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{\Large {Solution Derivations for Capa \#11\\[20pt]} }
\end{center}
Caution: The symbol $E$ is used interchangeably for energy and EMF.
1) DATA: $V_b = 5.0\ V$, $R = 155\ \Omega$, $L = 8.400 \times 10^{-2}\ H$.
In the diagram above, what is the voltage across the inductor in the instant
just after the switch is closed?
\begin{quote}
$V_{b}$ = Given\\
$R$ = Given\\
$L$ = Given\\
From Kirchhoff's Loop law, we can get an equation for the voltages around
the circuit.
\[
V_{b}-IR-E_{L}=0
\]
where $E_{L}$ is the EMF of the inductor. Thus,
\[
E_{L}=V_{b}-IR
\]
Immediately after the switch is closed, there is no current in the circuit.
The resistor and inductor are in series, and the inductor opposes the
change in current. Thus, there is no voltage drop across the resistor and
the voltage drop across the inductor is the initial voltage $V_{b}$.
\end{quote}
2) After the switch is closed for a long time, what is the energy stored in
the inductor?
\begin{quote}
The energy stored in an inductor is given by
\[
E=\frac{1}{2}LI^{2}
\]
The maximum current in this circuit is determined by the battery and the
resistor. From Ohm's law,
\[
I=\frac{V}{R}
\]
Thus,
\[
E=\frac{1}{2}L\left( \frac{V}{R}\right) ^{2}
\]
\end{quote}
3) The switch in the above diagram is closed after being open a long time.
The initial charge on the capacitor is zero. (For each statement select T
True, F False).
\begin{quote}
QUESTION:\\
A) In the instant after the switch is closed, the voltage across the
capacitor equals the voltage across the battery.\\
B) In the instant after the switch is closed, the voltage across the
inductor equals the voltage across the battery.\\
C) A long time after the switch is closed, the voltage across the capacitor
equals the voltage across the battery.\\
D) A long time after the switch is closed, the current through the resistor
is zero. \\
ANSWER:\\
A) False, initially capacitors act like wires without resistance, so there
is no voltage drop.\\
B) True, there is no current elsewhere in the circuit, so there are no
other voltage drops. See (\#1).\\
C) True, after a long time a capacitor acts like an infinite resistor, so
no current can flow. Thus, the voltage drop must be equal to the battery.\\
D) True, see (C).\\
CAPA is looking for an answer in the form FTTT
\end{quote}
4) The next 6 questions refer to this situation: An LR circuit is hooked up to
a battery as shown in the figure, with the switch initially open. The
resistance in the circuit is $R = 110\ \Omega$, the inductance is
$L = 3.40\ H$ and the battery maintains a voltage of $E = 30.0\ V$. At
time $t = 0$ the switch is closed. What is the current through the circuit
after the switch has been closed for $t = 4.57 \times 10^{-2}\ s$?
\begin{quote}
$R$ = Given\\
$L$ = Given\\
$E$ = Given\\
$t$ = Given\\
For a rising current in an RL circuit, the equation for the current at any
time is
\[
I=\frac{E}{R}\left( 1-e^{\frac{-Rt}{L}}\right)
\]
\end{quote}
5) What is the voltage across the inductor after the switch has been closed
for $t = 4.57 \times 10^{-2}\ s$?
\begin{quote}
This is essentially the same circuit as in \#1. Thus, the same
equation from Kirchhoff's loop law can be applied (or easily derived).
\[
E_{L}=V_{b}-IR
\]
However, there is a current though the resistor at this point which you just
calculated.
\end{quote}
6) What is the power dissipation in the resistor at
$t = 4.57 \times 10^{-2}\ s$?
\begin{quote}
The easiest way to calculate the answer is to use the formula for power
\[
P=I^{2}R
\]
Assuming the time is the same for these problems, the current $I$ was
calculated in (\#4) and the resistance was given.
\end{quote}
7) How much energy is stored in the inductor at $t = 4.57 \times 10^{-2}\ s$?
\begin{quote}
The energy stored in an inductor is given by
\[
U=\frac{1}{2}LI^{2}
\]
where the inductance $L$ was given and the current $I$ was calculated
previously.
\end{quote}
8) How much work has the battery done from the time the switch was closed
until $t = 4.57 \times 10^{-2}\ s$?
\begin{quote}
To solve this problem, you needed to integrate the power equation for the
time length specified (i.e. from $t=0$ to $t=t_{0}$ (Given)). Since this
is the work the battery has done, the logical choice for the power equation
would be
\[
P=IV=IE
\]
since the voltage is the EMF. The current is determined by the inductor and
changes with time according to the equation
\[
I=\frac{E}{R}\left( 1-e^{\frac{-Rt}{L}}\right)
\]
The resulting integral is
\begin{eqnarray*}
W & = & \int_{0}^{t_{0}}\left( E\frac{E}{R}\left( 1-e^{\frac{-Rt}{L}}\right)\right) dt\\
& = & \frac{E^{2}}{R^{2}}\left( t_{0}R+Le^{-t_{0}\frac{R}{L}}-L\right)
\end{eqnarray*}
I did this on a TI-92, but it can be verified by hand.
