--- /dev/null
+\begin{problem*}{29.1} % induction
+A flat, rectangular coil consisting of $50$ turns measures
+$25.0\U{cm}$ by $30.0\U{cm}$. It is in a uniform, $1.20\U{T}$,
+magnetic field, with the plane of the coil parallel to the field. In
+$0.222\U{s}$, it is rotated so that the plane of the coil is
+perpendicular to the field. \Part{a} What is the change in the
+magnetic flux through the coil due to this rotation? \Part{b} Find
+the magnitude of the average emf induced in the coil during this
+rotation.
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+Neither the area not the magnetic field changes, only the angle
+between them, so
+\begin{equation}
+ \Delta \Phi = AB[\cos(\theta_f)-\cos(\theta_i)]
+ = AB[\cos(0\dg)-\cos(90\dg)]
+ = AB = NLWB
+ = 50\cdot0.250\U{m}\cdot0.300\U{m}\cdot1.20\U{T}
+ = \ans{4.50\U{Wb}} \;.
+\end{equation}
+
+\Part{b}
+From Faraday's law of inductance (Eq.~29.3, p.~996)
+\begin{equation}
+ \EMF = -\pderiv{t}{\Phi_B} \;,
+\end{equation}
+so the magnitude of the average \EMF\ is
+\begin{equation}
+ |\EMF| = \frac{|\Delta \Phi|}{\Delta t}
+ = \frac{2.50\U{Wb}}{0.222\U{s}}
+ = \ans{20.3\U{V}} \;.
+\end{equation}
+
+\end{solution}
--- /dev/null
+\begin{problem*}{29.10} % induction
+A rectangle measuring $30.0\U{cm}$ by $40.0\U{cm}$ is located inside a
+region of a spatially uniform magnetic field of $1.25\U{T}$, with the
+field perpendicular to the plane of the coil (Fig.~29.29). The coil
+is pulled out at a steady rate of $2.00\U{cm/s}$ traveling
+perpendicular to the field lines. The region of field ends abruptly
+as shown. Find the emf induced in this coil when it is \Part{a} all
+inside the field; \Part{b} partly inside the field; \Part{c} all
+outside the field.
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 0.5cm;
+real height = 4u;
+real width = 3u;
+
+Vector Bs[];
+real dx = 1.1u;
+real dy = dx;
+int n = (int)(2*width / dx);
+int m = (int)(2*height / dy);
+real xstart = -width + (dx+fmod(2*width,dx))/2.0;
+real ystart = -height + (dy+fmod(2*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();
+}
+draw((width,-height)--(width,height), dashed);
+
+Vector v = Velocity((width/2,0), "$2.00\U{cm/s}$");
+v.draw();
+
+
+Distance Dw = Distance((-width/2,height/2), (width/2,height/2),
+ offset=-6pt, "$30.0\U{cm}$");
+Distance Dh = Distance((-width/2,-height/2), (-width/2,height/2),
+ offset=-6pt, "$40.0\U{cm}$");
+Dw.draw(labeloffset=8pt);
+Dh.draw(labeloffset=8pt);
+
+draw(scale(width/2,height/2)*((1,1)--(1,-1)--(-1,-1)--(-1,1)--cycle), yellow);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+The area, flux, and angle between them are all constant while the coil
+remains entirely with the $B$-field region, so the flux is constant.
+Therefore
+\begin{equation}
+ \EMF = -\pderiv{t}{\Phi_B} = \ans{0} \;.
+\end{equation}
+
+\Part{b}
+As the coil leaves the field, the area with non-zero magnetic field
+decreases, so
+\begin{equation}
+ \EMF = -\pderiv{t}{\Phi_B}
+ = -\pderiv{t}{}(HWB)
+ = -HB\pderiv{t}{W}
+ = HBv
+ = 0.400\U{m}\cdot1.25\U{T}\cdot2.00\E{-2}\U{m/s}
+ = \ans{10.0\U{mV}} \;.
+\end{equation}
+
+\Part{c}
+As in \Part{a}, the flux is constant, so
+\begin{equation}
+ \EMF = -\pderiv{t}{\Phi_B} = \ans{0} \;.
