From e2be9860999b39b303cb93f00a1e6925cef47d50 Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Fri, 1 Jun 2012 00:31:08 -0400 Subject: [PATCH] Add solutions for Serway and Jewett v8's chapter 31. --- .../Serway_and_Jewett_8/problem31.23.tex | 46 +++++++++- .../Serway_and_Jewett_8/problem31.30.tex | 39 ++++++++- .../Serway_and_Jewett_8/problem31.31.tex | 86 ++++++++++++++++++- .../Serway_and_Jewett_8/problem31.45.tex | 18 +++- .../Serway_and_Jewett_8/problem31.46.tex | 19 +++- 5 files changed, 202 insertions(+), 6 deletions(-) diff --git a/latex/problems/Serway_and_Jewett_8/problem31.23.tex b/latex/problems/Serway_and_Jewett_8/problem31.23.tex index b2618d2..80c2eec 100644 --- a/latex/problems/Serway_and_Jewett_8/problem31.23.tex +++ b/latex/problems/Serway_and_Jewett_8/problem31.23.tex @@ -39,5 +39,49 @@ Dl.draw(); \end{problem*} \begin{solution} -\end{solution} +\Part{a} +Moving the bar to the right increases the magnetic flux directed into +the page by increasing the area of magnetic field enclosed by the +loop. Increasing the flux induces a counter-clockwise current to +resist the change. +\begin{align} + \Phi_B &= AB = lxB \\ + |\EMF| &= |-\deriv{t}{\Phi_B}| = lB \deriv{t}{x} = lBv \\ + 0 &= \sum_\text{loop} V_i = \EMF - IR \\ + I &= \frac{\EMF}{R} = \frac{lBv}{R} \;. +\end{align} +The counter-clockwise current is moving up through the sliding bar, so +the magnetic force on the bar is directed to the left, with a +magnitude of +\begin{equation} + F_B = IlB\sin(90\dg) = IlB = \frac{l^2B^2v}{R} \;. +\end{equation} + +For the bar to move to the right at a constant speed, the net force in +the $\ihat$ direction should be zero. We'll have to apply an external +$F_\text{app}$ to the right to counter $F_B$. +\begin{align} + 0 &= \sum_i F_{i,x} = F_\text{app} - F_B \\ + F_\text{app} &= F_B = \frac{l^2B^2v}{R} + = \frac{(1.20\U{m}\cdot2.50\U{T})^2\cdot2.00\U{m/s}}{6.00\U{\Ohm}} + = \ans{3.00\U{N}} \;. +\end{align} +\Part{b} +The power delivered to a resister is +\begin{equation} + P_R = IV = I^2R = \p({\frac{lBv}{R}})^2 R = \frac{(lBv)^2}{R} + = \frac{(1.20\U{m}\cdot2.50\U{T}\cdot2.00\U{m/s})^2}{6.00\U{\Ohm}} + = \ans{6.00\U{W}} \;. +\end{equation} + +If you want to look at the problem in terms of energy conservation, +the external force is putting energy into the system at a rate of +\begin{equation} + P_\text{in} = \deriv{t}{F_\text{app} x} = F_\text{app}\deriv{t}{x} + = F_\text{app}v = \frac{l^2B^2v}{R} \cdot v = \frac{(lBv)^2}{R} \;. +\end{equation} +All this energy has to go somewhere, and the only place for it to go +is into resistor heat, so it's no surprise that we get the same +formula for $P_\text{in}$ that we got for $P_R$. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem31.30.tex b/latex/problems/Serway_and_Jewett_8/problem31.30.tex index b441fdd..d2fb136 100644 --- a/latex/problems/Serway_and_Jewett_8/problem31.30.tex +++ b/latex/problems/Serway_and_Jewett_8/problem31.30.tex @@ -1,7 +1,7 @@ \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 $\vect{v}$. +uniform magnetic field $\vect{B}$ 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? @@ -35,5 +35,40 @@ Dw.draw(); \end{problem*} \begin{solution} -\end{solution} +\Part{a} +If we define ``into the page'' as the positive direction, the flux +through the loop will be increasing as the coil enters the field, +which will induce a current in the counter-clockwise direction +opposing the changing flux. The right side of the coil and portions +of the top and bottom sides will be in the field regions, and because +of the current will be subject to magnetic forces directed to the +left, down, and up respectively. Because equal portions of the top +and bottom side will be in the field, there will be no vertical +component in the net force, which will be directed to the left. + +The magnitude of the induced \EMF\ is +\begin{equation} + |\EMF| = |-\deriv{t}{\Phi_B}| = \deriv{t}{AB} = B\deriv{t}{Nxw} + = NBw\deriv{t}{x} = NBwv \;. +\end{equation} +This leads to a current of +\begin{equation} + I = \frac{\EMF}{R} = \frac{Bwv}{R} \;, +\end{equation} +which causes a magnetic force of +\begin{equation} + F_B = NIwB\sin(90\dg) = \ans{\frac{N^2B^2w^2v}{R}} \;. +\end{equation} +\Part{b} +While the coil is completely inside the field, the flux remains +constant, so there is no induced current and \ans{no magnetic force}. + +\Part{c} +As the coil leaves the field, the flux drops back towards zero, +indicing a clockwise current trying to keep the flux up. Again, the +vertical components of the resulting magnetic force cancel out, but +the upward current in the left side will be subject to a magnetic +force directed to the left. The magnitude is the same as +for \Part{a}. