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\begin{center}
{\Large {Solution Derivations for Capa \#9\\[20pt]} }
\end{center}
1) Assume that a lightning bolt can be represented by a long straight line
of current. If $15.0\ C$ of charge passes by in a time of
$1.5 \times 10^{-3}\ s$, what is the magnitude of the magnetic field at a
distance of $28.0\ m$ from the bolt?
\begin{quote}
$Q$ = Given\\
$t$ = Given\\
$r$ = Given\\
Current is defined to be $I=\frac{dQ}{dt}$. We know that a total charge of
$Q$ passed by in a time $t$, so the current is $I=\frac{Q}{t}$. The
magnetic field of an infinite line is given by
\[
B=\frac{\mu _0 I}{2\pi r}=\frac{\mu _0 Q}{2\pi rt}
\]
\end{quote}
2) A small diameter, $21\ cm$ long solenoid has $263$ turns and is
connected in series with a resistor and battery. The size of the
magnetic field inside the solenoid is $6.662 \times 10^{-4}\ T$ when a
voltage of $80.0\ V$ is applied to the circuit. Calculate the resistance
of the circuit. Use units of ``Ohm''.
\begin{quote}
$l$ = Given\\
$N$ = Given\\
$B$ = Given\\
$V$ = Given\\
$R$ = ?\\
The B-field of a solenoid is given by
\[
B=\mu _{0}nI=\frac{\mu _{0}NI}{l}
\]
where the last step follows since $n=\frac{N}{l}$. From Ohm's Law,
\[
I=\frac{V}{R}
\]
Thus,
\[
B=\frac{\mu _{0}NV}{lR}
\]
So,
\[
R=\frac{\mu _{0}NV}{lB}
\]
\end{quote}
3) The wire in the above figure carries a current $I$ which travels into
the paper. North is up, East is right, etc. (For each statement select
T True, F False). WARNING: you have only 4 tries.
\begin{figure}[h]
\begin{center}
\includegraphics[width=4.4053in]{capa09-03.png}
\end{center}
\end{figure}
\begin{quote}
QUESTION:\\
A) The magnetic field at $p_{1}$ points east. \\
B) The magnetic field at $p_{1}$ points south. \\
C) The magnetic field at $p_{1}$ points out of the paper. \\
ANSWER:\\
A) False, Use the right hand rule. Point your thumb in the direction of
I, your fingers will curl in the direction of B. It is south.\\
B) True, See (A) \\
C) False, See (A)
\end{quote}
4) If the magnetic field at point $p_1$ is $2.5$ tesla, what is the magnetic
field at $p_2$?
\begin{quote}
$B_{p_{1}}$ = Given\\
$B_{p_{2}}$ = ?\\
The magnetic field due to a long, straight wire is
\[
B=\frac{\mu _{0}I}{2\pi r}
\]
We are told this field at point $p_{1}$.
\[
B_{p_{1}}=\frac{\mu _{0}I}{2\pi r}
\]
At point $p_{2}$, the distance $r$ is twice that at $p_{1}$. Thus,
\[
B_{p_{2}}=\frac{\mu _{0}I}{2\pi \left( 2r\right) }=\frac{1}{2}\frac{\mu
_{0}I}{2\pi r}=\frac{1}{2}B_{p_{1}}
\]
\end{quote}
5) In the figure below, the bottom wire carries a current of $4.0\ A$. The
current of the upper wire is directed in the opposite direction, and the
total magnetic field at point $p_1$ is zero. What is the current in the
top wire?
\begin{quote}
$I_{2}$ = Given\\
$B_{p_{1}}$ = Given (0)\\
Since we know the field at $p_{1}$, we can find other unknowns. Equating
the field due to each wire at $p_{1}$ and setting the equations equal to
each other,
\begin{eqnarray*}
B_{2} & = & \frac{\mu _{0}I_{2}}{2\pi a}\\
B_{1} & = & \frac{\mu _{0}I_{1}}{2\pi \left( 3a\right) }
\end{eqnarray*}
Thus,
\begin{eqnarray*}
\frac{\mu _{0}I_{2}}{2\pi a} & = & \frac{\mu _{0}I_{1}}{2\pi \left( 3a\right) }\\
I_{2} & = & \frac{I_{1}}{3}\\
I_{1} & = & 3I_{2}
\end{eqnarray*}
\end{quote}
6) Given that the wires are separated by $2a$, where $a = 0.021\ m$, what is
the magnitude of the magnetic field at point $p_2$?
\begin{quote}
$a$ = Given\\
Using the right hand rule, the B-fields due to each wire point into the
page at point $p_{2}$. Thus, we can add them together.
