--- /dev/null
+\begin{problem*}{27.39}
+A thin, $50.0\U{cm}$ long metal bar with mass $750\U{g}$ rests on, but
+is not attached to, two metallic supports in a uniform $0.450\U{T}$
+magnetic field, as shown in Fig.~27.51. A battery and a
+$25.0\U{\Ohm}$ resistor in series are connected to the
+supports. \Part{a} What is the highest voltage the battery can have
+without breaking the circuit at the supports? \Part{b} The battery
+voltage has the maximum value caculated in \Part{a}. If the resistor
+suddenly gets partially short-circuited, decreasing its resistance to
+$2.0\U{\Ohm}$, find the initial acceleration of the bar.
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+import Circ;
+
+real supRadius = 3mm;
+
+TwoTerminal V = source(ang=-180, type=DC, val="V");
+TwoTerminal R = resistor(V.beg, "R");
+
+real lBar = 0.6*abs(R.end-V.end);
+real hBar = 1mm;
+pair cBar = (V.end+R.end)/2 - (0,R.len);
+
+// top, left, and right points for the left support
+pair tLsup = cBar - (lBar/2, hBar/2);
+pair lLsup = tLsup + supRadius * dir(-120);
+pair rLsup = tLsup + supRadius * dir(-60);
+wire(V.end, lLsup, udsq);
+wire(lLsup, tLsup);
+wire(tLsup, rLsup);
+
+// top, left, and right points for the right support
+pair tRsup = cBar + (lBar/2, -hBar/2);
+pair lRsup = tRsup + supRadius * dir(-120);
+pair rRsup = tRsup + supRadius * dir(-60);
+wire(R.end, rRsup, udsq);
+wire(rRsup, tRsup);
+wire(tRsup, lRsup);
+
+Vector Bs[];
+real dBy = supRadius*2/sqrt(3);
+int i;
+Bs.push(BField(cBar-(0,dBy), phi=-90));
+Bs.push(BField(cBar+(0,dBy), phi=-90, "\vect{B}"));
+Bs.push(BField(cBar+(-lBar/2,dBy), phi=-90));
+Bs.push(BField(cBar+(+lBar/2,dBy), phi=-90));
+for (i=0; i<Bs.length; i+=1) {
+ Bs[i].out_of_plane_radius = 1mm;
+ Bs[i].draw();
+}
+
+Block bar = Block(cBar, width=lBar, height=hBar, fill=yellow);
+bar.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{27.64}
+A particle of charge $q>0$ is moving at speed $v$ in the
+$+z$-direction through a region of uniform magnetic field \vect{B}.
+The magnetic force on the particle is $\vect{F}=F_0(3\ihat+4\jhat)$,
+where $F_0$ is a positive constant. \Part{a} Determine the components
+$B_x$, $B_y$, and $B_z$, or at least as many of the three components
+as is possible from the information given. \Part{b} If it is given in
+addition that the magnetic field has magnitude $6F_0/qv$, determine as
+much as you can about the remaining components of \vect{B}.
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{27.68}
+A $3.00\U{N}$ metal bar, $1.50\U{m}$ long and having a resistance of
+$10.0\U{\Ohm}$, rests horizontally on conducting wires connecting it
+to the circuit shown in Fig.~27.62. The bar is in a uniform,
+horizontal, $1.60\U{T}$ magnnetic field and is not attached to the
+wires in the circuit. What is the acceleration of the bar just after
+the switch $S$ is closed?
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+import Circ;
+
+TwoTerminal Rv = resistor(ang=90, val="$10.0\U{\Ohm}$");
+TwoTerminal Rh = resistor(Rv.end, ang=-180, val="$25.0\U{\Ohm}$");
+TwoTerminal V = source(type=DC, "$120.0\U{V}$", draw=false);
+centerto(Rv, V, Rh.len); V.draw();
+TwoTerminal S = switchSPST(Rv.end, type=NO, "$S$");
+wire(V.beg, (2*Rh.len, Rv.beg.y), udsq);
+wire(V.end, Rh.end, udsq);
+wire(S.end, (2*Rh.len, S.end.y), nsq);
+
+real xBar = (S.end.x+2*Rh.len)/2;
+
+Vector Bs[];
+int i, n = 4;
+real dy = 1.4*Rv.len/(n-1);
+real yBstart = Rv.mid.y - (n-1)*dy/2;
+real xB = xBar - Rh.len/2;
+for (i=0; i<n; i+=1) {
+ Bs.push(BField((xB, yBstart+i*dy), mag=Rh.len));
+ if (i == n-1)
+ Bs[i].L = "\vect{B}";
+}
+for (i=0; i<Bs.length; i+=1) {
+ Bs[i].draw();
+}
+
+Block bar = Block((xBar, Rv.mid.y), width=1mm, height=1.2*Rv.len, fill=yellow);
+bar.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{27.73}
+A long wire carrying a $6.00\U{A}$ current reverses direction by means
+of two right-angle bends, as shown in Fig.~27.64. The part of the
+wire where the bend occurs is in a magnetic field of $0.666\U{T}$
+confined to the circular region of diameter $75\U{cm}$, as shown.
+Find the magnitude and direction of the net force that the magnetic
+field exerts on this wire.
+\begin{center}
+\begin{asy}
+import ElectroMag;
+import Circ;
+
+real u = 2cm;
+real dx = 2u;
+real dy = 0.45u;
+real r = 0.75u/2;
+
+pair pUL = (-dx,dy/2); // upper left
+pair pUR = (0,dy/2); // upper right
+pair pLL = (-dx,-dy/2); // lower left
+pair pLR = (0,-dy/2); // lower right
+
+wire(pUL, pUR);
+wire(pUR, pLR);
+wire(pLR, pLL);
+
+TwoTerminal I = current((0,0), 0, "$6.00\U{A}$", "", draw=false);
+I.centerto(pUL, (pUL+pUR)/2);
+I.draw();
+I.name = "";
+I.centerto((pLL+pLR)/2, pLL);
+I.draw();
+
+Vector Bs[];
+int i, j, n=2;
+real dxy = 0.7*r/n;
+pair p;
+for (i=-n; i<=n; i+=1) {
+ for (j=-n; j<=n; j+=1) {
+ p = dxy*(i,j);
+ Bs.push(BField(p, phi=90));
+ }
+}
+for (i=0; i<Bs.length; i+=1) {
+ Bs[i].out_of_plane_radius = 1mm;
+ if (abs(Bs[i].center)+Bs[i].out_of_plane_radius <= r)
+ Bs[i].draw();
+}
+draw(scale(r)*unitcircle, Bs[0].outline+dashed);
+
+Distance Dwire = Distance((pLL+2*pLR)/3, (pUL+2pUR)/3, "$45.0\U{cm}$");
+Dwire.draw(rotateLabel=false, labeloffset=24pt);
+Distance DdB = Distance((+r,-r), (-r,-r), offset=-3mm, L="$75.0\U{cm}$");
+DdB.draw(rotateLabel=false, labeloffset=12pt);
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
projects / course.git / commitdiff