-\begin{problem*}{29.3}
+\begin{problem*}{29.2}
Determine the initial direction of the deflection of charged particles
as they enter the magnetic fields shown in Figure~P29.2.
\begin{center}
v.draw();
a.draw();
-label("\Part{a}", (0,0.5*height), N);
+label("\Part{a}", (0,-0.5*height), S);
\end{asy}
\hspace{\stretch{1}}
\begin{asy}
v.draw();
a.draw();
-label("\Part{b}", (0,0.5*height), N);
+label("\Part{b}", (0,-0.5*height), S);
\end{asy}
\hspace{\stretch{1}}
\begin{asy}
v.draw();
a.draw();
-label("\Part{c}", (0,0.5*height), N);
+label("\Part{c}", (0,-0.5*height), S);
\end{asy}
\hspace{\stretch{1}}
\begin{asy}
v.draw();
a.draw();
-label("\Part{d}", (0,0.5*height), N);
+label("\Part{d}", a.lc.center, S);
\end{asy}
\hspace{\stretch{1}}
+\rule{0pt}{0pt}
\end{center}
\end{problem*}
\begin{solution}
+The force on a charged particle moving through a magnetic field is
+$\vect{F}=q\vect{v}\times\vect{B}$. From Newton second law,
+$\vect{a}=\vect{F}/m=q/m\cdot\vect{v}\times\vect{B}$. We can find the
+direction of deflection comes using this formula and the right hand
+rule.
+
+\Part{a}
+\ans{Up}.
+
+\Part{b}
+\ans{Out of the page}.
+
+\Part{c}
+\ans{No deflection}.
+
+\Part{d}
+\ans{Into the page}.
\end{solution}
label("\Part{c}", (0,0), S);
\end{asy}
\hspace{\stretch{1}}
+\rule{0pt}{0pt}
\end{center}
\end{problem*}
\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
+region 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''
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();
-}
+Vector v = BField(phi=-90);
+vector_field((0,0), width=width, height=height, v=v);
-Charge a = nCharge((-0.5*width-24pt, 0), "$e^-$");
+Charge a = nCharge((-0.5*width-24pt, 0), Label("$e^-$", align=S));
Vector v = Velocity(a.center(), dir=0,
L=Label("$\vect{v}$", position=EndPoint, align=RightSide));
v.draw();
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}} \;.
+ = \frac{(|q|Br)^2}{2m}
+ = \frac{(1.60\E{-19}\U{C}\cdot1.00\E{-3}\U{T}\cdot2.00\E{-2}\U{m})^2}
+ {2\cdot9.11\E{-31}\U{kg}}
+ = \ans{5.62\E{-18}\U{J}} \\
+ &= 5.62\E{-18}\U{J} \cdot \frac{1\U{eV}}{1.6\E{-19}\U{J}}
+ = \ans{35.1\U{eV}} \;.
\end{align}
\end{solution}
\end{problem*}
\begin{solution}
+\Part{a}
+Using the formula for magnetic force on a current carrying wire,
+\begin{align}
+ \vect{F} &= I\vect{L}\times\vect{B} \\
+ F &= ILB\sin(\theta) = ILB \;,
+\end{align}
+because $\theta=90\dg$. Solving for $B$,
+\begin{equation}
+ B = \frac{F}{IL} = \frac{F/L}{I}
+ = \frac{0.120\U{N/m}}{15.0\U{A}}
+ = \ans{8.00\U{mT}} \;.
+\end{equation}
+
+\Part{b}
+Because $\vect{B}$ is perpendicular to $\vect{L}$, $\vect{L}$ points
+along the positive $x$ axis, and $\vect{F}=I\vect{L}\times\vect{B}$,
+$\vect{B}$ points in the $-y$ direction, $\vect{B}$ must point \ans{in
+ the positive $z$ direction}.
\end{solution}
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?
