Minor typo corrections in chapter 27 problems

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}}
-            {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 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}
-  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 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}
-  -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}