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Minor typo corrections in chapter 27 problems
author
W. Trevor King
<wking@drexel.edu>
Wed, 12 Aug 2009 13:40:39 +0000
(09:40 -0400)
committer
W. Trevor King
<wking@drexel.edu>
Thu, 17 Sep 2009 16:50:15 +0000
(12:50 -0400)
latex/problems/Young_and_Freedman_12/problem27.22.tex
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latex/problems/Young_and_Freedman_12/problem27.39.tex
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latex/problems/Young_and_Freedman_12/problem27.64.tex
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diff --git
a/latex/problems/Young_and_Freedman_12/problem27.22.tex
b/latex/problems/Young_and_Freedman_12/problem27.22.tex
index c173740701b2ea75afcc3fed7b76d4e4b4ba47a5..789d09048c35a5619571d1a0715be010fd5fff27 100644
(file)
--- 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}}
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\cdot1.67
e-27
\U{kg}}
+ {12\cdot1.67
\E{-27}
\U{kg}}
= \ans{5.68\U{Mm/s}} \;.
\end{align}
= \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 3738ac145e33089315536ce28624e37146e223fc..38ffec89b2838e13421fb6813096744276c38da4 100644
(file)
--- 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}
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}
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}
= 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}
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}
\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}
\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}
= \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 095b2a4a9d80f77a766d5a964ebb0ade2c7dc5d5..0f7361ad80860c6dab2d021315968abe305fc1bf 100644
(file)
--- 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}
\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}.
\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
\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} \\
= 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}
\end{align}
However, we still cannot find the direction of $B_z$.
\end{solution}