--- /dev/null
+\begin{problem*}{24.11}
+Four closed surfaces, $S_1$ through $S_4$, together with the charges
+$-2Q$, $Q$, and $-Q$ are sketched in Figure P24.11. (The colored
+lines are the intersections of the surfaces with the page.) Find the
+electric flux through each surface.
+% -2Q
+%
+% Q
+%
+% -Q
+
+% S_1 contains -2Q, Q
+% S_2 contains Q, -Q
+% S_3 contains -2Q, Q, -Q
+% S_4 to the left, contains no charges
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{24.13}
+In the air over a particular region at an altitude of $500\U{m}$ above
+the ground, the electric field is $120\U{N/C}$ directed downward. At
+$600\U{m}$ above the ground, the electric field is $100\U{N/C}$
+downward. What is the average volume charge density in the layer of
+air between these two elevations? Is it positive or negative?
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{24.21}
+Figure P24.21 represents the top view of a cubic gaussian surface in a
+uniform electric field \vect{E} oriented parallel to the top and
+bottom faces of the cube. The field makes an angle $\theta$ with side
+$a$, and the area of each face is $A$. In sumbolic form, find the
+electric flux through \Part{a} face $a$, \Part{b} face $b$, \Part{c}
+face $c$, \Part{d} face $d$, and \Part{e} the top and bottom faces of
+the cube. \Part{f} What is the net electric flux through the
+cube? \Part{g} How much charge is enclosed within the gaussian
+surface?
+% ^
+% a E\th|
+% d b \ |
+% c \|
+
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+
+\Part{b}
+
+\Part{c}
+
+\Part{d}
+
+\Part{e}
+
+\Part{f}
+
+\Part{g}
+\end{solution}
--- /dev/null
+\begin{problem*}{24.31}
+A solid sphere of radius $40.0\U{cm}$ has a total positive charge of
+$26.0\U{$\mu$C}$ uniformly distributed throughout its volume.
+Calculate the magnitude of the electric field \Part{a}
+$0\U{cm}$, \Part{b} $10.0\U{cm}$, \Part{c} $40.0\U{cm}$, and \Part{d}
+$60.0\U{cm}$ from the center of the sphere.
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+The electric field at $0\U{cm}$ must be zero by symmetry. Because the
+situation is symmetric, there is no preferred direction for the
+electric field.
+
+\Part{b}
+The charge density inside the sphere is
+\begin{equation}
+ \rho = \frac{Q}{V(R)} = \frac{Q}{\frac{4}{3}\pi R^3} \;.
+\end{equation}
+
+Inside the sphere, the enclosed charge is
+\begin{equation}
+ q_\text{in} = \rho V(r) = \frac{Qr^3}{R^3} \l.
+\end{equation}
+
+By Gauss's law, the electric flux is
+\begin{equation}
+ \Phi_E = \frac{q_\text{in}}{\epsilon_0} = \frac{Qr^3}{\epsilon_0 R^3} \;,
+\end{equation}
+which is uniformly distributed over the surface of the gaussian
+sphere, so
+\begin{equation}
+ E = \frac{\Phi_E}{A}
+ = \frac{\frac{Qr^3}{\epsilon_0 R^3}}{4\pi r^2}
+ = \frac{Qr}{4\pi\epsilon_0 R^3} \;.
+ = \frac{kQr}{R^3} \;.
+\end{equation}
+The electric field at $r=10.0\U{cm}$ is therefore
+\begin{equation}
+ E(r=10.0\U{cm}) = \ans{365\U{kN/C}}
+\end{equation}
+
+\Part{c}
+On the surface of the sphere, the entire charge is enclosed, so
+\begin{equation}
+ E = \frac{kQ}{r^2} \;.
+\end{equation}
+For this particular $r$,
+\begin{equation}
+ E(r=40.0\U{cm}) = \ans{1.46\U{MN/C}} \;.
+\end{equation}
+
+\Part{d}
+Further outside the surface, we still enclose the same total charge, so
+\begin{equation}
+ E(r=60.0\U{cm}) = \ans{649\U{kN/C}} \;.
+\end{equation}
+
+All of these electric fields are directed radially outward.
+\end{solution}
--- /dev/null
+\begin{problem*}{24.33}
+Consider a long, cylindrical charge distribution of radius $R$ with a
+uniform charge density $\rho$. Find the electric field at distance
+$r$ from the axis, where $r<R$.
+\end{problem*}
+
+\begin{solution}
+The charge enclosed by a concentric gaussian cylinder of length $L$ is
+\begin{equation}
+ q_\text{in} = \rho V = \rho \pi r^2 L \;,
+\end{equation}
+because $r<R$.
+
+By Gauss's law, the electric flux is
+\begin{equation}
+ \Phi_E = \frac{q_\text{in}}{\epsilon_0}
+ = \frac{\pi r^2 \rho L}{\epsilon_0} \;.
+\end{equation}
+
+None of this flux comes through the end caps of the gaussian cylinder, so
+\begin{align}
+ EA &= E(2\pi r L) = \Phi_E = \frac{\pi r^2 \rho L}{\epsilon_0} \\
+ E &= \ans{\frac{\rho r}{2 \epsilon_0}} \;.
+\end{align}
+\end{solution}
--- /dev/null
+\begin{problem*}{24.41}
+Two identical conducting spheres each having a radius of $0.500\U{cm}$
+are connected by a light, $2.00\U{m}$ long conducting wire. A charge
+of $60.0\U{$\mu$C}$ is placed on one of the conductors. Assume the
+surface distribution of charge on each sphere is uniform. Determine
+the tension in the wire.
+\end{problem*}
+
+\begin{solution}
+There will be $Q/2$ on each of the conductors, which we are supposed
+to assume is distributed evenly along the surface. So from the
+outside each sphere will look like a point charge $Q/2$ located at the
+sphere center. The distance between the sphere centers is
+$R+L+R=L+2R$, so
+\begin{equation}
+ F = \frac{k\cdot\frac{Q}{2}\cdot\frac{Q}{2}}{(L+2R)^2}
+ = \frac{kQ^2}{4(L+2R)^2}
+ = \ans{2.00\U{N}}
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{24.51}
+A solid insulating sphere of radius $a=5.00\U{cm}$ carries a net
+positive charge of $Q=3.00\U{$\mu$C}$ uniformly distributed throughout
+its volume. Concentric with this sphere is a conducting spherical
+shell with inner radius $b=10.0\U{cm}$ and outer radius $c=15.0\U{cm}$
+as shown in Figure P24.51, having net charge $q=-1.00\U{$\mu$C}$.
+Prepare a graph of the magnitude of the electric field due to this
+configuration versus $r$ for $0<r<25.0\U{cm}$.
+\end{problem*}
+
+\begin{solution}
+
+\begin{center}
+\begin{asy}
+import graph;
+
+size(6cm, 4cm, IgnoreAspect);
+
+real k = 8.99e9;
+real a = 5;
+real b = 10;
+real c = 15;
+real r_max = 25;
+real q1 = 3e-6;
+real q2 = -1e-6;
+
+real E(real r) {
+ real q_enc = 0;
+
+ if (r == 0) {
+ return 0;
+ } else if (r < a) {
+ q_enc = q1 * r^3 / a^3;
+ } else if (r < b) {
+ q_enc = q1;
+ } else if (r < c) {
+ q_enc = 0;
+ } else {
+ q_enc = q1 + q2;
+ }
+
+ return k*q_enc/(r/100.)^2;
+}
+
+draw(graph(E, 0, r_max), red);
+xaxis("$r$ (cm)", xmin=0, xmax=r_max, LeftTicks);
+yaxis("$E$ (N/C)", ymin=0, ymax=max(E(a), E(b)), RightTicks);
+\end{asy}
+\end{center}
+
+At $r=b$ and $r=c$ there are sharp changes in electric field. To
+handle these properly, you'd need to know about the skin depth of the
+conductor, but for this course it's enough to say that the transition
+happens quickly.
