--- /dev/null
+\begin{problem*}{31.23}
+Figure~P31.23 shows a top view of a bar that can slide on two
+frictionless rails. The resistor is $R=6.00\U{\Ohm}$, and a
+$2.50\U{T}$ magnetic field is directed perpendicularly downward, into
+the paper. Let $l=1.20\U{m}$. \Part{a} Calculate the applied force
+required to move the bar to the right at a constant speed of
+$2.00\U{m/s}$. \Part{b} At what rate is energy delivered to the
+resistor?
+\begin{center}
+% +-------+--------
+% Z x x^x | x x Bin
+% R Z l | |--> F_app
+% Z x xvx | x x x x
+% +-------+--------
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+import Circ;
+
+real u = 1cm;
+real w = 1mm;
+
+MultiTerminal R = resistor(dir=90, "$R$", draw=false);
+real rlen = R.terminal[1].y - R.terminal[0].y;
+Vector B = BField(phi=-90);
+vector_field(R.center + (0.75*u, 0), width=1.5*u, height=rlen, v=B);
+R.draw();
+real yt = R.terminal[1].y;
+real yb = R.terminal[0].y;
+wire((0, yt), (1.5*u, yt));
+wire((0, yb), (1.5*u, yb));
+pair p = (u,(yt+yb)/2);
+Vector F = Force(p, Label("$F_\text{app}$", position=EndPoint)); F.draw();
+Block block = Block(p, width=w, height=rlen); block.draw();
+Distance Dl = Distance((u/2, yt), (u/2, yb), Label("$l$", embed=Shift));
+Dl.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
+
--- /dev/null
+\begin{problem*}{31.30}
+A rectangular coil with resistance $R$ has $N$ turns, each of length
+$l$ and width $w$ as shown in Figure~P31.30. The coil moves into a
+uniform magnetic fiield $\vect{B}$ with constant velocity $\vect{v}$.
+What are the magnitude and direction of the total magnetic force on
+the coil \Part{a} as it enters the magnetic field, \Part{b} as it
+moves within the field, and \Part{c} as it leaves the field?
+\begin{center}
+% x x x Bin
+% +--+ v x x x x
+% w| |-> x x x x
+% +--+ x x x x
+% l x x x x
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real w = 5mm;
+real l = 6mm;
+real u = 1.5cm;
+
+Vector B = BField(phi=-90);
+vector_field((l/2 + 7mm + u/2, 0), width=u, height=u, v=B, outline=dashed);
+Vector v = Velocity((l/2, 0), "$\vect{v}$"); v.draw();
+path p = scale(l, w)*shift((-0.5, -0.5))*unitsquare;
+draw(p);
+draw(shift(1.5pt*dir(-45))*p);
+Distance Dl = Distance((-l/2, -w/2), (l/2, -w/2), "$l$", offset=6pt);
+Dl.draw();
+Distance Dw = Distance((-l/2, w/2), (-l/2, -w/2), Label("$w$", embed=Shift),
+ offset=6pt);
+Dw.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
+
--- /dev/null
+\begin{problem*}{31.31}
+Two parallel rails with negligable resistance are $10.0\U{cm}$ apart
+and are connected by a resistor of resistance $R_3=5.00\U{\Ohm}$. The
+circuit also contains two metal rods having resistances of
+$R_1=10.0\U{\Ohm}$ and $R_2=15.0\U{\Ohm}$ sliding along the rails
+(Fig.~P31.31). The rods are pulled away from the resistor at constant
+speeds of $v_1=4.00\U{m/s}$ and $v_2=2.00\U{m/s}$, respectively. A
+uniform magnetic field of magnitude $B=0.0100\U{T}$ is applied
+perpendicular to the plane of the rails. Determine the current in
+$R_3$.
