=\allowbreak \frac{V_{peak}^{2}}{2R}
\]
\end{quote}
6) A source of electromagnetic waves radiates power uniformly in all
directions at a single frequency. At a distance of $6.50\ km$ from the
source, a detector measures the intensity of the wave to be
$20.0\ \mu W/m^2$. What is the peak value of the magnetic field at the
detector due to the waves?
\begin{quote}
$d$ = Given\\
$S$ = Given\\
From the lecture notes (pg. 34-6), we have several equations relating $S$ to
the rms and peak values of the electric and magnetic field. Choosing one
with the peak magnetic field and solving for this quantity,
\begin{eqnarray*}
S & = & \frac{B_{p}^{2}c}{2\mu _{0}}\\
B_{p} & = & \sqrt{\frac{2\mu _{0}S}{c}}
\end{eqnarray*}
\end{quote}
7) What is the average power radiated by the source?
\begin{quote}
The quantity $S$ is defined to be a power over an area. Since the
electromagnetic waves are emitted radially from the source, the total power
will be spread out over the surface of a sphere. Thus, by finding the
ratio of the detector's surface are to the sphere's surface area, it is
possible to determine the amount of power originally present.
\[
S=\frac{P_{0}}{4\pi R^{2}}
\]
where $R$ is the radius from the source emitting the waves. In \#6, the
detector was a distance $d$ from the detector. Thus, the original power is
\[
P_{0}=4\pi Sd^{2}
\]
\end{quote}
8) An electromagnetic wave from a wire antenna travels (from the reader)
toward the plane of the paper. At time $t=0\ s$ it strikes the paper at
normal incidence. At point O and $t=0$, the magnetic field vector has its
maximum value, $5.40 \times 10^{-8}\ T$ pointing in the negative y direction.
The frequency of this wave is $1.80 \times 10^{16}\ Hz$. What is the
x-component of the associated electric field
$\stackrel{\rightarrow}{\mathbf{E}}$ at time $t = 0\ s$? (Use the right-hand
rule to determine the direction of $\stackrel{\rightarrow}{\mathbf{E}}$, and
hence the sign of the x component.)
\begin{quote}
$B_{peak}$ = Given\\
$f$ = Given\\
Since the electromagnetic wave is in phase, both peak at the same time.
The relationship between the two is given by
\[
B_{peak}=\frac{E_{peak}}{c}
\]
Or, equivalently,
\[
E_{peak}=cB_{peak}
\]
Since the wave is traveling into the page, the intensity (or Poynting
vector) is also into the page. Using the right hand rule (with $S=E\times
B $), the direction of $E_{peak}$ is determined.
\end{quote}
9) What is the magnitude of the Poynting vector of the wave at time
$t = 0\ s$?
\begin{quote}
$t$ = Given\\
Since the electric field is always perpendicular to the magnetic field in an
electromagnetic wave, the intensity equation reduces to
\[
S=\frac{E\times B}{\mu _{0}}=\frac{EB\sin \theta }{\mu _{0}}=\frac{EB\sin
\left( 90^\circ \right) }{\mu _{0}}=\frac{EB}{\mu _{0}}
\]
At $t=0$, the magnetic field is at its maximum and so is the electric field.
\end{quote}
10) What is the y component of the magnetic field at point O at time
$0.278 \times 10^{-16}\ s$?
\begin{quote}
$t$ = Given\\
You must calculate the number of periods elapsed in the time given in order
to solve this problem. The frequency $f$ was given in \#8. The period
relates to the frequency by:
\[
T=\frac{1}{f}
\]
The number of periods that have passed by in time $t$ is calculated by
dividing by the time necessary for one period.
\[
N=\frac{t}{T}=tf
\]
Then, you must think about the graph of sine. From the lecture notes on
page 34-5, we know that the magnetic field oscillates according to:
\[
B=B_{peak}\sin (kx-\omega t)
\]
From the graph in \#8, we know this is in the y-direction. Knowing the
original magnitude and direction of $B_{peak}$, it is easy to calculate what
the graph of sine looks like after $N$ periods have passed.
\end{quote}
11) A linearly polarized electromagnetic wave has an average intensity of
$360\ W/m^2$. This wave is directed towards two ideal polarizers (in real
polarizers, transmission is also effected by reflection and absorption).