+\end{solution}
--- /dev/null
+\begin{problem*}{25.7}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{25.12}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{25.16}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{25.21}
+Two particles each with charge $+2.00\U{$\mu$C}$ are lopcated on the
+$x$ axis. One is at $x=1.00\U{m}$, and the other is at
+$x=-1.00\U{m}$. \Part{a} Determine the electric potential on the $y$
+axis at $y=0.500\U{m}$. \Part{b} Calculate the change in electric
+potential energy of the system as a third charged particle of
+$-3.00\U{$\mu$C}$ is brought in from infinitely far away to a positino
+on the $y$ axis at $y=0.500\U{m}$.
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+\begin{equation}
+ V_3 = k_e \frac{q_1}{r_13} + k_e \frac{q_2}{r_12}
+ = 2 k_e \frac{2\U{$\mu$C}}{\sqrt{(1.00\U{m})^2 + (0.500\U{m})^2}}
+ = \ans{32.2\U{kV}}
+\end{equation}
+
+\Part{b}
+\begin{equation}
+ \Delta U = q_3 V_3 - q_3 V_\infty = q_3 V_3 = \ans{-96.5\U{mJ}}
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{25.27}
+Two insulating spheres have radii $0.300\U{cm}$ and $0.500\U{cm}$,
+masses $0.100\U{kg}$ and $0.700\U{kg}$, and uniformly distributed
+charges $-2.00\U{$\mu$C}$ and $3.00\U{$\mu$C}$. They are released
+from rest when their centers are separated by $1.00\U{m}$. \Part{a}
+How fast will each be moving when they collide? \Part{b} \emph{What
+if?} If the spheres were conductors, would the speeds be greater
+or less than those calculated in part \Part{a}? Explain.
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+Conserving momentum
+\begin{align}
+ p_i = 0 &= p_f = m_1 v_1 - m_2 v_2 \\
+ m_2 v_2 &= m_1 v_1 \\
+ v_2 &= \frac{m_1}{m_2} v_1 \;.
+\end{align}
+
+The electric potential energy is initially
+\begin{equation}
+ U_i = qV_i = k_e \frac{q_1q_2}{r_{12,i}} \;,
+\end{equation}
+and the final electric potential energy is
+\begin{equation}
+ U_f = qV_f = k_e \frac{q_1q_2}{R_1+R_2} \;.
+\end{equation}
+Conserving energy
+\begin{align}
+ E_i = U_i &= E_f = U_f + K_{f,1} + K_{f,2} \\
+ K_{f,1} + K_{f,2} &= U_i - U_f \\
+ \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2
+ &= k_e \frac{q_1q_2}{r_{12,i}} - k_e \frac{q_1q_2}{R_1+R_2} \\
+ \frac{1}{2}\p({m_1 v_1^2 + m_2 \p({\frac{m_1}{m_2}v_1})^2})
+ &= k_e q_1 q_2 \p({\frac{1}{r_{12,i}} - \frac{1}{R_1+R_2}}) \\
+ \p({m_1 + \frac{m_1^2}{m_2}}) v_1^2
+ &= 2 k_e q_1 q_2 \p({\frac{1}{r_{12,i}} - \frac{1}{R_1+R_2}}) \\
+ \frac{m_1m_2 + m_1^2}{m_2} v_1^2
+ &= 2 k_e q_1 q_2 \p({\frac{1}{r_{12,i}} - \frac{1}{R_1+R_2}}) \\
+ v_1^2 &= \frac{2 k_e q_1 q_2 m_2}{m_1m_2 + m_1^2}
+ \p({\frac{1}{r_{12,i}} - \frac{1}{R_1+R_2}}) \\
+ v_1 &= \sqrt{\frac{2 k_e q_1 q_2 m_2}{m_1m_2 + m_1^2}
+ \p({\frac{1}{r_{12,i}} - \frac{1}{R_1+R_2}})}
+ = \ans{10.8\U{m/s}} \\
+ v_2 &= \frac{m_1}{m_2}v_1
+ = \sqrt{\frac{2 k_e q_1 q_2 m_1}{m_2^2 + m_1m_2}
+ \p({\frac{1}{r_{12,i}} - \frac{1}{R_1+R_2}})}
+ = \ans{1.55\U{m/s}}
+\end{align}
+
+\Part{b}
+The speeds for conductors would be greater, because as the spheres
+close the charge polarization will intensify, leading to a greater
+effective $\Delta r_{12}$ over a more sensitive portion of the $1/r$
+voltage curve.
+\end{solution}
--- /dev/null
+\begin{problem*}{25.36}
+An electric field in a region of space is parallel to the $x$ axis.
+The electric potential varies with position as shown in Figure P25.36.
+Graph the $x$ component of the electric field versus position in this
+region of space.
+\begin{center}
+\begin{asy}
+import graph;
+
+size(6cm, 4cm, IgnoreAspect);
+
+real xmin=0;
+real xmax=5;
+real Vmin=-30;
+real Vmax=30;
+
+draw((0,0)--(1,Vmax)--(2,0)--(3,0)--(4,Vmin)--(5,0), red);
+xaxis("$x$ (cm)", xmin=xmin, xmax=xmax, LeftTicks(NoZero));
+yaxis("$V$ (V)", ymin=Vmin, ymax=Vmax, RightTicks);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+$E_x=-\deriv{x}{V}$, so
+\begin{center}
+\begin{asy}
+import graph;
+
+size(6cm, 4cm, IgnoreAspect);
+
+real xmin=0;
+real xmax=5;
+real Emin=-3e3; // 30V/1cm = 3kV/m
+real Emax=3e3;
+
+draw((0,Emin)--(1,Emin), red);
+draw((1,Emax)--(2,Emax), red);
+draw((2,0)--(3,0), red);
+draw((3,Emax)--(4,Emax), red);
+draw((3,Emin)--(4,Emin), red);
+xaxis("$x$ (cm)", xmin=xmin, xmax=xmax, LeftTicks(NoZero));
+yaxis("$E$ (V/m)", ymin=Emin, ymax=Emax, RightTicks);
+\end{asy}
+\end{center}
+\end{solution}
--- /dev/null
+\begin{problem*}{26.4}
+An air-filled parallel-plate capacitor has plates of area
+$2.30\U{cm$^2$}$ separated by $1.50\U{mm}$. \Part{a} Find the value
+of its capacitance. The capacitor is connected to a $12.0\U{V}$
+battery. \Part{b} What is the charge on the capacitor? \Part{c} What
+is the magnitude of the uniform electric field between the plates?
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{26.16}
+Given a $2.50\U{$\mu$F}$ capacitor, a $6.25\U{$\mu$F}$ capacitor, and
+a $6.00\U{V}$ battery, find the charge on each capacitor if you
+connect them \Part{a} in series across the battery and \Part{b} in
+parallel across the battery.
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{26.18}
+Find \Part{a} the equivalent capacitance of the capacitors in
+Figure~P26.18, \Part{b} the charge on each capacitor, and \Part{c} the
+potential difference across each capacitor.
+% +-6uF-+
+% +-+ +-+
+% | +-2uF-+ |
+% 8uF 8uF
+% | 9V |
+% +---|i----+
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{26.31}
+A $12.0\U{V}$ battery is connected to a capacitor, resulting in
+$54.0\U{$\mu$C}$ of charge stored on the capacitor. How much energy
+is stored in the capacitor?