\end{quote}
9) How much energy has been dissipated in the resistor from the time the
switch was closed until $t = 4.57 \times 10^{-2}\ s$?
\begin{quote}
To find the energy dissipated by the resistor, we must again integrate the
power equation of the time interval specified (from $t=0$ to $t=t_{0}$
(Given)). This time, the best choice of the power equation is
\[
P=I^{2}R
\]
where $I$ is the current determined by the inductor. It is equal to
\[
I=\frac{E}{R}\left( 1-e^{\frac{-Rt}{L}}\right)
\]
Thus, the energy dissipated is
\begin{eqnarray*}
U & = & \int_{0}^{t_{0}}R\left( \frac{E}{R}\left( 1-e^{\frac{-Rt}{L}}\right)\right) ^{2}dt\\
& = & \frac{1}{2}\frac{E^{2}}{R^{2}}\left( -L+4Le^{t_{0}\frac{R}{L}}+2t_{0}
Re^{2t_{0}\frac{R}{L}}-3Le^{2t_{0}\frac{R}{L}}\right)e^{-2t_{0}\frac{R}{L}}
\end{eqnarray*}
\end{quote}
10) The next three questions refer to this situation. A very long solenoid
with a circular cross section and radius $r_1 = 1.90\ cm$ with
$n_s = 280\ turns/cm$ lies inside a short coil of radius
$r_2 = 5.00\ cm$ and $N_c = 35\ turns$. If the current in the solenoid is
ramped at a constant rate from zero to $I_s = 1.40\ A$ over a time interval
of $76.0\ ms$, what is the magnitude of the emf in the outer coil while the
current in the solenoid is changing?
\begin{quote}
$r_{1}$ = Given\\
$n_{s}$ = Given\\
$r_{2}$ = Given\\
$N_{c}$ = Given\\
$I_{s}$ = Given\\
$t$ = Given\\
This problem is easier if you do \#11 first. The magnitude of the EMF is
given by
\[
E_{2}=-M\frac{dI_{s}}{dt}
\]
Which we can calculate by
\[
E_{2}=-M\frac{\Delta I_{s}}{\Delta t}
\]
using $M$ calculated in \#11.
\end{quote}
11) What is the mutual inductance between the solenoid and the short coil?
\begin{quote}
The mutual inductance of a coil is given by
\[
M=\frac{\phi _{2}}{I_{1}}
\]
Since the solenoid is enclosed by the outer loop, the flux though the outer
loop will be the same as the flux through the solenoid. Flux is given by
\[
\phi =\int B\cdot dA=BA
\]
Basically this equation will be true as long as the magnetic field does not
vary over the area of the object. It may vary with time just as long as it
is uniform at any given instant. The total flux in the coil is the number
of turns in the coil multiplied by the flux in each turn. Thus,
\begin{eqnarray*}
M & = & \frac{N_{c}\phi _{2}}{I_{1}}=\frac{N_{c}B_{1}A_{1}}{I_{1}}=\frac{N_{c}\mu
_{0}nI_{1}A_{1}}{I_{1}}\\
& = & N_{c}\mu _{0}nA_{1}=N_{c}\mu _{0}n\left( \pi
r_{1}^{2}\right)
\end{eqnarray*}
Remember to convert $n$ from turns/cm to turns/m.
\end{quote}
12) Now reverse the situation. If the current in the short coil is ramped up
steadily from zero to $I_s = 2.90\ A$ over a time interval of $25.0\ ms$, what
is the magnitude of the emf in the solenoid while the current in the coil is
changing?
\begin{quote}
$I_{c}$ = Given\\
$t$ = Given\\
Realizing that the mutual inductance $M$ is a constant for a given
configuration of wire, the EMF is simply given by the equation
\[
E_{2}=-M\frac{dI_{c}}{dt}
\]
Which we can calculate by
\[
E_{2}=-M\frac{\Delta I_{c}}{\Delta t}
\]
\end{quote}
13) A very long solenoid with a circular cross section and radius
$r = 5.10\ cm$ with $n = 1.60 \times 10^4\ turns/m$ has a magnetic energy
density $u_B = 7.80\ mJ/m^3$. What is the current in the solenoid?
