\end{problem*}
\begin{solution}
+Start by labeling everything.
+\begin{center}
+\begin{verbatim}
+ 5.00=R5 I5 I4 8.00=R4
++------/\/\/--->--+--<---/\/\/------+
+|12.0V 1.00=R2 I2 a I1 1.00=R1 9.00V|
+b--|i---/\/\/--<--+-->--/\/\/---i|--c
+| I3 10.00=R3 |
++-------->------/\/\/---------------+
+\end{verbatim}
+\end{center}
+Then junction rules at $a$ and $b$ yield
+\begin{align}
+ 0 &= -I_1 - I_2 + I_5 + I_4 \\
+ 0 &= I_2 - I_3 - I_5
+\end{align}
+Loop rules for each sub-loop clockwise from the loop's upper left yield
+\begin{align}
+ 0 &= -I_5R_5 - I_2R_2 + V_2 \\
+ 0 &= +I_4R_4 - V_1 + I_1R_1 \\
+ 0 &= -V_2 + I_2R_2 - I_1R_1 + V_1 + I_3R_3
+\end{align}
+Which gives us five equations for the five unknown currents. Solve
+using your method of choice (e.g. linear algebra).
+\begin{align}
+ \begin{pmatrix}
+ 0 \\
+ 0 \\
+ -V_2 \\
+ V_1 \\
+ V_2-V_1
+ \end{pmatrix}
+ &=
+ \begin{pmatrix}
+ -1 & -1 & 0 & 1 & 1 \\
+ 0 & 1 & -1 & 0 & -1 \\
+ 0 & -R_2 & 0 & 0 & -R_5 \\
+ R_1 & 0 & 0 & R_4 & 0 \\
+ -R_1 & R_2 & R_3 & 0 & 0
+ \end{pmatrix}
+ \begin{pmatrix}
+ I_1 \\
+ I_2 \\
+ I_3 \\
+ I_4 \\
+ I_5
+ \end{pmatrix} \\
+ \begin{pmatrix}
+ I_1 \\
+ I_2 \\
+ I_3 \\
+ I_4 \\
+ I_5
+ \end{pmatrix}
+ &=
+ \begin{pmatrix}
+ -1 & -1 & 0 & 1 & 1 \\
+ 0 & 1 & -1 & 0 & -1 \\
+ 0 & -R_2 & 0 & 0 & -R_5 \\
+ R_1 & 0 & 0 & R_4 & 0 \\
+ -R_1 & R_2 & R_3 & 0 & 0
+ \end{pmatrix}^{-1}
+ \begin{pmatrix}
+ 0 \\
+ 0 \\
+ -V_2 \\
+ V_1 \\
+ V_2-V_1
+ \end{pmatrix}
+ =
+ \ans{
+ \begin{pmatrix}
+ 0.848 \\
+ 2.14 \\
+ 0.171 \\
+ 1.02 \\
+ 1.97
+ \end{pmatrix}
+ \U{A}}
+\end{align}
\end{solution}
\end{problem*}
\begin{solution}
+The energy stored in the capacitor is $1/e$ of its maximum value when
+\begin{align}
+ U = \frac{q^2}{2C} &= \frac{U_0}{e} = \frac{Q_0^2}{2Ce} \\
+ q^2 &= \frac{Q_0^2}{e} \\
+ q &= Q_0 e^{-0.5} \;.
+\end{align}
+This value is useful for both \Part{a} and \Part{b}.
+
+\Part{a}
+For a discharging circuit,
+\begin{equation}
+ q = Q_0 e^{\frac{-t}{RC}} \;,
+\end{equation}
+so the time to $U=U_0/e$ is given by
+\begin{align}
+ \frac{-t}{RC} &= -0.5 \\
+ t &= \ans{\frac{RC}{2}} \;.
+\end{align}
+
+\Part{b}
+For a charging circuit, the
+\begin{equation}
+ q = Q_0 \p({1-e^{\frac{-t}{RC}}}) \;,
+\end{equation}
+so the time to $U=U_0/e$ is given by
+\begin{align}
+ 1-e^{\frac{-t}{RC}} &= e^{-0.5} \\
+ e^{\frac{-t}{RC}} &= 1-e^{-0.5} \\
+ \frac{-t}{RC} &= \ln\p({1-e^{-0.5}}) \\
+ t &= -RC\ln\p({1-e^{-0.5}}) = \ans{0.933RC} \;.
+\end{align}
\end{solution}
\end{problem*}
\begin{solution}
+Following the hint, we note that
+\begin{align}
+ R_T &= R_1 + \p({\frac{1}{R_2} + \frac{1}{R_T}})^{-1} + R_1 \\
+ R_T - 2R_1 &= \p({\frac{1}{R_2} + \frac{1}{R_T}})^{-1} \\
+ (R_T - 2R_1) \cdot \p({\frac{1}{R_2} + \frac{1}{R_T}}) &= 1 \\
+ (R_T - 2R_1) \cdot (R_T + R_2) &= R_2R_T \\
+ R_T^2 - 2R_1R_T + R_2R_T - 2R_1R_2 &= R_2R_T \\
+ 0 &= R_T^2 - 2R_1R_T - 2R_1R_2 \;.