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{26.32}
+The immediate cause of many deaths is ventricular fibrillation, which
+is an uncoordinated quivering of the heart. An electric shock to the
+chest can cause momentary paralysis of the heart muscle, after which
+the heart sometimes resumes its proper beating. One type of
+\emph{defibrillator} (chapter opening photo, page 740) applies a stron
+electric shock to the chest over a time interval of a few
+milliseconds. This device contains a capacitor of several
+microfarads, charged to several thousand volts. Electrodes called
+paddles are held against the chest on both sides of the heart, and the
+capacitor is discharged through the patient's chest. Assume an energy
+of $300\U{J}$ is delivered from a $30.0\U{$\mu$F}$ capacitor. To what
+potential difference must it be charged?
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{26.57}
+A $2.00\U{nF}$ parallel-plate capacitor is charged to an initial
+potential difference of $\Delta V_i=100\U{V}$ and is then isolated.
+The dielectric material between the plates is mica, with a dielectric
+constant of $5.00$. \Part{a} How much work is required to withdraw
+the mica sheet? \Part{b} What is the potential difference across the
+capacitor after the mica is withdrawn?
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{26.67}
+Capacitors $C_1=6.00\U{$\mu$F}$ and $C_2=2.00\U{$\mu$F}$ are charged
+as a parallel combination across a $250\U{V}$ battery. The capacitors
+are disconnected from the battery and from each other. Then they are
+connected positive plate to negative plate and negative plate to
+positive plate. Calculate the resulting charge on each capacitor.
+\end{problem*}
+
+\begin{solution}
+After charging, the charges on the capacitors are
+\begin{align}
+ Q_1 &= C_1 V = 1.50\U{mC} \\
+ Q_2 &= C_2 V = 500\U{$\mu$C} \;.
+\end{align}
+
+After disconnecting the battery, flipping $C_2$, and reconnecting, the
+total charge on one side is $Q_t=Q_2+(-Q_2)=1.00\U{mC}$. This charge
+is divided into $Q_1'$ and $Q_2'$ such that
+\begin{align}
+ Q_t &= Q_1' + Q_2' \\
+ V_1' = \frac{Q_1'}{C_1} &= V_2' = \frac{Q_2'}{C_2} \\
+ Q_2' &= Q_1'\frac{C_2}{C_1} \\
+ Q_1' + Q_1'\frac{C_2}{C_1} &= Q_t \\
+ Q_1' &= \frac{Q_t}{1+\frac{C_2}{C_1}} = \ans{750.0\U{$\mu$C} \\
+ Q_2' &= Q_1'\frac{C_2}{C_1}
+ = \frac{Q_t}{\frac{C_1}{C_2} + 1} = \ans{250.0\U{$\mu$C}}
+\end{solution}
--- /dev/null
+\begin{problem*}{27.1}
+A proton beam in an accelerator carries a current of $125\U{$\mu$A}$
+If the beam is incident on a target, how many protons strike the
+target in a period of $23.0\U{s}$.
+\end{problem*}
+
+\begin{solution}
+Current is defined as $I\equiv\deriv{t}{Q}$. Therefore, strike rate is
+\begin{equation}
+ \deriv{t}{N} = \deriv{t}{Q} \cdot \frac{1\U{proton}}{q} = \frac{I}{q} \;,
+\end{equation}
+where $q=1.60\E{-19}\U{C}$ is the charge of a single proton. The
+number of strikes in the allotted time is thus
+\begin{equation}
+ \Delta N = \deriv{t}{N} \cdot \Delta t = \frac{I\Delta t}{q}
+ = \ans{17.9\E{15}\U{protons} = 17.9\U{petaprotons}} \;.
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{27.17}
+A portion of Nichrome wire of radius $2.50\U{mm}$ is to be used in
+winding a heating coil. If the coil must draw a current of
+$9.25\U{A}$ when a voltage of $120\U{V}$ is applied across its ends,
+find \Part{a} the required resistance of the coil and \Part{b} the
+length of wire you must use to wind the coil.
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+Using Ohm's law
+\begin{align}
+ V &= IR \\
+ R &= \frac{V}{I} = \ans{13.0\U{\Ohm}}
+\end{align}
+
+\Part{b}
+Using the formula for conductor resistance
+\begin{align}
+ R &= \rho \frac{l}{A} \\
+ l &= \frac{RA}{\rho} = \frac{R\pi r^2}{\rho}
+ = \ans{255\U{m}}
+\end{align}
+where the approximate resistivity of Nichrome at $20\celsius$
+($\rho=1.00\E{-6}\U{\Ohm$\cdot$m}$) comes from Table~27.2.
+\end{solution}
--- /dev/null
+\begin{problem*}{27.29}
+A certain waffle iron is rated at $1.00\U{kW}$ when connected to a
+$120\U{V}$ source. \Part{a} What current does the waffle iron
+carry? \Part{b} What is its resistance?
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+Using the power formula
+\begin{align}
+ P &= IV \\
+ I &= \frac{P}{V} = \ans{8.33\U{A}}
+\end{align}
+
+\Part{b}
+Using Ohm's law
+\begin{align}
+ V &= IR \\
+ R &= \frac{V}{I} = \frac{V^2}{P} = \ans{14.4\U{\Ohm}}
+\end{align}
+\end{solution}
--- /dev/null
+\begin{problem*}{27.31}
+Suppose your portable DVD player draws a current of $350\U{mA}$ at
+$6.00\U{V}$. How much power does the player require?
+\end{problem*}
+
+\begin{solution}
+Using the power formula
+\begin{equation}
+ P = IV = \ans{2.10\U{W}} \;.
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{27.32}
+The potential difference across a resisting neuron in the human body
+is about $75.0\U{mV}$ and carries a curent of about $0.200\U{mA}$.
+How much power does the neuron release?
+\end{problem*}
+
+\begin{solution}
+Using the power formula
+\begin{equation}
+ P = IV = \ans{15.0\U{$\mu$W}} \;.
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{27.49}
+A car owner forgets to turn off the headlights of his car while it is
+parked in his garage. If the $12.0\U{V}$ battery in his car is rated
+at $90.0\U{A$\cdot$h}$ and each headlight requires $36.0\U{W}$ of
+power, how long will it take the battery to completely discharge?
+\end{problem*}
+
+\begin{solution}
+The current drawn by each headlinght is given by
+\begin{align}
+ P &= IV \\
+ I_1 &= \frac{P_1}{V} = 3.00\U{A} \;.
+\end{align}
+So the total current drawn from the battery is
+\begin{equation}
+ I_t = 2I_1 = 6.00\U{A} \;.
+\end{equation}
+This will deplete the battery in
+\begin{equation}
+ \Delta t = \frac{\Delta Q}{\deriv{t}{Q}} = \frac{\Delta Q}{I_t}
+ = \frac{90.0\U{A$\cdot$h}}{6.00\U{A}}
+ = \ans{15.0\U{hours}} \;.
+\end{equation}
+If you want, you could plug in symbolically for $I_t$, giving
+\begin{equation}
+ \Delta t = \frac{\Delta Q V}{2P_1} \;.
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{28.2}
+Two $1.50\U{V}$ batteries---with their positive terminal in the same
+direction---are inserted in series into a flashlight. One battery has
+an internal resistance of $0.255\U{\Ohm}$, annd the other has an
+internal resistance of $0.153\U{\Ohm}$. When the switch is closed,
+the bulb carries a current of $600\U{mA}$. \Part{a} What is the
+bulb's reistance? \Part{b} What fraction of the chemical energy
+tranformed appears as internal energy in the batteries?