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{29.21} % magnetic force on charges, quasi-Hall effect
+In Fig.~29.37 a conducting rod of length $L=30.0\U{cm}$ moves in a
+magnetic field \vect{B} of magnitude $0.450\U{T}$ directed into the
+plane of the figure. The rod moves with speed $v=5.00\U{m/s}$ iun the
+direction show. \Part{a} What is the potential difference between the
+ends of the rod? \Part{b} Which point, $a$ or $b$, is at higher
+potential? \Part{c} When the charges in the rod are in equilibrium,
+what are the magnitude and direction of the electric field within the
+rod? \Part{d} When the charges in the rod are in equilibrium, which
+point, $a$ or $b$, has an excess of positive charge? \Part{e} What is
+the potential difference across the rod if it moves (i) parallel to
+$ab$ and (ii) directly out of the page?
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 1cm;
+real height = 2u;
+real width = 2u;
+real length = 1u;
+real angle = -45;
+
+Vector Bs[];
+real dx = 0.49u;
+real dy = dx;
+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 = Velocity((0,0), dir=-45, "$v$");
+v.draw();
+
+pair A = scale(length/2)*rotate(angle)*(0,-1);
+pair B = scale(length/2)*rotate(angle)*(0,1);
+Distance L = Distance(A, B, offset=-6pt, "$L$");
+L.draw();
+
+draw(A--B, yellow);
+label("$a$", A, SW);
+label("$b$", B, NE);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+The conducting rod has mobile charges of charge $q$. These charges
+feel a magnetic force from passing through the magnetic field
+\begin{equation}
+ F_B = |\vect{F}_B| = | q\vect{v}\times\vect{B} | = qvB \;,
+\end{equation}
+because \vect{v} and \vect{B} are perpendicular.
+
+This force pushes the charges from one end of the rod to the other
+which causes $Q=n|q|$ to build up on one side of the rod ($b$) and
+$-Q$ to be exposed on the other side ($a$). The charge separation
+increases until the electric force
+\begin{equation}
+ F_E = qE
+\end{equation}
+balances the magnetic force
+\begin{align}
+ qE &= qvB \\
+ E &= vB \;.
+\end{align}
+Because the electric field in the bar is constant, the electric
+potential difference $V_{ab}$ is given by
+\begin{equation}
+ V_{ab} = EL = vBL
+ = 5.00\U{m/s}\cdot0.450\U{T}\cdot0.300\U{m}
+ = \ans{675\U{mV}} \;.
+\end{equation}
+The potential energy difference for moving a single charge between $a$
+and $b$ (considering both $B$ and $E$) is zero, because the net
+external force on the charge is zero.
+
+\Part{b}
+The positive charge build up at $b$ gives a higher electric potential
+at $\ans{b}$.
+
+\Part{c}
+The electric field points from $b$ to $a$, resisting further charge
+separation. As we showed above, $E=vB=\ans{2.25\U{V/m}}$
+
+\Part{d}
+As shown above, $b$ has the excess positive charge.
+
+\Part{e}
+If the rod moves parallel to $ab$
+\begin{equation}
+ V_{ab} = EL = \ans{0} \;,
+\end{equation}
+because the rod is thin ($L\approx0$).
+
+If the rod moves directly out of the page
+\begin{equation}
+ \vect{F}_B = q\vect{v}\times\vect{B} = 0 \;,
+\end{equation}
+because \vect{v} and \vect{B} would be in the same direction, and no
+charge separation occurs, so $V_{ab}=\ans{0}$.
+\end{solution}
--- /dev/null
+\begin{problem*}{29.25} % rail guns
+The conducting rod $ab$ shown in Fig.~29.38 makes contact with metal
+rails $ca$ and $db$. The apparatus is in a uniform magnetic field of
+$0.800\U{T}$, perpendicular to the plnae of the figure. \Part{a} Find
+the magnitude of the emf induced in the rod when it is moved toward
+the right with speed $7.50\U{m/s}$. \Part{b} In what direction does
+the current flow in the rod? \Part{c} If the resistance of the
+circuit $abdc$ is $1.50\U{\Ohm}$ (assumed to be constant), find the
+force (magnitude and direction) required to keep the rod moving to the
+right with a constant speed of $7.50\U{m/s}$. You can ignore
+friction. \Part{d} Compare the rate at which mechanical work is done
+by the force ($Fv$) with the rate at which thermal energy is developed
+in the circuit ($I^2R$).