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+import Circ;
+
+real d = 2cm;
+real L = 3cm;
+real r = 6pt;
+
+Vector v = BField(phi=-90);
+vector_field(((L+r)/2,0), width=L-r, height=d, v=v);
+wire((0, d/2), (L, d/2));
+wire((0, -d/2), (L, -d/2));
+Block bar = Block(width=2*r, height=d, fill=yellow); bar.draw();
+
+Distance Dr = Distance((0, -d/2), (r, -d/2), offset=3pt,
+ Label("$r$", align=S));
+Dr.draw();
+Distance Dd = Distance((0, -d/2), (0, d/2), offset=-24pt,
+ Label("$d=12.0\U{cm}$", align=W));
+Dd.draw();
+Distance DL = Distance((0, d/2), (L, d/2), offset=-12pt,
+ Label("$L=45.0\U{cm}$", align=N));
+DL.draw();
+
+draw((0, d/4)--(0, -d/4), CurrentPen, Arrow);
+label("$I$", (-r,0), align=W);
+\end{asy}
+\end{center}
\end{problem*}
\begin{solution}
+The magnetic force on the rod is
+\begin{align}
+ \vect{F}_B &= I\vect{d}\times{B} \\
+ F_B &= IdB\sin(\theta) = IdB \;.
+\end{align}
+After the rod rolls a distance $\vect{L}$ along the rails, this force
+has done
+\begin{equation}
+ W_B = \vect{F}_B \cdot \vect{L} = F_B L = IdBL
+\end{equation}
+of work on the rod. Conserving energy,
+\begin{equation}
+ W_B = IdBL = K = \frac{1}{2} m v^2 + \frac{1}{2}I_m\omega^2 \;.
+\end{equation}
+To solve this, we need the moment of inertia for a cylinder about it's
+axis ($I_m=\frac{1}{2}mR^2$) and a relationship between $\omega$ and
+$v$ ($v=R\omega$). The rest is algrebra:
+\begin{align}
+ IdBL &= \frac{1}{2} m v^2 + \frac{1}{4} m R^2 \cdot \p({\frac{v}{R}})^2
+ = \frac{3}{4} m v^2 \\
+ v^2 &= \frac{4IdBL}{3m} \\
+ v &= \sqrt{\frac{4IdBL}{3m}} = \ans{1.07\U{m/s}} \;.
+\end{align}
+This speed doesn't depend on the mass of the rod.
\end{solution}
\begin{problem*}{29.40}
-Consider the system pictured in Figure P29.40. A $15.\U{cm}$
+Consider the system pictured in Figure P29.40. A $15.0\U{cm}$
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
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?
+\begin{center}
+% | | 15.0 cm between verticals
+% +--<--+
+% | 5A |
+% | |
+% v 5A ^ 5A
+\begin{asy}
+import Mechanics;
+import Circ;
+
+real u = 1cm;
+
+pair ul = (-u, u); // upper left
+pair ml = (-u, 0); // middle left
+pair ll = (-u, -2u); // lower left
+pair ur = (u, u); // upper right
+pair mr = (u, 0);
+pair lr = (u, -2u);
+
+MultiTerminal I = current(label=Label("$5.00\U{A}$"), draw=false);
+I.centerto(lr, mr); I.label.align = E; I.draw();
+wire(lr, I.terminal[0]);
+wire(I.terminal[1], ur);
+I.centerto(mr, ml); I.label.align = S; I.draw();
+wire(mr, I.terminal[0]);
+wire(I.terminal[1], ml);
+I.centerto(ml, ll); I.label.align = W; I.draw();
+wire(ul, I.terminal[0]);
+wire(I.terminal[1], ll);
+dot(mr);
+dot(ml);
+
+Distance D = Distance(ml, mr, offset=-u/2, Label("$15.0\U{cm}$", align=N));
+D.draw();
+\end{asy}
+\end{center}
\end{problem*}
\begin{solution}
+\Part{a}
+The only forces acting on the horizontal wire are \ans{gravity and the
+force of the magnetic field on the current carried by the wire}.