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{28.11}
+A battery with $\varepsilon=6.00\U{V}$ and no internal resistance
+supplies current to the circuit shown in Figure~P28.11. When the
+double-throw switch $S$ is open as shown in the figure, the current in
+the battery is $1.00\U{mA}$. When the switch is closed in position
+$a$, the current in the battery is $1.20\U{mA}$. When the switch is
+closed in position $b$, the current in the battery is $2.00\U{mA}$.
+Find the resistances \Part{a} $R_1$, \Part{b} $R_2$, and \Part{c}
+$R_3$.
+% +---R1---+----R2----+
+% | | |
+% | R2 |
+% _ a |
+% ___ S ----------+
+% | b R3
+% | | |
+% +--------+----------+
+\end{problem*}
+
+\begin{solution}
+
+\end{solution}
--- /dev/null
+\begin{problem*}{28.21}
+The circuit shown in Figure~P28.21 is connected for
+$2.00\U{min}$. \Part{a} Determine the current in each branch of the
+circuit. \Part{b} Find the energy delivered by each
+battery. \Part{c} Find the energy delivered to each
+resistor. \Part{d} Identify the type of energy storage transformation
+that occurs in the operation of the circuit. \Part{e} Find the total
+amount of energy transformed into internal energy in the resistors.
+% +-----+----3O--+
+% | 5O |
+% | | 1O
+% | 1O |
+% | | |
+% 8O ___ ___
+% | - 4V - 12V
+% +-----+--------+
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{28.26}
+The following equations describe an electric circuit:
+\begin{align}
+ -I_1 (220\U{\Ohm}) + 5.80\U{V} - I_2 (370\U{\Ohm}) &= 0 \\
+ +I_2 (370\U{\Ohm}) + I_3 (150\U{\Ohm}) - 3.10\U{V} &= 0 \\
+ I_1 + I_3 - I_2 &= 0
+\end{align}
+\Part{a} Draw a [possible] diagram of the circuit. \Part{b}
+Calculate the unknowns sand identify the physical meaning of each
+unknown.
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{28.59}
+The pair of capacitors in Figure~P28.59 are fully charged by a
+$12.0\U{V}$ battery. The battery is disconnected, and the switch is
+then closed. After $1.00\U{ms}$ have elapsed, \Part{a} how much
+charge remains on the $3.00\U{$\mu$F}$ capacitor? \Part{b} How much
+charge remains on the $2.00\U{$\mu$F}$ capacitor? \Part{c} What is
+the current in the resistor at this time?
+% +---||---+
+% | 3 |
+% +---||---+
+% S 2 |
+% | |
+% +--5O----+
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{28.71}
+In Figure~P28.71, suppose the switch has been closed for a time
+interval sufficiently long for the capacitor to become fully charged.
+Find \Part{a} the steady state current in each resistor and \Part{b}
+the charge $Q$ on the capacitor. \Part{c} The switch is now opened at
+$t=0$. Write an equation for the current in $R_2$ as a function of
+time and \Part{d} find the time interval required for the charge on
+the capacitor to fall to one-fifth its initial value.
+\begin{center}
+\begin{asy}
+% +---S----12kO--+-----+
+% | | ___
+%___ | ___ 10uF
+% - R2=15kO |
+% | | 3kO
+% | | |
+% +--------------+-----+
+import Circ;
+
+real u = 1cm;
+real dx = u/2;
+real dy = u/2;
+
+TwoTerminal S = switchSPST((0,0), ang=0, name="$S$");
+TwoTerminal R1 = resistor(S.end, ang=0, name="$R_1$", val="$12.0\U{k\Ohm}$");
+pair Pu = R1.end + (dx,0); // top junction
+pair Pul = S.beg - (dx,0); // upper-left corner
+pair Pur = Pu + (2u,0); // upper-right corner
+TwoTerminal C = capacitor(
+ Pur-(0,dy), ang=-90, name="$C$", val="$10.0\U{$\mu$F}$");
+TwoTerminal R3 = resistor(
+ C.end+(0,-dy), -90, name="$R_3$", val="$3.00\U{k\Ohm}$");
+TwoTerminal R2 = resistor(name="$R_2$", val="$15.0\U{k\Ohm}$", draw=false);
+R2.centerto(R3.end, C.beg, offset=Pur.x - Pu.x);
+R2.draw();
+pair Pb = (R2.beg.x, R3.end.y - dy);
+TwoTerminal V = source(type=DC, name="$V$", val="$9.00\U{V}$", draw=false);
+V.centerto(R3.end, C.beg, offset=Pur.x - Pul.x);
+V.draw();
+
+wire(R1.end, R2.end, rlsq);
+wire(Pu, C.beg, rlsq);
+wire(C.end, R3.beg);
+wire(R3.end, Pb, udsq);
+wire(R2.beg, Pb);
+wire(Pb, V.beg, rlsq);
+wire(V.end, S.beg, udsq);
+dot(Pu);
+dot(Pb);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+Label the currents $I_1$, $I_2$, and $I_3$ from left to right with
+each current moving up in its vertical wire. In the steady state,
+$I_3=\ans{0}$ (otherwise $Q$ would be changing, which would not be
+``steady state''). Applying the Kirchhoff's junction rule to the top
+junction.
+\begin{center}
+\begin{asy}
+import Circ;
+
+real dx = 6pt;
+
+TwoTerminal I1 = current((0, 0), ang=0, name="$I_1$");
+pair P = I1.end + (dx, 0);
+TwoTerminal I2 = current(P - (0, dx), ang=90, name="$I_2$", draw=false);
+I2.shift((0, -I2.len-dx));
+I2.draw();
+TwoTerminal I3 = current(P + (dx, 0), ang=180, name="$I_3$", draw=false);
+I3.shift((I3.len, 0));
+I3.draw();
+
+wire(I1.end, I3.end);
+wire(P, I2.end);
+dot(P);
+\end{asy}
+\end{center}
+\begin{align}
+ 0 &= I_1 + I_2 + I_3 = I_1 + I_2 \\
+ I_2 &= - I_1 \;.
+\end{align}
+
+Applying Kirchhoff's loop rule to the left-hand loop moving clockwise
+from the lower-left corner, we have
+\begin{center}
+\begin{asy}
+import Circ;
+
+real u = 2cm;
+
+wire((0, 0), (2u, u), rlsq);
+wire((0, 0), (2u, u), udsq);
+wire((u, 0), (u, u));
+dot((u, 0));
+dot((u, u));
+
+TwoTerminal I1 = current(name="$I_1$", draw=false);
+I1.centerto((0, 0), (0, u));
+I1.draw();
+TwoTerminal I2 = current(name="$I_2$", draw=false);
+I2.centerto((u, 0), (u, u));
+I2.draw();
+TwoTerminal I3 = current(name="$I_3$", draw=false);
+I3.centerto((2u, 0), (2u, u));
+I3.draw();
+
+pair[] points = {(0, 0), (0, u), (u, u), (u, 0)};
+kirchhoff_loop(points);
+\end{asy}
+\end{center}
+\begin{align}
+ 0 &= V - I_1 R_1 + I_2 R_2 \\
+ V &= I_1 R_1 - I_2 R_2 = I_1 R_1 + I_1 R_2 = I_1 (R_1+R_2) \\
+ I_1 &= \frac{V}{R_1+R_2} = \ans{333\U{$\mu$A}} \\
+ I_2 &= -I_1 = \ans{-333\U{$\mu$A}} \;.
+\end{align}
+So we have $333\U{$\mu$A}$ of current flowing clockwise through the
+left loop, and no current in the right loop.