\begin{eqnarray*}
B_{p_{2}} & = & B_{1}+B_{2}\\
B_{1} & = & \frac{\mu _{0}I_{1}}{2\pi a}\\
B_{2} & = & \frac{\mu _{0}I_{2}}{2\pi a}\\
B_{p_{2}} & = & \frac{\mu _{0}I_{1}}{2\pi a}+\frac{\mu _{0}I_{2}}{2\pi a}=\frac{%
\mu _{0}}{2\pi a}\left( I_{1}+I_{2}\right)
\end{eqnarray*}
\end{quote}
7) In the figure below, a long straight wire carries a current of
$I_a = 5.0\ A$. A square loop of edge length $0.25\ m$ is placed a distance
of $0.10\ m$ away from the wire. The square loop carries a current
$I_b = 2.5\ A$. Find the magnitude of the net force on the square loop.
\begin{quote}
$I_a$ = Given\\
$I_b$ = Given\\
$L$ = Given (edge length)\\
$d$ = Given (loop separation)\\
From symmetry, it can be argued that the force on the sides of the loop
below will cancel each other out. Using the right hand rule, the B-Field
everywhere on the loop is into the page. The total force on the loop will
be that due to the top portion and that of the bottom portion. From the
right hand rule, using the equation $F=IL\times B$, the bottom portion of
the loop will feel an attractive force while the top portion will be
repelled. Choosing up to be a positive direction,
\[
F_{net}=F_{bot}-F_{top}
\]
Since the wire, the B-Field, and the magnetic force are orthogonal (all
perpendicular),
\[
F=IL\times B=ILB
\]
Thus,
\[
F_{net}=I_{b}LB_{bot}-I_{b}LB_{top}
\]
Since the wire is long and straight,
\begin{eqnarray*}
B_{bot} & = & \frac{\mu _{0}I_{a}}{2\pi r}=\frac{\mu _{0}I_{a}}{2\pi \left(
d+L\right) }\\
B_{top} & = & \frac{\mu _{0}I_{a}}{2\pi r}=\frac{\mu _{0}I_{a}}{2\pi d}\\
F_{net} & = & I_{b}L\frac{\mu _{0}I_{a}}{2\pi \left( d+L\right) }-I_{b}L\frac{\mu
_{0}I_{a}}{2\pi d}=I_{b}L\frac{\mu _{0}I_{a}}{2\pi }\left( \frac{1}{d+L}-
\frac{1}{d}\right)
\end{eqnarray*}
\end{quote}
8) A single-coil loop of radius $r = 7.10\ mm$, shown below, is formed in
the middle of an infinitely long, thin, insulated straight wire carrying the
current $i = 47.0\ mA$. What is the magnitude of the magnetic field at the
center of the loop?
\begin{quote}
$r$ = Given\\
$I$ = Given\\
Since the magnetic field obeys the laws of superposition, we can add the
fields due to each of the components (the line and the loop). From the book,
the B-Field of an infinite line is
\[
B_{line}=\frac{\mu _{0}I}{2\pi r}.
\]
From the lecture notes, the B-Field of a loop is
\begin{eqnarray*}
B_{loop} & = & \frac{\mu _{0}I}{2r}\\
B_{tot} & = & B_{line}+B_{loop}\\
& = & \frac{\mu _{0}I}{2\pi r}+\frac{\mu _{0}I}{2r}=\frac{\mu _{0}I}{2r}
\left( \frac{1}{\pi }+1\right)
\end{eqnarray*}
\end{quote}
9) The four wires that lie at the corners of a square of side $a = 3.80\ cm$
are carrying equal currents $i = 2.30\ A$ into (+) or out of (-) the page, as
shown in the picture. Calculate the $y$ component of the magnetic field at
the center of the square.
\begin{quote}
$a$ = Given\\
$I$ = Given\\
Using the right hand rule, we can see that the magnetic field from the top
left and bottom right wires points in the upper right (the (+) direction is
into the page). The other two corners have their magnetic field pointing
in the upper left. The x-components for each B-Field vector will cancel
leaving only the y-component determining the magnitude. From symmetry, the
magnetic field from each wire is equal at the center. Thus, the total
B-Field is four times that due to one of the wires. Also from symmetry,
the angle each vector makes with the y-axis is $45^\circ$. Thus, it does
not matter whether we use the sine or cosine to find the vertical component.