+\begin{center}
+% --+-----+-----+--
+% x | x x Z x x | x
+% <-| x x ZR3 x |->
+% v1| x x Z x x |v2
+% --+-----+-----+--
+% R1 R2
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+import Circ;
+
+real u = 1cm;
+real w = 1mm;
+
+MultiTerminal R3 = resistor(dir=90, "$R_3$", draw=false);
+real rlen = R3.terminal[1].y - R3.terminal[0].y;
+Vector B = BField(phi=-90);
+vector_field(R3.center, width=3*u, height=rlen, v=B);
+R3.draw();
+real yt = R3.terminal[1].y;
+real yb = R3.terminal[0].y;
+wire((-1.5*u, yt), (1.5*u, yt));
+wire((-1.5*u, yb), (1.5*u, yb));
+dot(R3.terminal[0]);
+dot(R3.terminal[1]);
+pair p1 = (-u,(yt+yb)/2);
+Vector v1 = Velocity(p1, dir=180, "$v_1$"); v1.draw();
+Block R1 = Block(p1, width=w, height=rlen); R1.draw();
+label("$R_1$", (p1.x, yb), align=S);
+pair p2 = (-p1.x, p1.y);
+Vector v1 = Velocity(p2, "$v_2$"); v1.draw();
+Block R2 = Block(p2, width=w, height=rlen); R2.draw();
+label("$R_2$", (p2.x, yb), align=S);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
+
--- /dev/null
+\begin{problem*}{31.45}
+A circular coil enclosing an area $A=0.0100\U{m$^2$}$ is made of $200$
+turns of copper wire as shown in Figure~P31.45. Initially, a uniform
+magnetic field of magnitude $B=1.10\U{T}$ points upward in a direction
+perpendicular to the plane of the coil. The direction of the field
+then reverses in a time interval $\Delta t$. Determine how much
+charge enters one end of the resistor during this time interval if
+$R=5.00\U{Ohm}$.
+\begin{center}
+% B ^ ^ ^
+% | | | (from back of coil)
+% +-+-+---+
+% +-+-+ Z R
+% +-+-+ Z
+% +-+-+---+
+% | | | (from font of coil)
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+import Circ;
+
+real dx = 5mm;
+real r = 5mm;
+int n = 10; // number of coil loops
+real tense = 8; // increasing tension flattens the coil loops
+
+MultiTerminal R = resistor(dir=90, Label("$R$", align=E));
+real rlen = R.terminal[1].y - R.terminal[0].y;
+real xr = R.center.x - dx;
+real xl = xr - 2*r;
+real dy = rlen / n;
+for (int i = 0; i < n; i += 1) { // back sides of coil
+ pair dout = S;
+ if (i == n-1) {
+ dout = E;
+ }
+ real y0 = R.terminal[0].y + i * dy;
+ draw((xl, y0 + dy/2){N}..tension tense ..{dout}(xr, y0 + dy));
+}
+Vector B = BField(dir=90);
+B.outline += linewidth(0.5mm);
+vector_field(((xl+xr)/2, R.center.y+rlen/3), width=2*r, height=2*rlen, v=B);
+for (int i = 0; i < n; i += 1) { // front sides of coil
+ pair din = S;
+ if (i == 0) {
+ din = W;
+ }
+ real y0 = R.terminal[0].y + i * dy;
+ draw((xr, y0){din}..tension tense ..{N}(xl, y0 + dy/2));
+}
+draw((xr, R.terminal[0].y) -- R.terminal[0]);
+draw((xr, R.terminal[1].y) -- R.terminal[1]);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
+
--- /dev/null
+\begin{problem*}{31.46}
+A circular loop of wire of resistance $R=0.500\U{\Ohm}$ and radius
+$r=8.00\U{cm}$ is in a uniform magnetic field directed out of the page
+as in Figure~P31.46. If a clockwise current of $I=2.50\U{mA}$ is
+induced in the loop, \Part{a} is the magnetic field increasing or
+decreasing in time? \Part{b} Find the rate at which the field is
+changing in time.
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real r = 1cm;
+real dr = 6pt;
+
+Vector B = BField(phi=90);
+vector_field(width=2.5*r, height=2.5*r, v=B);
+draw(scale(r)*unitcircle);
+Distance Dr = Distance((0,0), (r,0), "$r$"); Dr.draw();
+draw(arc((0,0), r+dr, angle1=10, angle2=-10), CurrentPen, ArcArrow);
+label("$I$", (r+dr, 0), align=E);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
+
projects / course.git / commitdiff