From 9277598e22a83f930f8e5a54d2300cbc130ca507 Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Tue, 4 Aug 2009 21:11:00 -0400 Subject: [PATCH] Added solutions for rec7. --- .../Young_and_Freedman_12/problem26.61.tex | 80 +++++++++++++++++++ .../Young_and_Freedman_12/problem26.86.tex | 31 +++++++ .../Young_and_Freedman_12/problem26.91.tex | 21 +++++ .../Young_and_Freedman_12/problem27.22.tex | 27 ++++++- .../Young_and_Freedman_12/problem27.30.tex | 12 +++ .../Young_and_Freedman_12/problem27.35.tex | 24 ++++++ 6 files changed, 194 insertions(+), 1 deletion(-) diff --git a/latex/problems/Young_and_Freedman_12/problem26.61.tex b/latex/problems/Young_and_Freedman_12/problem26.61.tex index 39e7011..f3ef226 100644 --- a/latex/problems/Young_and_Freedman_12/problem26.61.tex +++ b/latex/problems/Young_and_Freedman_12/problem26.61.tex @@ -14,4 +14,84 @@ circuit diagram shown in Fig.~26.65. \end{problem*} \begin{solution} +Start by labeling everything. +\begin{center} +\begin{verbatim} + 5.00=R5 I5 I4 8.00=R4 ++------/\/\/--->--+--<---/\/\/------+ +|12.0V 1.00=R2 I2 a I1 1.00=R1 9.00V| +b--|i---/\/\/--<--+-->--/\/\/---i|--c +| I3 10.00=R3 | ++-------->------/\/\/---------------+ +\end{verbatim} +\end{center} +Then junction rules at $a$ and $b$ yield +\begin{align} + 0 &= -I_1 - I_2 + I_5 + I_4 \\ + 0 &= I_2 - I_3 - I_5 +\end{align} +Loop rules for each sub-loop clockwise from the loop's upper left yield +\begin{align} + 0 &= -I_5R_5 - I_2R_2 + V_2 \\ + 0 &= +I_4R_4 - V_1 + I_1R_1 \\ + 0 &= -V_2 + I_2R_2 - I_1R_1 + V_1 + I_3R_3 +\end{align} +Which gives us five equations for the five unknown currents. Solve +using your method of choice (e.g. linear algebra). +\begin{align} + \begin{pmatrix} + 0 \\ + 0 \\ + -V_2 \\ + V_1 \\ + V_2-V_1 + \end{pmatrix} + &= + \begin{pmatrix} + -1 & -1 & 0 & 1 & 1 \\ + 0 & 1 & -1 & 0 & -1 \\ + 0 & -R_2 & 0 & 0 & -R_5 \\ + R_1 & 0 & 0 & R_4 & 0 \\ + -R_1 & R_2 & R_3 & 0 & 0 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 \\ + I_4 \\ + I_5 + \end{pmatrix} \\ + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 \\ + I_4 \\ + I_5 + \end{pmatrix} + &= + \begin{pmatrix} + -1 & -1 & 0 & 1 & 1 \\ + 0 & 1 & -1 & 0 & -1 \\ + 0 & -R_2 & 0 & 0 & -R_5 \\ + R_1 & 0 & 0 & R_4 & 0 \\ + -R_1 & R_2 & R_3 & 0 & 0 + \end{pmatrix}^{-1} + \begin{pmatrix} + 0 \\ + 0 \\ + -V_2 \\ + V_1 \\ + V_2-V_1 + \end{pmatrix} + = + \ans{ + \begin{pmatrix} + 0.848 \\ + 2.14 \\ + 0.171 \\ + 1.02 \\ + 1.97 + \end{pmatrix} + \U{A}} +\end{align} \end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem26.86.tex b/latex/problems/Young_and_Freedman_12/problem26.86.tex index 3be73a8..a070d07 100644 --- a/latex/problems/Young_and_Freedman_12/problem26.86.tex +++ b/latex/problems/Young_and_Freedman_12/problem26.86.tex @@ -7,4 +7,35 @@ maximum value? \end{problem*} \begin{solution} +The energy stored in the capacitor is $1/e$ of its maximum value when +\begin{align} + U = \frac{q^2}{2C} &= \frac{U_0}{e} = \frac{Q_0^2}{2Ce} \\ + q^2 &= \frac{Q_0^2}{e} \\ + q &= Q_0 e^{-0.5} \;. +\end{align} +This value is