+
+\Part{b}
+Applying Kirchhoff's loop rule to the right-hand loop moving clockwise
+from the lower-right corner, we have
+\begin{center}
+\begin{asy}
+import Circ;
+
+real u = 2cm;
+
+wire((0, 0), (2u, u), rlsq);
+wire((0, 0), (2u, u), udsq);
+wire((u, 0), (u, u));
+dot((u, 0));
+dot((u, u));
+
+pair[] points = {(2u, 0), (u, 0), (u, u), (2u, u)};
+kirchhoff_loop(points);
+\end{asy}
+\end{center}
+\begin{align}
+ 0 &= - I_2 R_2 - \frac{Q}{C} + I_3 R_3 \\
+ I_2 R_2 &= -\frac{Q}{C} \\
+ Q &= -C I_2 R_2 = -(10.0\U{$\mu$F})\cdot(-333\U{$\mu$A})\cdot(15.0\U{k\Ohm})
+ = \ans{50\U{$\mu$C}} \;.
+\end{align}
+You could jump to the final equation for $Q$ by using the capacitor
+equation $Q=CV$ and noting that the voltage drop over the capacitor
+must match the voltage drop $V=|I_2R_2|$ over $R_2$ (effectively doing
+Kirchhoff's loop rule in your head).
+
+The top capacitor plate will hold the positive charge, because when
+the switch was first closed, the capacitor was completely discharged.
+The current $I_1$ split and flowed down through both the middle and
+right wire. As time went on, the charge $Q$ built up on the
+capacitor, and current through the right wire slowed, stopping at
+equilibrium with the capacitor fully charged.
+
+\Part{c}
+With the switch opened, $I_1=0$. Resolving our earlier junction equation
+\begin{align}
+ 0 &= I_1 + I_2 + I_3 = I_2 + I_3 \\
+ I_2 &= - I_3 \;,
+\end{align}
+so current will flow up through the right-hand wire ($I_3>0$) and down
+through the center wire ($I_2<0$) as the capacitor discharges.
+Repeating our Kirchhoff loop from \Part{b},
+\begin{align}
+ 0 &= - I_2 R_2 - \frac{q}{C} + I_3 R_3 \\
+ \frac{q}{C} &= (R_2+R_3)I_3 = R'I_3 \;,
+\end{align}
+where $R'\equiv R_2+R_3$ is shorthand to allow cleaner formulas for
+the rest of the problem, and $q$ is the charge on the capacitor
+($q(t=0)=Q$).
+
+As the capacitor discharges, $\deriv{t}{q}<0$, so $I_3=-\deriv{t}{q}$.
+We can then solve for $q(t)$ by integrating with respect to time.
+\begin{align}
+ \frac{q}{C} &= -R'\deriv{t}{q} \\
+ \frac{-\dd t}{R'C} &= \frac{\dd q}{q} \\
+ \int_{t=0}^t \frac{-\dd t}{R'C} &= \int_{t=0}^t \frac{\dd q}{q} \\
+ \frac{-t}{R'C} &= \ln(q) - \ln(Q) = \ln\p({\frac{q}{Q}}) \;,
+\end{align}
+where $-\ln(Q)$ is a constant of integration which is determined by
+the conditions at $t=0$. Raising $e$ to either side this equation, we
+have
+\begin{align}
+ e^{\frac{-t}{R'C}} &= e^{\ln\p({\frac{q}{Q}})} = \frac{q}{Q} \\
+ q &= Q e^{\frac{-t}{R'C}} \\
+ I_3 &= -\deriv{t}{q}
+ = -\p({\frac{-1}{R'C}}) Q e^{\frac{-t}{R'C}}
+ = \frac{V_0}{R'} e^{\frac{-t}{R'C}} \;,
+\end{align}
+where $V_0=Q/C$ is the initial voltage across the capacitor. You can
+see that both the charge on the capacitor and the current through the
+loop will drop off exponentially with a time constant $R'C$ as the
+system discharges.
+
+Now we can put together our knowledge of the switch-closed and
+switch-open cases to write an equation for $I_2$.
+
+\begin{equation}
+ I_2 = \begin{cases}
+ \frac{-V}{R_1+R_2} = \ans{-333\U{$\mu$A}}
+ & t<0 \\
+ -\frac{Q}{R'C} e^{\frac{-t}{R'C}}
+ = \frac{C I_2(t<0) R_2}{R'C} e^{\frac{-t}{R'C}}
+ = \frac{-V R_2}{(R_1+R_2)\cdot(R_2+R_3)} e^{\frac{-t}{R'C}}
+ = \ans{-(278\U{$\mu$A})\cdot e^{\frac{-t}{180\U{ms}}}}
+ & t>0
+ \end{cases}
+\end{equation}
+
+\Part{d}
+Plugging into our formula for $q(t)$
+\begin{align}
+ \frac{Q}{5} &= q(t) = Q e^{\frac{-t}{R'C}} \\
+ \frac{1}{5} &= e^{\frac{-t}{R'C}} \\
+ \ln\p({\frac{1}{5}}) &= \frac{-t}{R'C} \\
+ t &= -R'C\ln\p({\frac{1}{5}}) = R'C\ln(5) = \ans{290\U{ms}}
+\end{align}
+\end{solution}
--- /dev/null
+\begin{problem*}{29.3}
+Determine the initial direction of the deflection of charged particles
+as they enter the magnetic fields shown in Figure~P29.2.
+\begin{center}
+\hspace{\stretch{1}}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 1cm;
+real height = 2u;
+real width = 2u;
+real dx = 0.49u;
+real dy = dx;
+
+Vector Bs[];
+int n = (int)(width / dx);
+int m = (int)(height / dy);
+real xstart = -width/2 + (dx+fmod(width,dx))/2.0;
+real ystart = -height/2 + (dy+fmod(height,dy))/2.0;
+for (int i=0; i<n; i+=1) {
+ for (int j=0; j<m; j+=1) {
+ Bs.push(BField((xstart+i*dx, ystart+j*dy), phi=-90));
+ }
+}
+for (int i=0; i<Bs.length; i+=1) {
+ Bs[i].draw();
+}
+
+Charge a = pCharge((-0.5*width-12pt, 0), "$+$");
+Vector v = Velocity(a.center, dir=0);
+v.draw();
+a.draw();
+
+label("\Part{a}", (0,0.5*height), N);
+\end{asy}
+\hspace{\stretch{1}}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 1cm;
+real height = 2u;
+real width = 2u;
+real dx = 0.49u;
+
+Vector Bs[];
+int n = (int)(width / dx);
+real xstart = -width/2 + (dx+fmod(width,dx))/2.0;
+real ystart = -height/2;
+for (int i=0; i<n; i+=1) {
+ Bs.push(BField((xstart+i*dx, -height/2), mag=height, dir=90));
+}
+for (int i=0; i<Bs.length; i+=1) {
+ Bs[i].draw();
+}
+
+Charge a = nCharge((0.5*width+12pt, 0), "$-$");
+Vector v = Velocity(a.center, dir=180);
+v.draw();
+a.draw();
+
+label("\Part{b}", (0,0.5*height), N);
+\end{asy}
+\hspace{\stretch{1}}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 1cm;
+real height = 2u;
+real width = 2u;
+real dx = 0.49u;
+real dy = dx;
+
+Vector Bs[];
+int m = (int)(height / dy);
+real xstart = -width/2;
+real ystart = -height/2 + (dy+fmod(width,dy))/2.0;
+for (int i=0; i<m; i+=1) {
+ Bs.push(BField((-width/2,ystart+i*dy), mag=width, dir=0));
+}
+for (int i=0; i<Bs.length; i+=1) {
+ Bs[i].draw();
+}
+
+Charge a = pCharge((0.5*width+12pt, 0), "$+$");
+Vector v = Velocity(a.center, dir=180);
+v.draw();
+a.draw();
+
+label("\Part{c}", (0,0.5*height), N);
+\end{asy}
+\hspace{\stretch{1}}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 1cm;
+real height = 2u;
+real width = 2u;
+real dx = 0.49u;
+real dy = dx;
+
+Vector Bs[];
+int n = (int)(width / dx);
+int m = (int)(height / dy);
+real xstart = -width/2 + (dy+fmod(width,dx))/2.0;
+real ystart = -height/2 + (dy+fmod(width,dy))/2.0;
+for (int i=0; i<n; i+=1) {
+ Bs.push(BField((xstart,ystart+i*dy), mag=sqrt(2)*(height/2-(ystart+i*dy)), dir=45));
+}
+for (int i=0; i<m; i+=1) {
+ Bs.push(BField((xstart+i*dx,ystart), mag=sqrt(2)*(width/2-(xstart+i*dx)), dir=45));
+}
+for (int i=0; i<Bs.length; i+=1) {
+ Bs[i].draw();
+}
+
+draw((0, -height/2)--(0, height/2), dashed);
+Angle t = Angle(dir(90)-(0,dy), (0,0)-(0,dy), dir(45)-(0,dy), "$\theta=45\dg$");
+t.draw();
+
+Charge a = pCharge((0, -0.5*height-12pt), "$+$");
+Vector v = Velocity(a.center, dir=90);
+v.draw();
+a.draw();
+
+label("\Part{d}", (0,0.5*height), N);
+\end{asy}
+\hspace{\stretch{1}}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{29.3}
+Find the direction of the magnetic field acting on a positively
+charged particle moving in the various situations shown in
+Figure~P29.3 if the direction of the magnetic force acting on it is as
+indicated.