\[
B=4\frac{\mu _{0}I}{2\pi r}\sin 45^\circ=4\frac{\mu _{0}I}{2\pi r}
\frac{1}{\sqrt{2}}=4\frac{\mu _{0}I}{2\sqrt{2}\pi r}
\]
Again from symmetry, $r$ is the length of the hypotenuse whose side lengths
are both $\frac{a}{2}$. Thus,
\begin{eqnarray*}
r & = & \sqrt{2}\frac{a}{2}\\
B & = & 4\frac{\mu _{0}I}{2\sqrt{2}\pi \sqrt{2}\frac{a}{2}}=4\frac{2\mu _{0}I}{%
4\pi a}=\frac{2\mu _{0}I}{\pi a}
\end{eqnarray*}
\end{quote}
10) In the picture below, the two wires carry current $i_1$ and $i_2$
respectively, with positive current to the right. The charge q, midway
between the wires, is positive and has velocity v to the right. North is
up, East is to the right, etc. Which of the following statements for
magnetism are true? (Give ALL correct answers in alphabetical order, i.e.,
B, AC, BCD...)
\begin{figure}[h]
\begin{center}
\includegraphics[width=3.2638in]{capa09-10.png}
\end{center}
\end{figure}
\begin{quote}
QUESTION:\\
A) If $i_1 > i_2$, then the force on $q$ is north. \\
B) If $i_2=0$ and $i_1 > 0$, then the magnetic field
near charge $q$ points out of the page. \\
C) If $v=0$, then the force on $q$ is zero. \\
D) If $i_1=-i_2$, then the force on $q$ is zero. \\
E) If $i_1=0$ and $i_2 > 0$, then the force on $q$
points out of the page. \\
F) If $i_1=0$ and $i_2 > 0$, then the force on $q$ is
south. \\
G) If $i_1=i_2$, then the wires are attracted to each other. \\
ANSWER:\\
A) True, The top wire produces a magnetic field into the page at $q$. The
bottom wire produces a magnetic field pointing out of the page at $q$. If
the top current is stronger, the force will be north (from $qv\times B$).\\
B) False, The top wire produces a magnetic field into the page at $q$ (right
hand rule).\\
C) True, Magnetic forces are created by moving charges ($qv\times B$).\\
D) False, In this case, both wires produce a magnetic field out of the page
at $q$ and the force is downward.\\
E) False, The bottom wire produces a magnetic field pointing out of the page
at $q$. Thus, the force is south ($qv\times B$).\\
F) True, See (E).\\
G) True, parallel currents attract (see lecture notes 30-2).
\end{quote}
11) A solid rod of radius R carries a uniformly distributed current
$i_0 = 3.00\ A$ into the page. A wire lies a distance 2R away from the
surface of the rod. What is the magnitude and direction (into (+) or out
of (-) the page) of the current in the wire so that the magnetic field,
$\stackrel{\rightarrow}{\mathbf{B}}$, at a point P, half-way between the two, equals
the field at the center of the rod? (The two fields must be equal in
magnitude and direction.)
\begin{quote}
$I$ = Given\\
The magnetic field at the center of the rod is only due to the wire (since
there is no current of the rod contained in any amperian loop, there is no
B-Field due to it). Choose the direction of $I$ to be positive (into the
page) to avoid any imbedded negative signs.
\[
B_{center}=-\frac{\mu _{0}I_{wire}}{2\pi \left( 3R\right) }
\]
This is negative since the right hand rule gives the direction as down. At
point $P$, the B-Field is influenced by both the wire and the rod.
\[
B_{P,wire}=-\frac{\mu _{0}I_{wire}}{2\pi R}
\]
Where again it is negative since the direction is down.
\[
B_{P,rod}=\frac{\mu _{0}I_{rod}}{2\pi \left( 2R\right) }
\]
Where the radius is $2R$ since it is measured from the center of the rod
(just like the radius from a solid sphere of charge was taken from the
center). \ It is positive because the right hand rule gives the direction as
up. \ Equating the B-Field at point $P$ to that at the center of the rod
allows us to solve for the unknown current.
\begin{eqnarray*}
B_{center} & = & B_{P,wire}+B_{P,rod}\\
-\frac{\mu _{0}I_{wire}}{2\pi \left( 3R\right) } & = & -\frac{\mu _{0}I_{wire}}
{2\pi R}+\frac{\mu _{0}I_{rod}}{2\pi \left( 2R\right) }\\
-\frac{\mu _{0}}{2\pi R}\frac{I_{wire}}{3} & = & \frac{\mu _{0}}{2\pi R}\left(
-I_{wire}+\frac{I_{rod}}{2}\right) \\
-\frac{I_{wire}}{3} & = & -I_{wire}+\frac{I_{rod}}{2}\\
\frac{2I_{wire}}{3} & = & \frac{I_{rod}}{2}\\
I_{wire}& = & \frac{3}{4}I_{rod}
\end{eqnarray*}
\end{quote}
12) Two very long solenoids have the same length, but solenoid A has 17 times
the number of turns, 1/9 the radius, and 4 times the current of solenoid B.