+
+\Part{b}
+If the wire is moving at a constant velocity,
+\begin{align}
+ 0 &= \deriv{t}{\vect{v}} = \vect{a} = \frac{\vect{F}}{m} \\
+ \ans{\vect{F}} &= \ans{0} \;.
+\end{align}
+In English, this means the net force on the wire must be zero.
+Because there are only two forces acting on the wire, this means these
+forces must be equal in magnitude and opposite in direction.
+
+\Part{c}
+The gravitational force on the wire will be $F_g=mg$ directed
+downwards, and the magnetic force on the wire will be
+$F_B=I\vect{L}\times\vect{B}$ directed upwards. Because the magnetic
+field is perpendicular to the page, and the current is directed along
+the page, the angle between $\vect{L}$ and $\vect{B}$ is $90\dg$, so
+$F_B=ILB\sin{90\dg}=ILB$. Summing forces in the vertical direction,
+\begin{align}
+ 0 &= \sum_i F_{i,y} = F_B - F_g = ILB - mg \\
+ ILB &= mg \\
+ B &= \frac{mg}{IL}
+ = \frac{0.0150\U{kg} \cdot 9.80\U{m/s$^2$}}{5.00\U{A} \cdot 0.150\U{m}}
+ = \ans{196\U{mT}} \;.
+\end{align}
+In order to generate an upwards force on a leftwards current, the
+magnetic field must be directed \ans{out of the page}.
+
+\Part{d}
+If the magnetic field exceeds this value, \ans{the rod will accelerate
+ upwards}, moving up faster and faster. If the magnetic field is
+less than this value, the rod will accelerate downwards, moving up
+slower and slower, eventually pausing, and then moving back down
+faster and faster. At the critical magnetic field, where the forces
+balance, the rod will keep moving forever at it's initial velocity (or
+hang in place if the initial velocity was zero).
\end{solution}
\begin{problem*}{29.57}
-The upper poretion of
+The upper portion of the circuit in Figure~P29.57 is fixed. The
+horizontal wire at the bottom has a mass of $10.0\U{g}$ and is
+$5.00\U{cm}$ long. This wire hangs in the gravitational field of the
+Earth from identical light springs connected to the upper portion of
+the circuit. The springs stretch $0.500\U{cm}$ under the weight of
+the wire, and the circuit has a total resistance of $12.0\U{\Ohm}$.
+When a magnetic field is turned on, directed out of the page, the
+springs stretch an additional $0.300\U{cm}$. Only the horizontal wire
+at the bottom of the circuit is in the magnetic field. What is the
+magnitude of the magnetic field?
+\begin{center}
+% 24.0V
+% +--/\/\/--|i--+
+% | |
+% Sp Sp
+% |<-- 5.00 cm->|
+% +-------------+
+% Bout
+%
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+import Circ;
+
+real u = 1cm;
+
+MultiTerminal Bat = battery(dir=180, label=Label("$24.0\U{V}$", align=N));
+MultiTerminal R = resistor(Bat.terminal[1], dir=180);
+Spring Sl = Spring(R.terminal[1] - (0,u), R.terminal[1] - (0,2u),
+ deadLength=1mm, unstretchedLength=u, L=Label("$k$", embed=Shift));
+Sl.draw();
+Spring Sr = Spring(Bat.terminal[0] - (0,u), Bat.terminal[0] - (0,2u),
+ deadLength=1mm, unstretchedLength=u, L=Label("$k$", embed=Shift));
+Sr.draw();
+
+Vector v = BField(phi=90);
+real w = Sr.pTo.x - Sl.pTo.x;
+vector_field((Sl.pTo + Sr.pTo)/2 + (0, -u), width=1.5*w, height=u, v=v);
+
+Distance DL = Distance(Sl.pTo, Sr.pTo, offset=u/3,
+ Label("$5.00\U{cm}$", align=N));
+DL.draw();
+
+wire(R.terminal[1], Sl.pFrom);
+wire(Bat.terminal[0], Sr.pFrom);
+wire(Sl.pTo, Sr.pTo, udsq, dist=-u);
+\end{asy}
+\end{center}
\end{problem*}
\begin{solution}
+We have two situations here, one where $B=0$ and $\Delta
+x=0.500\U{cm}$ (call this $B_1$ and $\Delta x_1$), and another where
+$B$ is unknown and $\Delta x=\Delta x_1 + 0.300\U{cm}=0.800\U{cm}$
+(call this $B_2$ and $\Delta x_2$). Because both are in equilibrium,
+the net force on the hanging wire must be zero in each case. The only
+forces acting on the wire are gravity pulling down with $F_g=mg$, two
+springs pulling up with $F_s=k\Delta x$ (for each spring), and the magnetic
+field pulling down with $F_B=ILB\cos(90\dg)$. Summing forces in the
+vertical direction for the first situation,
+\begin{align}
+ 0 &= \sum_i F_{i,y} = 2k\Delta x_1 - mg \\
+ 2k\Delta x_1 &= mg \\
+ k &= \frac{mg}{2\Delta x_1} \;.