+\begin{center}
+\hspace{\stretch{1}}
+\begin{asy}
+import Mechanics;
+
+real u=1cm;
+
+Vector f = Force(mag=u, dir=90, "$\vect{F}_B$");
+f.draw();
+Vector v = Velocity(mag=u, dir=0, "$\vect{v}$");
+v.draw();
+label("\Part{a}", (0,0), S);
+\end{asy}
+\hspace{\stretch{1}}
+\begin{asy}
+import Mechanics;
+
+real u=1cm;
+
+Vector f = Force(mag=u, dir=90, "$\vect{F}_B$");
+f.draw();
+Vector v = Velocity(mag=u, phi=90, "$\vect{v}$");
+v.draw();
+label("\Part{b}", (0,0), S);
+\end{asy}
+\hspace{\stretch{1}}
+\begin{asy}
+import Mechanics;
+
+real u=1cm;
+
+Vector f = Force(mag=u, dir=180, "$\vect{F}_B$");
+f.draw();
+Vector v = Velocity(mag=u, phi=-90, "$\vect{v}$");
+v.draw();
+label("\Part{c}", (0,0), S);
+\end{asy}
+\hspace{\stretch{1}}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+Using the magnetic force formula $\vect{F}_B=q\vect{v}\times\vect{B}$,
+the magnetic field must point \ans{into the page}.
+
+\Part{b}
+\ans{To the right}.
+
+\Part{c}
+\ans{Down}.
+\end{solution}
--- /dev/null
+\begin{problem*}{29.22}
+Assume the region to the right of a certain plain contains a uniform
+magnetic field of magnitude $1.00\U{mT}$ and the field is zero in the
+fregion to the left of the plane as shown in Figure~P29.22. An
+electron, originally traveling perpendicular to the boundary plane,
+passes into the region of the field. \Part{a} Determine the time
+interval required for the electron to leave the ``field-filled''
+region, noting that the electron's path is a semicircle. \Part{b}
+Assuming the maximum depth of penetration into the field is
+$2.00\U{cm}$, find the kinetic energy of the electron.
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 1cm;
+real height = 4u;
+real width = 2u;
+real dx = 0.49u;
+real dy = dx;
+
+Vector Bs[];
+int n = (int)(width / dx);
+int m = (int)(height / dy);
+real xstart = -width/2 + (dx+fmod(width,dx))/2.0;
+real ystart = -height/2 + (dy+fmod(height,dy))/2.0;
+for (int i=0; i<n; i+=1) {
+ for (int j=0; j<m; j+=1) {
+ Bs.push(BField((xstart+i*dx, ystart+j*dy), phi=-90));
+ }
+}
+for (int i=0; i<Bs.length; i+=1) {
+ Bs[i].draw();
+}
+
+Charge a = nCharge((-0.5*width-24pt, 0), "$e^-$");
+Vector v = Velocity(a.center, dir=0, "$\vect{v}$");
+v.draw();
+a.draw();
+
+draw((-width/2, -height/2)--(-width/2, height/2), dashed);
+label("$B=0$", (-width,-0.5*height), S);
+label("$B=1.00\U{mT}$", (0,-0.5*height), S);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+The magnetic force on the particle in the field-filled region is
+\begin{equation}
+ F_B = qvB\cos(\theta) = qvB \;.
+\end{equation}
+The acceleration that force induces is
+\begin{align}
+ F_B &= ma \\
+ a &= \frac{F_B}{m} = \frac{qvB}{m} \;.
+\end{align}
+This force (perpendicular to the direction of motion) turns the
+electron in a semicircle. We can use our usual centerward
+acceleration of circling objects to get a handle on the radius.
+\begin{align}
+ |a| &= \frac{v^2}{r} \\
+ r &= \frac{v^2}{|a|} = \frac{v^2m}{|q|vB} = \frac{vm}{|q|B} \;.
+\end{align}
+
+The time it takes to make the half turn is thus
+\begin{align}
+ \Delta t = \frac{\Delta t}{\Delta x} \cdot \Delta x
+ = \frac{\Delta x}{v} = \frac{\pi r}{v}
+ = \frac{\pi vm}{v\cdot |q|B} = \frac{\pi m}{|q|B}
+ = \ans{1.79\E{-8}\U{s}} \;.
+\end{align}
+Note that the velocity $v$ cancels out of the final $\Delta t$
+calculation, which is good because we don't know what it is.
+
+\Part{b}
+Now that they've given us $r$, we can find $v$ and $K$.
+\begin{align}
+ r &= \frac{vm}{|q|B} \\
+ v &= \frac{|q|Br}{m} \\
+ K &= \frac{1}{2} m v^2 = \frac{1}{2} m \p({\frac{|q|Br}{m}})^2
+ = \frac{q^2 B^2 r^2}{2m}
+ = \ans{5.63\E{-18}\U{J}}
+ = 5.63\E{-18}\U{J} \cdot \frac{1\U{eV}}{1.6\E{-19}\U{J}}
+ = \ans{35.2\U{eV}} \;.
+\end{align}
+\end{solution}
--- /dev/null
+\begin{problem*}{29.31}
+A conductor carrying a current $I=15.0\U{A}$ is directed along the
+positive $x$ axis and perpendicular to a uniform magnetic field. A
+magnetic force per unit length of $0.120\U{N/m}$ acts on the conductor
+in the negative $y$ direction. Determine \Part{a} the magnitude
+and \Part{b} the direction of the magnetic field in the region through
+which the current passes.
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{29.37}
+A rod of mass $0.720\U{kg}$ and radius $6.00\U{cm}$ rests on two
+parallel rails (Fig.~P29.37) that are $d=12.0\U{cm}$ apart and
+$L=45.0\U{cm}$ long. The rod carries a current of $I=48.0\U{A}$ in
+the direction shown and rolls along the rails without slipping. A
+uniform magnetic field of magnitude $0.240\U{T}$ is directed
+perpendicular to the rod and rails. If it starts from rest, what is
+the speed of the rod as it leaves the rails?