useful for both \Part{a} and \Part{b}. + +\Part{a} +For a discharging circuit, +\begin{equation} + q = Q_0 e^{\frac{-t}{RC}} \;, +\end{equation} +so the time to $U=U_0/e$ is given by +\begin{align} + \frac{-t}{RC} &= -0.5 \\ + t &= \ans{\frac{RC}{2}} \;. +\end{align} + +\Part{b} +For a charging circuit, the +\begin{equation} + q = Q_0 \p({1-e^{\frac{-t}{RC}}}) \;, +\end{equation} +so the time to $U=U_0/e$ is given by +\begin{align} + 1-e^{\frac{-t}{RC}} &= e^{-0.5} \\ + e^{\frac{-t}{RC}} &= 1-e^{-0.5} \\ + \frac{-t}{RC} &= \ln\p({1-e^{-0.5}}) \\ + t &= -RC\ln\p({1-e^{-0.5}}) = \ans{0.933RC} \;. +\end{align} \end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem26.91.tex b/latex/problems/Young_and_Freedman_12/problem26.91.tex index 495eec2..5e74a0f 100644 --- a/latex/problems/Young_and_Freedman_12/problem26.91.tex +++ b/latex/problems/Young_and_Freedman_12/problem26.91.tex @@ -22,4 +22,25 @@ b-/\/\/-d-/\/\/-+-/\/\/-+-... \end{problem*} \begin{solution} +Following the hint, we note that +\begin{align} + R_T &= R_1 + \p({\frac{1}{R_2} + \frac{1}{R_T}})^{-1} + R_1 \\ + R_T - 2R_1 &= \p({\frac{1}{R_2} + \frac{1}{R_T}})^{-1} \\ + (R_T - 2R_1) \cdot \p({\frac{1}{R_2} + \frac{1}{R_T}}) &= 1 \\ + (R_T - 2R_1) \cdot (R_T + R_2) &= R_2R_T \\ + R_T^2 - 2R_1R_T + R_2R_T - 2R_1R_2 &= R_2R_T \\ + 0 &= R_T^2 - 2R_1R_T - 2R_1R_2 \;. +\end{align} +Plugging this into the quadratic formula +\begin{equation} + R_T = \frac{2R_1 \pm \sqrt{4R_1^2 - 4\cdot1\cdot(-2R_1R_2)}}{2} + = R_1 \pm \sqrt{R_1^2 + 2R_1R_2} + = \ans{R_1 + \sqrt{R_1^2 + 2R_1R_2}} \;, +\end{equation} +which is what we set out to show. Note that we chose the $+$ case +from $\pm$ because +\begin{equation} + R_1 < \sqrt{R_1^2 + 2R_1R_2} \;, +\end{equation} +and $R_T$ must be greater than zero. \end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem27.22.tex b/latex/problems/Young_and_Freedman_12/problem27.22.tex index 76916fd..c173740 100644 --- a/latex/problems/Young_and_Freedman_12/problem27.22.tex +++ b/latex/problems/Young_and_Freedman_12/problem27.22.tex @@ -1,6 +1,6 @@ \begin{problem*}{27.22} In an experiment with cosmic rays, a verticle beam of particles that -have chagre of magnitude $3e$ and mass $12$ times the proton mass +have charge of magnitude $3e$ and mass $12$ times the proton mass enters a uniform horizontal magnetic field of $0.250\U{T}$ and is bent in a semicircle of diameter $95.0\U{cm}$, as shown in Fig.~27.47. \Part{a} Find the speed of the particles and the sign of @@ -10,4 +10,29 @@ they enter the field compare to their speed as they exit the field? \end{problem*} \begin{solution} +\Part{a} +From the figure, the particle rotates counterclockwise with the +\vect{B} field pointing out of the page. From the righthand rule and +$\vect{F}=q\vect{v}\times\vect{B}$, the particle must have a negative +charge. Because this centerward force makes the particle move in a +circle, +\begin{align} + F_c &= qvB = ma_c = m\frac{v^2}{r} \\ + v &= \frac{rqB}{m} + = \frac{0.950\U{m}\cdot3\cdot1.6\E{-19}\U{C}\cdot0.250\U{T}} + {12\cdot1.67e-27\U{kg}} + = \ans{5.68\U{Mm/s}} \;. +\end{align} + +\Part{b} +The