+\end{align}
+Plugging this into the quadratic formula
+\begin{equation}
+ R_T = \frac{2R_1 \pm \sqrt{4R_1^2 - 4\cdot1\cdot(-2R_1R_2)}}{2}
+ = R_1 \pm \sqrt{R_1^2 + 2R_1R_2}
+ = \ans{R_1 + \sqrt{R_1^2 + 2R_1R_2}} \;,
+\end{equation}
+which is what we set out to show. Note that we chose the $+$ case
+from $\pm$ because
+\begin{equation}
+ R_1 < \sqrt{R_1^2 + 2R_1R_2} \;,
+\end{equation}
+and $R_T$ must be greater than zero.
\end{solution}
\begin{problem*}{27.22}
In an experiment with cosmic rays, a verticle beam of particles that
-have chagre of magnitude $3e$ and mass $12$ times the proton mass
+have charge of magnitude $3e$ and mass $12$ times the proton mass
enters a uniform horizontal magnetic field of $0.250\U{T}$ and is bent
in a semicircle of diameter $95.0\U{cm}$, as shown in
Fig.~27.47. \Part{a} Find the speed of the particles and the sign of
\end{problem*}
\begin{solution}
+\Part{a}
+From the figure, the particle rotates counterclockwise with the
+\vect{B} field pointing out of the page. From the righthand rule and
+$\vect{F}=q\vect{v}\times\vect{B}$, the particle must have a negative
+charge. Because this centerward force makes the particle move in a
+circle,
+\begin{align}
+ F_c &= qvB = ma_c = m\frac{v^2}{r} \\
+ v &= \frac{rqB}{m}
+ = \frac{0.950\U{m}\cdot3\cdot1.6\E{-19}\U{C}\cdot0.250\U{T}}
+ {12\cdot1.67e-27\U{kg}}
+ = \ans{5.68\U{Mm/s}} \;.
+\end{align}
+
+\Part{b}
+The gravitational force $mg=1.97\E{-25}\U{N}$, while the magnetic
+force $qvB=6.82\E{-13}\U{N}$. Therefore, the gravitational force
+can be safely ignored.
+
+\Part{c}
+Because the magnetic force $\vect{F}=q\vect{v}\times\vect{B}$ is
+always perpendicular to the particle's velocity \vect{v}, the magnetic
+force will never change the speed of a particle. Only the direction
+of motion is affected. So the particle leaves the field with the same
+speed it entered with.
\end{solution}
\end{problem*}
\begin{solution}
+\Part{a}
+This is a straigtforward application of the Lorentz force.
+\begin{align}
+ \vect{F} &= 0 = q(\vect{E} + \vect{v}\times\vect{B}) \\
+ \vect{E} &= -\vect{v}\times\vect{B}
+ = 5.85\E{3}\U{m/s}\cdot1.35\U{T}(\jhat \times \khat)
+ = \ans{7.90\U{V/m}\ihat} \;.
+\end{align}
+
+\Part{b}
+Because the charge canceled out in \Part{a}, the electric field should
+be the same: $\vect{E}=\ans{7.90\U{V/m}\ihat}$.
\end{solution}
Distance Dhorizontal = Distance((0,0),(u,0), offset=2mm, L="$60.0\U{cm}$");
Distance Dvertical = Distance((u,0),(u,u), offset=2mm, L="$60.0\U{cm}$");
Distance Dbend = Distance((.25u,.25u),(.25u,.75u), offset=2mm, L="$30.0\U{cm}$");
+Vector I = Vector((-.25u,.25u), .125u, red, "$I$");
draw(scale(u)*((0,0)--(1,0)--(1,1)--(0,1)--cycle), blue);
draw(scale(u)*((-.25,.25)--(.25,.25)--(.25,.75)--(1.25,.75)), red);
+I.draw();
Dhorizontal.draw(labelangle=-90, labeloffset=8pt);
Dvertical.draw(labelangle=-90, labeloffset=8pt);
Dbend.draw(rotateLabel=false, labelangle=-90, labeloffset=22pt);
\end{problem*}
\begin{solution}
+The verticle portion of the wire feels a force of
+\begin{equation}
+ \vect{F}_v = I\vect{l}_v\times\vect{B} = Il_vB \ihat
+ = 4.50\U{A}\cdot0.300\U{m}\cdot0.240\U{T} \ihat
+ = 0.324\U{N}\ihat \;.
+\end{equation}
+The two horizontal legs combined feel
+\begin{equation}
+ \vect{F}_h = I\vect{l}_h\times\vect{B} = -Il_hB \jhat
+ = -4.50\U{A}\cdot0.600\U{m}\cdot0.240\U{T} \jhat
+ = -0.648\U{N}\jhat \;.
+\end{equation}
+This gives the total force a magnitude of
+\begin{equation}
+ F = \sqrt{F_v^2+F_h^2} = \ans{0.724\U{N}}
+\end{equation}
+in the direction
+\begin{equation}
+ \theta = \arctan\p({\frac{0.648\U{N}}{0.324\U{N}}})
+ = \arctan(2) = \ans{63.4\dg}
+\end{equation}
+below the horizontal.
\end{solution}
projects / course.git / commitdiff