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem31.31.tex b/latex/problems/Serway_and_Jewett_8/problem31.31.tex index 5cbcc58..654f3e5 100644 --- a/latex/problems/Serway_and_Jewett_8/problem31.31.tex +++ b/latex/problems/Serway_and_Jewett_8/problem31.31.tex @@ -47,5 +47,89 @@ label("$R_2$", (p2.x, yb), align=S); \end{problem*} \begin{solution} -\end{solution} +The magnetic flux through the loop will increase as the area enclosed +increases, inducing \EMF\ in each rod. +\begin{align} + \EMF_1 &= BLv_1 = 4.00\U{mV} \\ + \EMF_2 &= BLv_2 = 2.00\U{mV} \;. +\end{align} +The direction of the induced \EMF{}s will try to resist the increasing +flux, so $\EMF_1$ will be directed downward and $\EMF_2$ will be +directed upward. The situation is then equivalent to the following +circuit: +\begin{center} +\begin{asy} +import Circ; + +real u = 1cm; +MultiTerminal V1 = source(dir=-90, type=DC, Label("$\EMF_1$", align=W)); +MultiTerminal R1 = resistor(V1.terminal[1], dir=-90, Label("$R_1$", align=W)); +MultiTerminal R3 = resistor(dir=90, "$R_3$", draw=false); +R3.centerto(V1.terminal[0], R1.terminal[1], offset=u); R3.draw(); +MultiTerminal V2 = source(R1.terminal[1] + (2*u, 0), dir=90, type=DC, + Label("$\EMF_2$", align=E)); +MultiTerminal R2 = resistor(V2.terminal[1], dir=90, Label("$R_2$", align=E)); +wire(V1.terminal[0], R2.terminal[1]); +wire(R1.terminal[1], V2.terminal[0]); +pair top = (R3.center.x, V1.terminal[0].y); +pair bot = (top.x, R1.terminal[1].y); +wire(top, R3.terminal[0]); +wire(bot, R3.terminal[1]); +dot(top); +dot(bot); +\end{asy} +\end{center} +Solve this circuit using the usual Kirchhoff approach. Label the +current through the three branches $I_1$, $I_2$, and $I_3$ each +pointing up, and you have +\begin{align} + I_1 + I_2 + I_3 &= 0 \\ + \EMF_1 + I_1 R_1 - I_3 R_3 &= 0 \\ + \EMF_2 - I_2 R_2 + I_3 R_3 &= 0 \;. +\end{align} +This gives three equations with three unknowns. Solve using your +favorite method. +\begin{align} + \begin{pmatrix} + 0 \\ + \EMF_1 \\ + \EMF_2 + \end{pmatrix} + &= + \begin{pmatrix} + 1\U{\Ohm} & 1\U{\Ohm} & 1\U{\Ohm} \\ + -R_1 & 0 & R_3 \\ + 0 & R_2 & -R_3 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} \\ + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} + =& + \begin{pmatrix} + 1\U{\Ohm} & 1\U{\Ohm} & 1\U{\Ohm} \\ + -R_1 & 0 & R_3 \\ + 0 & R_2 & -R_3 + \end{pmatrix}^{-1} + \begin{pmatrix} + 0 \\ + \EMF_1 \\ + \EMF_2 + \end{pmatrix} + = + \begin{pmatrix} + -327 \\ + 182 \\ + 145 + \end{pmatrix} \U{$\mu$A} + \;. +\end{align} +Therefore, \ans{$I_3=145\U{$\mu$A}$ upwards}. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem31.45.tex b/latex/problems/Serway_and_Jewett_8/problem31.45.tex index 5acdf74..183e4c2 100644 --- a/latex/problems/Serway_and_Jewett_8/problem31.45.tex +++ b/latex/problems/Serway_and_Jewett_8/problem31.45.tex @@ -55,5 +55,21 @@ draw((xr, R.terminal[1].y) -- R.terminal[1]); \end{problem*} \begin{solution} +Picking up as the positive flux direction, the induced \EMF\ is +\begin{equation} + \EMF = -\deriv{t}{\Phi_B} = -\deriv{t}{NAB} = -NA\deriv{t}{B} + = -NA\frac{-2B}{\Delta t} = \frac{2NAB}{\Delta t} \;. +\end{equation} +This \EMF\ drives a current through the resistor +\begin{align} + 0 &= \EMF - IR \\ + I &= \frac{\EMF}{R} = \frac{-2NAB}{R\Delta t} \;. +\end{align} +Current is defined as the charge passing through a cross section of +your circuit per unit time, so the charge entering one end of the +resistor is +\begin{equation} + \Delta q = \frac{\Delta q}{\Delta t} \Delta t = I \Delta t + = \frac{2NAB}{R} = \ans{0.880\U{C}} \;. +\end{equation} \end{solution} - diff --git a/latex/problems/Serway_and_Jewett_8/problem31.46.tex b/latex/problems/Serway_and_Jewett_8/problem31.46.tex index e03c4ab..6a63351 100644 --- a/latex/problems/Serway_and_Jewett_8/problem31.46.tex +++ b/latex/problems/Serway_and_Jewett_8/problem31.46.tex @@ -24,5 +24,22 @@ label("$I$", (r+dr, 0), align=E); \end{problem*} \begin{solution} -\end{solution} +\Part{a} +A clockwise current induces an inward magnetic field inside the loop, +so the external flux must be increasingly out of the page. Therefore, +the magnetic field is \ans{increasing}. +\Part{b} +The induced \EMF\ must be +\begin{align} + 0 &= \EMF - IR \\ + \EMF &= IR \;, +\end{align} +This is related to the changing field via magnetic flux. +\begin{align} + |\EMF| &= |-\deriv{t}{\Phi_B}| = \deriv{t}{AB} = A\deriv{t}{B} + = \pi r^2 \deriv{t}{B} \\ + \deriv{t}{B} &= \frac{\EMF}{\pi r^2} = \frac{IR}{\pi r^2} + = \ans{62.2\U{T/s}} \;. +\end{align} +\end{solution} -- 2.26.2