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 1cm;
+real height = 2u;
+real width = 2u;
+real ySep = u;
+real xSep = 0.5u;
+real xBar = 0.5u;
+
+Vector Bs[];
+real dx = 0.49u;
+real dy = dx;
+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 = Velocity((xBar,0), "$v$");
+v.draw();
+
+draw((width/2,ySep/2)--(-width/2+xSep,ySep/2)
+ --(-width/2+xSep,-ySep/2)--(width/2,-ySep/2));
+Block bar = Block((xBar, 0), width=1mm, height=ySep, fill=yellow);
+bar.draw();
+
+label("$a$", (xBar, ySep/2), N);
+label("$b$", (xBar,-ySep/2), S);
+label("$c$", (-width/2+xSep, ySep/2), N);
+label("$d$", (-width/2+xSep,-ySep/2), S);
+
+Distance D = Distance((width/2,ySep/2), (width/2,-ySep/2),
+ offset=-6pt, "50.0\U{cm}");
+D.draw(labeloffset=8pt);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+As the bar moves to the right, the flux increases
+\begin{align}
+ \EMF &= -\pderiv{t}{\Phi_B}
+ = -\pderiv{t}{}(HWB)
+ = -HB\pderiv{t}{W}
+ = -HBv
+ = -0.500\U{m}\cdot0.800\U{T}\cdot7.50\U{m/s}
+ = -3.00\U{V} \\
+ |\EMF| &= \ans{3.00\U{V}} \;.
+\end{align}
+
+\Part{b}
+From \Part{a}, we see that $\EMF<0$ when we pick coordinates such that
+$B>0$. Thus the current direction is \ans{counter-clockwise}
+($b\rightarrow a$).
+% solutions say "clockwise"
+
+\Part{c}
+The current in the circuit will be
+\begin{align}
+ V &= IR \\
+ I &= \frac{V}{R}
+ = \frac{3.00\U{V}}{1.50\U{\Ohm}} = 2.00\U{A}
+\end{align}
+You must then pull the bar with enough force to balance the magnetic
+force
+\begin{equation}
+ F = IHB = \frac{V}{R}HB = \frac{HBv}{R}HB = \frac{H^2B^2v}{R}
+ = \ans{0.800\U{N}}
+\end{equation}
+
+\Part{d}
+The mechanical power you put into the bar is given by
+\begin{equation}
+ P_M = Fv = \frac{H^2B^2v^2}{R} \;.
+\end{equation}
+The electrical power dissipated by the resistance is
+\begin{equation}
+ P_R = IV = \frac{V^2}{R} = \frac{H^2B^2v^2}{R} = P_M \;.
+\end{equation}
+This is good, because we expect energy to be conserved, and the
+circuit contains no capacitors or other means of storing energy.
+Therefore, what goes in must come out.
+\end{solution}
--- /dev/null
+\begin{problem*}{29.28} % ?
+A long, thin solenoid has $900$ turns per meter and radius
+$2.50\U{cm}$. The current in the solenoid is increasing at a uniform
+rate of $60.0\U{A/s}$. What is the magnitude of the induced electric
+field at a point near the center of the solenoid and \Part{a}
+$0.500\U{cm}$ from the axis of the solenoid; \Part{b} $1.00\U{cm}$
+from the axis of the solenoid?
+\end{problem*}
+
+\begin{solution}
+For any position inside the solenoid ($r<R=2.50\U{cm}$), the magnetic
+field inside a solenoid is
+\begin{equation}
+ B = \mu_0 nI
+\end{equation}
+neglecting fringe effects, which we can do because the solenoid is
+long and thin.
+
+Because the system has spherical symmetry, we know that the electric
+field mustalso have cylindrical symmetry, so \vect{E} can only be a
+function of the distance $r$ from the solenoid's axis.