+\end{align}
+We can plug this into the sum of vertical forces for the second
+situation,
+\begin{align}
+ 0 &= \sum_i F_{i,y} = 2k\Delta x_2 - mg - ILB_2 \\
+ ILB_2 &= 2k\Delta x_2 - mg
+ = 2\frac{mg}{2\Delta x_1}\Delta x_2 - mg
+ = mg\p({\frac{\Delta x_2}{\Delta x_1} - 1}) \\
+ B_2 &= \frac{mg}{IL}\p({\frac{\Delta x_2}{\Delta x_1} - 1}) \;.
+\end{align}
+This looks good, but we don't have a value for the current $I$. Using
+a Kirchhoff loop starting behind the battery and going counter
+clockwise,
+\begin{align}
+ 0 &= V - IR \\
+ I &= \frac{V}{R} \;.
+\end{align}
+Plugging in for $B_2$,
+\begin{equation}
+ B_2 = \frac{mgR}{VL}\p({\frac{\Delta x_2}{\Delta x_1} - 1})
+ = \frac{0.0100\U{kg} \cdot 9.80\U{m/s$^2$} \cdot 12.0\U{\Ohm}}
+ {24.0\U{V} \cdot 0.0500\U{m}}
+ \p({\frac{0.800\U{cm}}{0.500\U{cm}} - 1})
+ = \ans{588\U{mT}} \;.
+\end{equation}
\end{solution}
\end{problem*}
\begin{solution}
+This is just like Problem~29.40, except the force balancing the
+magnetic force is friction instead of gravity. Drawing a free-body
+diagram looking from the side,
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real mag = 1cm;
+real mu = 0.3;
+
+Vector F_B = Force(mag=mu*mag, Label("$F_B$", position=EndPoint));
+Vector F_g = Force(mag=mag, dir=-90, Label("$F_g$", position=EndPoint));
+Vector F_N = Force(mag=mag, dir=90, Label("$F_N$", position=EndPoint));
+Vector F_f = Force(mag=mu*mag, dir=180, Label("$F_f$", position=EndPoint));
+F_B.draw();
+F_g.draw();
+F_N.draw();
+F_f.draw();
+dot((0,0));
+\end{asy}
+\end{center}
+
+Summing the forces in the vertical direction,
+\begin{align}
+ 0 &= \sum_i F_{i,y} = F_N - F_g = F_N - mg \\
+ F_N &= mg \\
+\end{align}
+Now that we know the normal force, we can find the force of friction
+and sum the forces in the horizontal direction.
+\begin{align}
+ 0 &= \sum_i F_{i,x} = F_B - F_f = ILB - \mu F_N = ILB\sin(90\dg) - \mu mg \\
+ ILB &= \mu mg \\
+ B &= \frac{\mu mg}{IL}
+ = \frac{0.100 \cdot 0.200\U{kg} \cdot 9.80\U{m/s$^2$}}
+ {10.0\U{A} \cdot 0.500\U{m}}
+ = \ans{39.2\U{mT}} \;.
+\end{align}
\end{solution}
projects / course.git / commitdiff