\end{problem*}
\begin{solution}
+\Part{a}
+The current flowing to the right in the bar feels a lifting force from
+the magnetic field $\vect{F}_B=I\vect{l}\times\vect{B}$ which balances
+the gravitational force $F_g=mg$. Because the current and magnetic
+field are perpendicular to each other, we can focus on the magnitudes
+\begin{align}
+ F_B &= IlB = F_g = mg \
+ I &= \frac{mg}{lB}
+ = \frac{0.750\U{kg}\cdot9.8\U{m/s}}{0.500\U{m}\cdot0.450\U{T}}
+ = 32.7\U{A} \;.
+\end{align}
+This maximum current would when the voltage $V$ from the battery
+balanced an $IR$ drop across the resistor, so
+\begin{equation}
+ V = IR = \ans{817\U{V}} \;.
+\end{equation}
+
+\Part{b}
+When the resistor shorts, the current jumps to
+\begin{equation}
+ I' = \frac{V}{R'} = 408\U{A} \;,
+\end{equation}
+because the resistor voltage still has to match the battery voltage.
+This creates a net lifting force and acceleration on the bar.
+\begin{align}
+ F &= F_B-F_g = I'lB - mg = I'lB - IlB = (I'-I)lB = ma \\
+ a &= (I'-I)\frac{lB}{m}
+ = 376\U{A}\cdot\frac{0.500\U{m}\cdot0.450\U{T}}{0.750\U{kg}}
+ = \ans{113\U{m/s$^2$}} \;.
+\end{align}
\end{solution}
\end{problem*}
\begin{solution}
+\Part{a}
+Writing $\vect{F}_B=q\vect{v}\cdot\vect{B}$ in terms of components
+\begin{align}
+ \vect{F}_B
+ &= q[\ihat(v_yB_z-v_zB_y)-\jhat(v_xB_z-v_zB_x)+\khat(v_xB_y-v_yB_x)]
+ = q(-v_zB_y\ihat+v_zB_x\jhat) = F_0(3\ihat+4\jhat) \;.
+\end{align}
+Matching components and writing $v_z=v$ we have
+\begin{align}
+ -qvB_y &= 3F_0 & qvB_x &= 4F_0 \\
+ B_y &= \ans{\frac{-3F_0}{qv}} & B_x &= \ans{\frac{4F_0}{qv}} \;.
+\end{align}
+We can't find $B_z$ because it does not contribute to the force felt
+by the charge, which is currently our only handle on \vect{B}.
+
+\Part{b}
+With $|\vect{B}|$, we can solve for $|B_z|$.
+\begin{align}
+ |\vect{B}|^2 &= B_x^2 + B_y^2 + B_z^2 \\
+ B_z^2 &= |\vect{B}^2| - B_x^2 - B_y^2
+ = 36\frac{F_0^2}{q^2v^2} - 9\frac{F_0^2}{q^2v^2} - 16\frac{F_0^2}{q^2v^2}
+ = (36-9-16)\frac{F_0^2}{q^2v^2}
+ = 11\frac{F_0^2}{q^2v^2} \\
+ B_z &= \ans{\sqrt{11}\frac{F_0}{qv}} \;.
+\end{align}
+However, we still cannot find the direction of $B_z$.
\end{solution}
\end{problem*}
\begin{solution}
+This problem is extremely similar to Prob.~27.39, but with a more
+complicated circuit driving current through the bar, and a top-down
+view in the figure.
+
+When $S$ is closed, current begins to flow, creating a magnetic force
+$\vect{F}_b=I_b\vect{l}\times\vect{B}$ on the bar, where $I_b$ is the
+current through the bar. Using our cross product right-hand rule and
+magnitude formula, we see that the force on the bar is out of the page
+with a magnitude $F_b=I_blB$. We just need to find $I_b$.
+
+Because the bar and vertical resistor are in parallel, the equivalent
+resistance for the vertical resistor and bar is
+\begin{equation}
+ R' = \p({\frac{1}{10.0\U{\Ohm}} + \frac{1}{10.0\U{\Ohm}}})^{-1}
+ = 5.00\U{\Ohm} \;,
+\end{equation}
+and the resistance for the entire circuit is
+\begin{equation}
+ R = 25.0\U{\Ohm} + 5.00\U{\Ohm}
+ = 30.0\U{\Ohm} \;.
+\end{equation}
+This effective resistance must drop all the voltage generated by the
+battery, so we can manipulate Ohm's law $V=IR$ to yield
+\begin{equation}
+ I = \frac{V}{R} = \frac{120.0\U{V}}{30\U{\Ohm}}
+ = 4.00\U{A} \;.
+\end{equation}
+This is the total current through the battery. Now we need to find
+the current through the bar. The voltage across the
+vertical-resistor/bar portion is
+\begin{equation}
+ V_b = IR' = 4.00\U{A}\cdot5.00\U{\Ohm} = 20.0\U{V} \;,
+\end{equation}
+and current through the bar is
+\begin{equation}
+ I_b = \frac{V_b}{R_b} = \frac{20.0\U{V}}{10.0\U{\Ohm}}
+ = 2.00\U{A}
+\end{equation}
+
+The lifting force and acceleration are then
+\begin{align}
+ F &= F_b - F_g = I_blB - mg = ma \\
+ a &= (\frac{I_blB}{mg} - 1)g
+ = \p({\frac{2.00\U{A}\cdot1.50\U{m}\cdot1.60\U{T}}{3.00\U{N}} - 1})\cdot9.8\U{m/s$^2$}
+ = \ans{5.88\U{m/s$^2$}} \;.
+\end{align}
+Note that I factored out $g$ to $mg$ in the denominator, because they
+give the \emph{weight} of the bar ($mg=3.00\U{N}$), not the
+\emph{mass} $m$. Obviously, you could get a number for $m$ if you
+want, but keeping $mg$ together allows us to avoid the trouble.
\end{solution}
wire(pUR, pLR);
wire(pLR, pLL);
-TwoTerminal I = current((0,0), 0, "$6.00\U{A}$", "", draw=false);
+TwoTerminal I = current("$6.00\U{A}$", "", draw=false);
I.centerto(pUL, (pUL+pUR)/2);
I.draw();
I.name = "";
\end{problem*}
\begin{solution}
+The horizontal portions of wire feel equal and opposite magnetic
+force, because $\vect{F}_B=I\vect{l}\cdot\vect{B}$ and the only
+differences are in the sign of \vect{l}. Therefore we only need worry
+about the vertical bit of wire, where the force is \ans{to the left}
+from the right hand rule. The magnitude of the force is
+\begin{equation}
+ F_B = IlB = 6.00\U{A}\cdot0.450\U{m}\cdot0.666\U{T}
+ = \ans{1.80\U{N}} \;.
+\end{equation}
\end{solution}
projects / course.git / commitdiff