+
+From Gauss' law
+\begin{equation}
+ \oint \vect{E}\cdot\dd\vect{A} = \frac{q_\text{enc}}{\epsilon_0} \;,
+\end{equation}
+we know that the radial component of the electric field must be zero,
+since there is no excess of charge in the system, and any radial
+component to the \vect{E}-field would lead to a net flux through a
+Gaussian cylinder.
+
+We can find the tangential component of the electric field from
+Faraday's law
+\begin{equation}
+ \oint \vect{E}\cdot\dd\vect{l} = -\deriv{t}{\Phi_B} \;.
+\end{equation}
+We choose the path to be a circle of radius $r$ centered on the
+cylinder axis. Because all the requested radii are inside the
+solenoid, the flux-area is the entire $\pi r^2$ area within the loop.
+\begin{align}
+ E \oint \dd l &= -\deriv{t}{}(\pi r^2 B) \\
+ 2\pi rE &= -\pi r^2 \deriv{t}{B} = -\pi r^2 \deriv{t}{}(\mu_0 nI) \\
+ 2E &= -r\mu_0 n \deriv{t}{I} \\
+ E &= -\frac{1}{2} r\mu_0 n \deriv{t}{I}
+ =-\frac{1}{2}\cdot4\pi\E{-7}\U{Wb/A$\cdot$m}\cdot900\cdot60.0\U{A/s}\cdot r
+ = -0.0339\U{V/m$^2$} \cdot r \;.
+\end{align}
+Plugging in the three requested radii
+\begin{align}
+ E(0) &= \ans{0} \\
+ E(0.500\U{cm}) &= \ans{-170\U{$\mu$V/m}} \\
+ E(1.00\U{cm}) &= \ans{-339\U{$\mu$V/m}} \;,
+\end{align}
+where the $-$ sign denotes counter-clockwise electric field when
+looking along the direction of the solenoid's magnetic field.
+\end{solution}
--- /dev/null
+\begin{problem*}{29.53} % induction
+A flexible, circular loop $6.50\U{cm}$ in diameter lies in a magnetic
+field with magnitude $0.950\U{T}$, directed into the plane of the
+page, as shown in Fig.~29.46. The loop is pulled at the points
+indicate by the arrows, forming a loop of zero area in
+$0.250\U{s}$. \Part{a} Find the average induced emf in the
+circuit. \Part{b} What is the direction of the current in $R$: from
+$a$ to $b$ or from $b$ to $a$? Explain your reasoning.
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+import Circ;
+
+real r = 1cm;
+real abSep = 0.2r;
+real RSep = 1r;
+real height = 3r;
+real width = height;
+
+Vector Bs[];
+real dx = height/5.1;
+real dy = dx;
+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 vt = Velocity((0,r), dir=90);
+Vector vb = Velocity((0,-r), dir=-90);
+vt.draw();
+vb.draw();
+
+pair A = (width/2, abSep/2);
+pair B = (width/2, -abSep/2);
+
+real abAngle = aSin((abSep/2)/r);
+path arcpath = arc((0,0), r, abAngle, -abAngle, CCW);
+draw(arcpath);
+wire(relpoint(arcpath, 0), A);
+wire(relpoint(arcpath, 1), B);
+
+TwoTerminal R = resistor("$R$", draw=false);
+R.centerto(A, B, offset=RSep);
+R.draw();
+wire(A, R.beg, rlsq, RSep/2);
+wire(B, R.end, rlsq, RSep/2);
+
+dot("$a$", A, N);
+dot("$b$", B, S);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+\begin{equation}
+ \Delta \Phi = \Delta A \cdot B = (A_f - A_i) B = -A_iB \;,
+\end{equation}
+so the average \EMF is
+\begin{equation}
+ \EMF = -\frac{\Delta \Phi}{\Delta t}
+ = \frac{A_iB}{\Delta t}
+ = \frac{\pi(d/2)^2B}{\Delta t}
+ = \frac{\pi\cdot(3.25\E{-2}\U{m})^2\cdot0.950\U{T}}{0.250\U{s}}
+ = \ans{12.6\U{mV}} \;. % solutions: 0.126 V
+\end{equation}
+
+
+\Part{b}
+The positive emf drives a current in the $B$-field direction:
+clockwise, so current passes through the resistor \ans{from $a$ to
+ $b$}.
+\end{solution}
projects / course.git / commitdiff