Calculate the ratio of the magnetic field inside A to that inside B.
\begin{quote}
Given:\\
$L_A=L_B$\\
$N_A=xN_B$\\
$r_A=yr_B$\\
$I_A=zI_B$\\
where $x,y,z$ are proportionality constants.\\
The B-Field of a solenoid is given by
\[
B=\mu _{0}nI
\]
where $n=N/L$. Note that the radius is nowhere in the equation. The ratio
wanted is that of the field in A to that in B. Thus,
\[
\frac{B_{A}}{B_{B}}=\frac{\mu _{0}\frac{N_{A}}{L}I_{A}}{\mu _{0}\frac{N_{B}%
}{L}I_{B}}=\frac{\mu _{0}\frac{xN_{B}}{L}zI_{B}}{\mu _{0}\frac{N_{B}}{L}I_{B}%
}=x\ast z
\]
\end{quote}
13) The figure shows a hollow cylindrical conductor with radii $a = 1.1\ cm$
and $b = 4.2\ cm$ which carries a current $5.0\ A$ uniformly spread over its
cross-section. Find the magnitude of the magnetic field at a distance of
$2.6\ cm$ from the axis.
\begin{quote}
$a$ = Given\\
$b$ = Given\\
$I$ = Given\\
$r$ = Given (distance at which to calculate the magnetic field)\\
This is similar to concept test 30-3. From Amp\`{e}re's Law,
\[
\oint B\cdot dl=\mu _{0}I_{enc}
\]
Since the B-field is constant at a given radius and perpendicular to the
length vector, it can be taken outside the integral.
\begin{eqnarray*}
B\oint dl & = & \mu _{0}I_{enc}\\
B\ast 2\pi r & = & \mu _{0}I_{enc}\\
B & = & \frac{\mu _{0}I_{enc}}{2\pi r}
\end{eqnarray*}
So the hollow rod behaves like a wire. Now we need to know the current
enclosed. We can find the current density of the rod and then multiply by
the area to radius $r$ to find the current enclosed. This is only possible
since the current is uniformly distributed.
\[
J=\frac{I}{A}=\frac{I}{\pi b^{2}-\pi a^{2}}
\]
Where the area is just the total area of the conductor minus the area of the
hollow part. The area of the conductor enclosed to the radius $r$ is
\[
A_{enc}=\pi r^{2}-\pi a^{2}
\]
Thus,
\begin{eqnarray*}
I_{enc} & = & J\ast A_{enc}\\
& = & \frac{I}{\pi b^{2}-\pi a^{2}}\left( \pi r^{2}-\pi a^{2}\right)\\
& = & \frac{I\pi \left( r^{2}-a^{2}\right) }{\pi \left(
b^{2}-a^{2}\right) }=\frac{I\left( r^{2}-a^{2}\right) }{\left(
b^{2}-a^{2}\right) }
\end{eqnarray*}
So,
\[
B=\frac{\mu _{0}I_{enc}}{2\pi r}=\frac{\mu _{0}I\left( r^{2}-a^{2}\right) }{%
2\pi r\left( b^{2}-a^{2}\right) }
\]
\end{quote}
14) A long hairpin is formed by bending an infinitely long wire, as shown.
If a current of $4.8\ A$ is set up in the wire, what is the magnitude of the
magnetic field at the point $a$? Assume $R = 6.0\ cm$.
\begin{quote}
$I$ = Given\\
$R$ = Given\\
This problem can be thought of as consisting of half a loop of wire and two
half-infinite lengths of wire (is there such a thing?). Thus,
\begin{eqnarray*}
B_{a} & = & \frac{1}{2}B_{loop}+2\left( \frac{1}{2}B_{wire}\right)\\
& = & \frac{1}{2}\frac{\mu _{0}I}{2R}+\frac{\mu _{0}I}{2\pi R}\\
& = & \frac{\mu _{0}I}{2R}\left(\frac{1}{2}+\frac{1}{\pi }\right)
\end{eqnarray*}
\end{quote}
15) Decide from the following list of possibilities what is the
appropriate direction of the force on the wire for each of the diagrams
1 through 4: WARNING: you have only 8 tries for this problem.
\begin{quote}
Using the right hand rule and the equation $IL\times B$,\\
A) out of the page\\
B) zero force. The vectors are antiparallel, $IL\times B=ILB\sin \theta $,
$\sin 180^\circ=0$\\
C) into the page\\
D) out of the page
\end{quote}
\end{document}