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{29.40}
+Consider the system pictured in Figure P29.40. A $15.\U{cm}$
+horizontal wire of mass $15.0\U{g}$ is placed between two thin,
+vertical conductors, and a uniform magnetic field acts perpendicular
+to the page. The wire is free to move vertically without friction on
+the two vertical conductors. When a $5.00\U{A}$ curent is directed as
+shown in the figure, the horizontal wire moves upward at a constant
+velocity in the presence of gravity. \Part{a} What forces act on the
+horizontal wire, and \Part{b} under what condition is the wire able to
+move upward at a constant velocity. \Part{c} Find the magnitude and
+direction of the minimum magnetic field required to move the wire at a
+constant speed. \Part{d} What happens if the magnetic field exceeds
+this minimum value?
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{29.57}
+The upper poretion of
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{29.61}
+A $0.200\U{kg}$ metal rod carrying a current of $10.0\U{A}$ glides on
+two horizontal rails $0.500\U{m}$ apart. If the coefficient of
+kinetic friction between the rails is $0.100$, what vertical magnetic
+field is required to keep the rod moving at a constant speed?
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{30.18}
+Two long, parallel wires carry currents of $I_1=3.00\U{A}$ and
+$I_2=5.00\U{A}$ in the directions indicated in
+Figure~P30.18. \Part{a} Find the magnitude and direction of the
+magnetic field at a point midway between the wires. \Part{b} Find the
+magnitude and direction of the magnetic field at point $P$, located
+$d=20.0\U{cm}$ above the wire carrying the $5.00\U{A}$ current.
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real d = 2cm;
+pair a = (0,0);
+pair b = a + (d,0);
+pair M = (a+b)/2;
+pair P = b + (0,d);
+
+Vector Ia = CurrentWire(a, phi=90, "$I_1$");
+Ia.draw();
+Vector Ib = CurrentWire(b, phi=90, "$I_2$");
+Ib.draw();
+Distance ab = Distance(a, b, "$d$", offset=18pt);
+ab.draw();
+Distance bP = Distance(b, P, "$d$", offset=18pt);
+bP.draw();
+dot("$M$", M, N);
+dot("$P$", P, E);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+Using the formula for magnetic field from a long straight wire, the
+total magnetic field at each point is the sum of the contributions
+from each source wire. Selecting \emph{up} as the positive direction,
+\begin{equation}
+ B_M = \frac{\mu_0 I_1}{2\pi\frac{d}{2}} - \frac{\mu_0 I_2}{2\pi\frac{d}{2}}
+ = \frac{\mu_0}{\pi d} (I_1-I_2)
+ = \ans{-4.00\U{$\mu$T}} \;,
+\end{equation}
+so the magnetic field at $M$ points \ans{down} with a magnitude of
+\ans{4.00\U{$\mu$T}}.
+
+\Part{b}
+This time we have to use vector addition.
+\begin{align}
+ \vect{B}_P
+ &= \frac{\mu_0 I_1}{2\pi(\sqrt{2}d)}\cdot\frac{-\ihat+\jhat}{\sqrt{2}}
+ + \frac{\mu_0 I_2}{2\pi d}\cdot(-\ihat)
+ = \frac{\mu_0}{2\pi d}
+ \cdot \p[{\frac{I_1}{2} \cdot (-\ihat+\jhat) - I_2\ihat}]
+ = \frac{\mu_0}{2\pi d}
+ \cdot \p[{-\p({\frac{I_1}{2}+I_2})\ihat + \frac{I_1}{2}\jhat}]
+ = (-6.50\ihat + 1.50\jhat)\U{$\mu$T} \\
+ |\vect{B}_P| &= \sqrt{(-6.50)^2 + 1.50^2}\U{$\mu$T}
+ = \ans{6.67\U{$\mu$T}} \\
+ \theta_P &= \arctan\p({\frac{1.50}{-6.50}})
+ = \ans{167\dg} \;,
+\end{align}
+where we used the backside $180\dg$ rule to adjust the output from
+$\arctan$. You could also use \verb+atan2(y, x)+ and let your
+calculator handle the backside $180\dg$ automatically if your
+calculator supports that function.
+\end{solution}
--- /dev/null
+\begin{problem*}{30.30}
+Niobium metal becomes a superconductor when cooled below $9\U{K}$.
+Its superconductivity is destroyed when the surface magnetic field
+exceeds $0.100\U{T}$. In the absence of any external magnetic field,
+determine the maximum current a $2.00\U{mm}$ diameter niobium wire can
+carry and remain superconducting.
+\end{problem*}
+
+\begin{solution}
+Using our formula for the electric field from a long, straight wire,
+the magnetic field at the surface of this wire will be
+\begin{equation}
+ B = \frac{\mu_0 I}{2\pi r} \;.
+\end{equation}
+Solving for $I$,
+\begin{equation}
+ I = \frac{2\pi r B}{\mu_0}
+ = \frac{2\pi (1.00\E{-3}\U{m}) \cdot (0.100\U{T})}{\mu_0}
+ = \ans{500\U{A}} \;.
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{30.32}
+Four long, parallel conductors carry equal currents of $I=5.00\U{A}$.
+Figure~P30.32 is an end view of the conductors. The current direction
+is into the page at points $A$ and $B$ and out of the page at points
+$C$ an $D$. Calculate \Part{a} the magnitude and \Part{b} direction
+of the magnetic field at point $P$, located at the center of the
+square of edge length $l=0.200\U{m}$.
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real d = 2cm;
+
+Vector Ill = CurrentWire((0,0), phi=-90, "$B$");
+Ill.draw();
+Vector Ilr = CurrentWire((d,0), phi=90, "$D$");
+Ilr.draw();
+Vector Iul = CurrentWire((0,d), phi=-90, "$A$");
+Iul.draw();
+Vector Iur = CurrentWire((d,d), phi=90, "$C$");
+Iur.draw();
+
+Distance dx = Distance((0,0), (d,0), "$l$", offset=18pt);
+dx.draw();
+Distance dy = Distance((0,0), (0,d), "$l$", offset=-18pt);
+dy.draw();
+dot("$P$", (d,d)/2, S);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+Using right hand rules, vector addition, and the forumula for magnetic
+field from a long, straight wire, we can find the magnetic field at
+$P$. Because each wire carries the same current $I$ and is the same
+distance $r=l/\sqrt{2}$ from $P$, the only thing that will change
+between per-wire contributions is the direction of the generated
+field.
+\begin{equation}
+ \vect{B}_P
+ = \vect{B}_{AP} + \vect{B}_{BP} + \vect{B}_{CP} + \vect{B}_{DP}
+ = \frac{\mu_0 I}{2\pi r} \cdot \p({
+ \frac{-\ihat-\jhat}{\sqrt{2}} + \frac{\ihat-\jhat}{\sqrt{2}}
+ + \frac{\ihat-\jhat}{\sqrt{2}} + \frac{-\ihat-\jhat}{\sqrt{2}}
+ })
+ = \frac{\mu_0 I}{2\pi\frac{l}{\sqrt{2}}} \cdot \frac{-4\jhat}{\sqrt{2}}
+ = \frac{-2 \mu_0 I}{\pi l}\jhat
+ = \ans{-20.0\jhat\U{$\mu$T}}
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{30.61}
+Two long, straight wires cross each other perpendicularly as shown in
+Figure P30.61. The wires do not touch. Find \Part{a} the magnitude
+and \Part{b} the direction of the magnetic field at point $P$, which
+is in the same plane as the two wires. \Part{c} Find the magnetic
+field at a point $30.0\U{cm}$ above the point of intersection of the
+wires along the $z$ axis; that is, $30.0\U{cm}$ out of the page,
+toward you.