Polarizer A is oriented with its transmission axis at an angle of
$\theta_1 = 26.5^\circ$ with the incident electric field. Polarizer B has
its axis at an angle of $\theta_2 = 76.5^\circ$ with the incident electric
field, as in the figure below. What is the average intensity of the wave
after it goes through polarizer A?
\begin{quote}
$S_{0}$ = Given\\
$\theta _{1}$ = Given\\
$\theta _{2}$ = Given\\
The Law of Malus, stated on page 878 of the textbook and 34-8 of the lecture
notes, says that the transmitted intensity is given by
\[
S=S_{0}\cos ^{2}\theta
\]
After passing through polarizer $A$ which has the polarizing axis an angle
$\theta _{1}$ from perpendicular, the intensity is
\[
S_{A}=S_{0}\cos ^{2}\theta _{1}
\]
\end{quote}
12) What is the average intensity of the wave after it goes through polarizer
B?
\begin{quote}
After the light passes through the second filter, its intensity drops again.
However, its initial intensity is that of the light after it passes
through filter $A$. Also, the angle between that which the light is
traveling and that which it must pass through is $\theta _{2}-\theta _{1}$.
This is because the first polarizing filter only let light through at that
angle. Thus,
\[
S_{B}=S_{A}\cos ^{2}\left( \theta _{2}-\theta _{1}\right) =S_{0}\cos
^{2}\theta _{1}\cos ^{2}\left( \theta _{2}-\theta _{1}\right)
\]
\end{quote}
13) Suppose that the two polarizers A and B are interchanged. What would the
average intensity be after passing through both polarizers?
\begin{quote}
If the filters are reversed, the intensity is not the same. The
calculation is nearly the same, however. The intensity after passing
through the first filter is now
\[
S_{B}=S_{0}\cos ^{2}\theta _{2}
\]
The angle between the pass axis of the two filters is the same regardless of
their orientation. That is, it does not matter which way the light is
oriented toward the other filter as it will be the same angle difference.
Thus, the intensity after passing through the second filter is then
\[
S_{A}=S_{B}\cos ^{2}\left( \theta _{2}-\theta _{1}\right) =S_{0}\cos
^{2}\theta _{2}\cos ^{2}\left( \theta _{2}-\theta _{1}\right)
\]
\end{quote}
14) The average intensity of solar radiation incident on Earth is
$1390\ W/m^2$. What is the total power radiated by the sun? DATA: The mean
distance between Earth and the sun is $1.5 \times 10^{11} m$.
\begin{quote}
$S$ = Given\\
$R$ = Given\\
This is nearly the same question as \#7. We know that
\[
S=\frac{P_{0}}{4\pi R^{2}}
\]
so
\[
P_{0}=4\pi SR^{2}
\]
\end{quote}
15) The average intensity of light from an industrial laser is
$\bar{S} = 1.23 \times 10^9\ W/m^2$. What is the rms value of the electric
field?
\begin{quote}
$S$ = Given\\
Using an equation from page 34-6 which has the rms value of the electric
field, we can solve directly for the answer.
\begin{eqnarray*}
S & = & \frac{E_{rms}^{2}}{\mu _{0}c}\\
E_{rms} & = & \sqrt{S\mu _{0}c}
\end{eqnarray*}
\end{quote}
16) For the same laser, what is the rms value of the magnetic field?
\begin{quote}
Recalling that $B_{peak}=\frac{E_{peak}}{c}$, we can solve for the rms value
of the magnetic field by dividing both sides by $\sqrt{2}$.
\[
\frac{B_{peak}}{\sqrt{2}}=\frac{E_{peak}}{\sqrt{2}c}
\]
thus,
\[
B_{rms}=\frac{E_{rms}}{c}
\]
\end{quote}
17) Select ALL the correct statements (AC, ADEF, etc.)
\begin{quote}
QUESTION:\\
A) The sun's radiation is most intense in the infrared region.\\
B) Gamma rays can be produced through electrons oscillating in wires of
electric circuits.\\
C) Visible light is often emitted when valence electrons change their state.\\
D) Blue light has a longer wavelength than red.\\
E) Ultraviolet radiation causes common sunburn.\\
F) A vertical automobile antenna is sensitive to electric fields polarized
vertically.\\
G) X-rays can be produced in transitions involving inner electrons in an
atom.\\
ANSWER:\\
A) False, since we can see the sun, the most intense radiation is in the
visible spectrum.\\
B) False, gamma rays can not be produced this way. Nuclear state
transitions will produce gamma rays. See the diagram in the lecture notes
after page 34-4.\\
C) True, valence electrons are the outer electrons of an atom. When these
change states, visible light is emitted. See the diagram in the lecture
notes after page 34-4.\\
D) False, from an electromagnetic spectrum chart, red has a longer
wavelength than blue.\\
E) True.\\
F) True, if it were not, the atoms in the antenna would not have very far
to move in order to pick up the signal.\\
G) True. See the diagram in the lecture notes after page 34-4.
CAPA is looking for an answer in the form CEFG.
\end{quote}
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