\begin{quote}
$r$ = Given\\
$n$ = Given\\
$u_{B}$ = Given\\
The equation for magnetic energy density is
\[
u_{B}=\frac{B^{2}}{2\mu _{0}}
\]
which we can use to solve for $B$:
\[
\sqrt{2\mu _{0}u_{B}}=B
\]
Knowing the magnetic field of a solenoid, we can now find the current:
\begin{eqnarray*}
\sqrt{2\mu _{0}u_{B}} & = & B=\mu _{0}nI\\
I & = & \frac{\sqrt{2\mu _{0}u_{B}}}{\mu _{0}n}
\end{eqnarray*}
\end{quote}
14) What is the total energy stored in the solenoid if its length is
$0.780\ m$? (Neglect end effects.)
\begin{quote}
$l$ = Given\\
The easiest way to do this problem is to realize the magnetic energy density
given in \#13 is just the magnetic field energy per volume. Thus, if we
can find the volume of a solenoid of length $l$, we can find the energy
stored in it. A solenoid is basically a cylinder. The volume of which is
\[
V=\pi r^{2}l
\]
Thus, the energy is
\[
U=u_{B}V=u_{B}\pi r^{2}l
\]
\end{quote}
15) The switch in the following circuit has been open for a long time. The
switch is then closed. The current through the battery immediately after the
switch is closed is $I(t=0)$ and the value at very large $t$ is $I(t=\infty)$.
Calculate the ratio $I(t=0) / I(t=\infty)$.
\begin{quote}
$E$ = Given\\
$R_{1}$ = Given\\
$R_{2}$ = Given\\
$L$ = Given\\
An inductor has mainly opposite properties than those of a capacitor.
Thus, initially it acts like an infinite resistance and after a long time it
acts like a piece of wire. The ratio of the current initially ($I_{0}$) to
the final current ($I_{\infty }$) can be calculated in two steps.
1) Since no current initially flows through the inductor, it must flow
through the other two resistors. Imagine the circuit as being open where
the inductor is so that the two resistors are in series. Using Ohm's law,
\[
I_{0}=\frac{V}{R}=\frac{E}{R_{1}+R_{2}}
\]
2) After a long time, the inductor acts like a piece of wire (effectively
zero resistance). Thus, all the current will flow through it. Imagine
$R_{2}$ has been removed from the circuit. Then, only $R_{1}$ determines
the current.
\[
I_{\infty }=\frac{V}{R}=\frac{E}{R_{1}}
\]
Taking the ratio of these two yields the final answer
\[
\frac{I_{0}}{I_{\infty }}=\frac{\frac{E}{R_{1}+R_{2}}}{\frac{E}{R_{1}}}=%
\frac{R_{1}}{R_{1}+R_{2}}
\]
\end{quote}
16) A circular coil with one turn is in a perpendicular (time dependent)
magnetic field given by $B = 1.500 - 0.0800t\ Tesla$, where time t is in
seconds. The induced voltage in the loop is $2.56\ V$. Calculate the radius
of the coil.
\begin{quote}
$B(t)$ = Given\\
$E$ = Given\\
From Faraday's law, we know the induced EMF is given by
\begin{eqnarray*}
E & = & -\frac{d\phi }{dt}=-\frac{d}{dt}\left( \int B\cdot dA\right) =-\frac{d}{dt}BA\\
E & = & -A\frac{d}{dt}B\\
A & = & \pi r^{2}=-\frac{E}{\frac{dB}{dt}}\\
r & = & \sqrt{-\frac{E}{\pi \frac{dB}{dt}}}
\end{eqnarray*}
The signs work out okay since the derivative of $B(t)$ is negative.
\end{quote}
17) A 230 turn conducting coil with a radius of $15.7\ cm$ rotates at a
frequency of $f = 70.0\ Hz$ in a magnetic field $B = 0.210\ T$. Calculate
the generated {\em rms emf}. The rms (root-mean-square) value of a
sinusoidal quantity is the amplitude divided by root of 2.
\begin{quote}
$N$ = Given\\
$r$ = Given\\
$f$ = Given\\
$B$ = Given\\
Example 31-6 on page 792 of the text is exactly the same as this problem.
The spinning coil is a generator. The equation for the induced EMF as a
function of time is derived. However, we are only concerned with the peak
voltage. The equation given is
\[
E=2\pi NfBA\sin \left( 2\pi ft\right)
\]
The EMF will be maximum when $\sin $ is 1, so we can ignore the last term.
Simply plugging into the equation (where $A$ is the area of the coil)
\[
E=2\pi NfB\pi r^{2}=2\pi ^{2}NfBr^{2}
\]
The rms EMF is the maximum voltage divided by the root of 2. Thus,
\[
E_{rms}=\frac{2\pi ^{2}NfBr^{2}}{\sqrt{2}}=\sqrt{2}\pi ^{2}NfBr^{2}
\]
\end{quote}
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