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real d = 6cm;
+
+real low = 0.3d;
+real high = 0.5d;
+
+draw_ijhat(0.8*low*(-1,-1));
+
+Vector Ix = CurrentWire((-low, 0), (high+low)+12pt, dir=0, "$5.00\U{A}$");
+Ix.draw();
+Vector Iy = CurrentWire((0, -low), (high+low), dir=90, "$3.00\U{A}$");
+Iy.draw();
+
+pair P = (0.4d, 0.3d);
+Distance Dh = Distance((0, P.y), P, "$30.0\U{cm}$");
+Dh.draw();
+Distance Dv = Distance((P.x, 0), P, "$40.0\U{cm}$");
+Dv.draw();
+
+dot("$P$", P, NE);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+We can use the formula for the magnetic field generated by a long,
+straight wire to calculate the contribution from each wire. Letting
+\emph{out of the page} be positive,
+\begin{equation}
+ B_P = \frac{\mu_0 (5.00\U{A})}{2\pi(0.400\U{m})}
+ - \frac{\mu_0 (3.00\U{A})}{2\pi(0.300\U{m})}
+ = \ans{500\U{nT}} \;. \label{eq:30.61.a}
+\end{equation}
+
+\Part{b}
+Because we picked \emph{out of the page} as the positive direction,
+the $B_P>0$ in \cref{eq:30.61.a} means the magnetic field points
+\ans{out of the page}.
+
+\Part{c}
+We take the same approach for point $Q$, $30.0\U{cm}$ above the
+intersection, but using the right hand rule to determine the direction
+of the magnetic field contributions, we can see that we'll need to use
+vector addition to determine the resulting magnetic field.
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real d = 0.5cm;
+
+Vector Bx = BField(mag=5d, dir=-90, "$\vect{B}_{xP}$");
+Vector By = BField(mag=3d, dir=0, "$\vect{B}_{yP}$");
+pair b = Bx.mag*dir(Bx.dir) + By.mag*dir(By.dir);
+Vector B = BField(mag=length(b), dir=degrees(b), "$\vect{B}_P$");
+Angle t = Angle(dir(0), (0,0), b, "$\theta_P$");
+t.draw();
+Bx.draw();
+By.draw();
+B.draw();
+dot("$P$", (0,0), W);
+
+draw_ijhat((-3d, -2.5d));
+\end{asy}
+\end{center}
+The contribution from the $x$-aligned wire will be along $-\jhat$,
+while the contribution from the $y$-aligned wire will be along
+$\ihat$. The total magnetic field is
+\begin{align}
+ \vect{B}_P
+ &= \frac{\mu_0 I_x}{2\pi r} \cdot (-\jhat)
+ + \frac{\mu_0 I_y}{2\pi r}\ihat
+ = \frac{\mu_0}{2\pi r}\cdot(I_y\ihat-I_x\jhat)
+ = (2.00\ihat - 3.33\jhat)\U{$\mu$T} \\
+ |\vect{B}_P| &= \sqrt{2.00^2 + (-3.33)^2}\U{$\mu$T}
+ = \ans{3.89\U{$\mu$T}} \\
+ \theta_P &= \arctan\p({\frac{-3.33}{2.00}})
+ = \ans{-59.0\dg} \;.
+\end{align}
+\end{solution}
--- /dev/null
+\begin{problem*}{30.64}
+Two coplanar and concentric circular loops of wire carry currents of
+$I_1=5.00\U{A}$ and $I_2=3.00\U{A}$ in opposite directions as in
+Figure P30.64. If $r_1=12.0\U{cm}$ and $r_2=9.00\U{cm}$, what
+are \Part{a} the magnitude and \Part{b} the direction of the magnetic
+field at the center of the two loops? \Part{c} Let $r_1$ remain fixed
+at $12.0\U{cm}$ and let $r_2$ be variable. Determine the value of
+$r_2$ such that the net field at the center of the loops is zero.
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+import Circ;
+
+real u = 0.2cm;
+real r1 = 12u;
+real r2 = 9u;
+real dr = 6pt;
+
+draw(scale(r1)*unitcircle, line);
+draw(scale(r2)*unitcircle, line);
+Distance d1 = Distance((0,0), r1*dir(-45), "$r_1$");
+d1.draw();
+Distance d2 = Distance((0,0), r2*dir(45), "$r_2$");
+d2.draw();
+
+draw(arc((0,0), r1+dr, angle1=10, angle2=-10), CurrentPen, ArcArrow);
+label("$I_1$", (r1+dr)*dir(0), E);
+draw(arc((0,0), r2-dr, angle1=170, angle2=190), CurrentPen, ArcArrow);
+label("$I_2$", (r2-dr)*dir(180), E);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+We're interested in the magnetic field generated at the center of the
+ring. There is no path for an Amperian loop that takes advantage of
+the problem's symmetry, so we'll use the Biot--Savart law directly.
+Integrating around a single ring of radius $r$, the magnetic field at
+the center is given by
+\begin{align}
+ \vect{B} &= \oint_S \dd\vect{B}
+ = \oint_S \frac{\mu_0}{4\pi}\cdot\frac{I \dd\vect{s} \times \rhat}{r^2}
+ = \frac{\mu_0 I}{4\pi r^2} \oint_S |\dd\vect{s}|\cdot|\rhat|\cdot\sin(\theta)
+ = \frac{\mu_0 I}{4\pi r^2} \oint_S |\dd\vect{s}|
+ = \frac{\mu_0 I}{4\pi r^2} \cdot 2\pi r
+ = \frac{\mu_0 I}{2 r} \;,
+\end{align}
+where we took advantage of the following facts:
+\begin{itemize}
+ \item The current $I$ and distance $r$ from the wire to the point we
+ care about are the same for every segment $\dd\vect{s}$ in the
+ loop, so we can pull these constants outside the integral.
+ \item The cross product of two vectors can be written
+ $\vect{A}\times\vect{B}=|\vect{A}|\cdot|\vect{B}|\cdot\sin(\theta)$.
+ \item \rhat is a unit vector ($|\rhat|=1$).
+ \item \rhat always points centerward and $\dd\vect{s}$ is always
+ tangent to the ring, so $\theta=90\dg$ and $\sin(\theta)=1$.
+ \item $\oint_S |\dd\vect{s}|$ is just the total length of the path
+ $S$, which is the circumfernce of the ring ($2\pi r$).
+\end{itemize}
+Using the right hand rule, it is clear that a counterclockwise current
+will generate a magnetic field at the center of the ring which points
+out of the page.
+
+Now that we have a formula for the magnetic field at the center of the
+ring, we can calculate the magnetic field at the center of the two
+rings using superposition. Letting \emph{out of the page} be the
+positive direction,
+\begin{equation}
+ B = \frac{\mu_0 (-I_1)}{2 r_1} + \frac{\mu_0 I_2}{2 r_2}
+ = -5.24\U{$\mu$T} \;, \label{eq:30.64.a}
+\end{equation}
+so the magnetic field has a magnitude of $\ans{5.24\U{$\mu$T}}$.
+
+\Part{b}
+Because we picked \emph{out of the page} as the positive direction,
+the $-$ sign in \cref{eq:30.64.a} means that the magnetic field points
+\ans{into the page}.
+
+\Part{c}
+If we allow $r_2$ to vary, we have no magnetic field when
+\begin{align}
+ 0 &= B = \frac{\mu_0 (-I_1)}{2 r_1} + \frac{\mu_0 I_2}{2 r_2} \\
+ \frac{\mu_0 I_1}{2 r_1} &= \frac{\mu_0 I_2}{2 r_2} \\
+ \frac{I_1}{r_1} &= \frac{I_2}{r_2} \\
+ r_2 &= \frac{I_2}{I_1} r_1 = \ans{7.20\U{cm}} \;.
+\end{align}
+\end{solution}
projects / course.git / commitdiff