From 4a407add1d64cd928756faf9edf663af19bf3004 Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Wed, 12 Aug 2009 09:42:18 -0400 Subject: [PATCH] Added solutions to rec8 problems --- .../Young_and_Freedman_12/problem27.39.tex | 30 +++++++++++ .../Young_and_Freedman_12/problem27.64.tex | 26 ++++++++++ .../Young_and_Freedman_12/problem27.68.tex | 50 +++++++++++++++++++ .../Young_and_Freedman_12/problem27.73.tex | 11 +++- 4 files changed, 116 insertions(+), 1 deletion(-) diff --git a/latex/problems/Young_and_Freedman_12/problem27.39.tex b/latex/problems/Young_and_Freedman_12/problem27.39.tex index a052690..3738ac1 100644 --- a/latex/problems/Young_and_Freedman_12/problem27.39.tex +++ b/latex/problems/Young_and_Freedman_12/problem27.39.tex @@ -58,4 +58,34 @@ bar.draw(); \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} diff --git a/latex/problems/Young_and_Freedman_12/problem27.64.tex b/latex/problems/Young_and_Freedman_12/problem27.64.tex index f436ea7..095b2a4 100644 --- a/latex/problems/Young_and_Freedman_12/problem27.64.tex +++ b/latex/problems/Young_and_Freedman_12/problem27.64.tex @@ -10,4 +10,30 @@ much as you can about the remaining components of \vect{B}. \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} diff --git a/latex/problems/Young_and_Freedman_12/problem27.68.tex b/latex/problems/Young_and_Freedman_12/problem27.68.tex index 39201a3..d89c89f 100644 --- a/latex/problems/Young_and_Freedman_12/problem27.68.tex +++ b/latex/problems/Young_and_Freedman_12/problem27.68.tex @@ -43,4 +43,54 @@ bar.draw(); \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} diff --git a/latex/problems/Young_and_Freedman_12/problem27.73.tex b/latex/problems/Young_and_Freedman_12/problem27.73.tex index 271d42e..2659a66 100644 --- a/latex/problems/Young_and_Freedman_12/problem27.73.tex +++ b/latex/problems/Young_and_Freedman_12/problem27.73.tex @@ -24,7 +24,7 @@ wire(pUL, pUR); 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 = ""; @@ -57,4 +57,13 @@ DdB.draw(rotateLabel=false, labeloffset=12pt); \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} -- 2.26.2