gravitational force $mg=1.97\E{-25}\U{N}$, while the magnetic +force $qvB=6.82\E{-13}\U{N}$. Therefore, the gravitational force +can be safely ignored. + +\Part{c} +Because the magnetic force $\vect{F}=q\vect{v}\times\vect{B}$ is +always perpendicular to the particle's velocity \vect{v}, the magnetic +force will never change the speed of a particle. Only the direction +of motion is affected. So the particle leaves the field with the same +speed it entered with. \end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem27.30.tex b/latex/problems/Young_and_Freedman_12/problem27.30.tex index e0d8e29..3825683 100644 --- a/latex/problems/Young_and_Freedman_12/problem27.30.tex +++ b/latex/problems/Young_and_Freedman_12/problem27.30.tex @@ -9,4 +9,16 @@ ignore the weight of the particle. \end{problem*} \begin{solution} +\Part{a} +This is a straigtforward application of the Lorentz force. +\begin{align} + \vect{F} &= 0 = q(\vect{E} + \vect{v}\times\vect{B}) \\ + \vect{E} &= -\vect{v}\times\vect{B} + = 5.85\E{3}\U{m/s}\cdot1.35\U{T}(\jhat \times \khat) + = \ans{7.90\U{V/m}\ihat} \;. +\end{align} + +\Part{b} +Because the charge canceled out in \Part{a}, the electric field should +be the same: $\vect{E}=\ans{7.90\U{V/m}\ihat}$. \end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem27.35.tex b/latex/problems/Young_and_Freedman_12/problem27.35.tex index bd1a6b1..4c33eed 100644 --- a/latex/problems/Young_and_Freedman_12/problem27.35.tex +++ b/latex/problems/Young_and_Freedman_12/problem27.35.tex @@ -13,9 +13,11 @@ real u = 2.5cm; Distance Dhorizontal = Distance((0,0),(u,0), offset=2mm, L="$60.0\U{cm}$"); Distance Dvertical = Distance((u,0),(u,u), offset=2mm, L="$60.0\U{cm}$"); Distance Dbend = Distance((.25u,.25u),(.25u,.75u), offset=2mm, L="$30.0\U{cm}$"); +Vector I = Vector((-.25u,.25u), .125u, red, "$I$"); draw(scale(u)*((0,0)--(1,0)--(1,1)--(0,1)--cycle), blue); draw(scale(u)*((-.25,.25)--(.25,.25)--(.25,.75)--(1.25,.75)), red); +I.draw(); Dhorizontal.draw(labelangle=-90, labeloffset=8pt); Dvertical.draw(labelangle=-90, labeloffset=8pt); Dbend.draw(rotateLabel=false, labelangle=-90, labeloffset=22pt); @@ -24,4 +26,26 @@ Dbend.draw(rotateLabel=false, labelangle=-90, labeloffset=22pt); \end{problem*} \begin{solution} +The verticle portion of the wire feels a force of +\begin{equation} + \vect{F}_v = I\vect{l}_v\times\vect{B} = Il_vB \ihat + = 4.50\U{A}\cdot0.300\U{m}\cdot0.240\U{T} \ihat + = 0.324\U{N}\ihat \;. +\end{equation} +The two horizontal legs combined feel +\begin{equation} + \vect{F}_h = I\vect{l}_h\times\vect{B} = -Il_hB \jhat + = -4.50\U{A}\cdot0.600\U{m}\cdot0.240\U{T} \jhat + = -0.648\U{N}\jhat \;. +\end{equation} +This gives the total force a magnitude of +\begin{equation} + F = \sqrt{F_v^2+F_h^2} = \ans{0.724\U{N}} +\end{equation} +in the direction +\begin{equation} + \theta = \arctan\p({\frac{0.648\U{N}}{0.324\U{N}}}) + = \arctan(2) = \ans{63.4\dg} +\end{equation} +below the horizontal. \end{solution} -- 2.26.2