From 1325a92fe000b685a3d76689b1fc5cefdd688942 Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Wed, 12 Aug 2009 09:40:39 -0400 Subject: [PATCH] Minor typo corrections in chapter 27 problems --- .../Young_and_Freedman_12/problem27.22.tex | 2 +- .../Young_and_Freedman_12/problem27.39.tex | 17 +++++++++-------- .../Young_and_Freedman_12/problem27.64.tex | 10 +++++----- 3 files changed, 15 insertions(+), 14 deletions(-) diff --git a/latex/problems/Young_and_Freedman_12/problem27.22.tex b/latex/problems/Young_and_Freedman_12/problem27.22.tex index c173740..789d090 100644 --- a/latex/problems/Young_and_Freedman_12/problem27.22.tex +++ b/latex/problems/Young_and_Freedman_12/problem27.22.tex @@ -20,7 +20,7 @@ circle, 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}} + {12\cdot1.67\E{-27}\U{kg}} = \ans{5.68\U{Mm/s}} \;. \end{align} diff --git a/latex/problems/Young_and_Freedman_12/problem27.39.tex b/latex/problems/Young_and_Freedman_12/problem27.39.tex index 3738ac1..38ffec8 100644 --- a/latex/problems/Young_and_Freedman_12/problem27.39.tex +++ b/latex/problems/Young_and_Freedman_12/problem27.39.tex @@ -64,28 +64,29 @@ 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 \ + 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}} + = \frac{0.750\U{kg}\cdot9.8\U{m/s$^2$}}{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 +This maximum current would occur when the voltage $V$ from the battery balanced an $IR$ drop across the resistor, so \begin{equation} - V = IR = \ans{817\U{V}} \;. + V = IR = 32.7\U{A}\cdot25.0\U{\Ohm} = \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} \;, + I' = \frac{V}{R'} \;, \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}} + F &= F_B-F_g = I'lB - mg = \frac{VlB}{R'} - mg = ma \\ + a &= \frac{VlB}{R'm} - g + = \frac{817\U{V}\cdot0.500\U{m}\cdot0.450\U{T}} + {2.00\U{\Ohm}\cdot0.750\U{kg}} - 9.8\U{m/s$^2$} = \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 095b2a4..0f7361a 100644 --- a/latex/problems/Young_and_Freedman_12/problem27.64.tex +++ b/latex/problems/Young_and_Freedman_12/problem27.64.tex @@ -19,8 +19,8 @@ Writing $\vect{F}_B=q\vect{v}\cdot\vect{B}$ in terms of components \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}} \;. + -qvB_y &= 3F_0 & qvB_x &= 4F_0 \\ + B_y &= \ans{\frac{-3F_0}{qv}} < 0 & B_x &= \ans{\frac{4F_0}{qv}} > 0 \;. \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}. @@ -30,10 +30,10 @@ 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} + = 36\frac{F_0^2}{q^2v^2} - 16\frac{F_0^2}{q^2v^2} - 9\frac{F_0^2}{q^2v^2} + = (36-16-9)\frac{F_0^2}{q^2v^2} = 11\frac{F_0^2}{q^2v^2} \\ - B_z &= \ans{\sqrt{11}\frac{F_0}{qv}} \;. + B_z &= \ans{\pm\sqrt{11}\frac{F_0}{qv}} \;. \end{align} However, we still cannot find the direction of $B_z$. \end{solution} -- 2.26.2