\begin{problem*}{22.34}
Two long, parallel conductors, separated by $r = 10.0\U{cm}$, carry
current in the same direction.  The first wire carries current $I_1 =
5.00\U{A}$, and the second carries $I_2 = 8.00\U{A}$.
\Part{a} What is the magnitude of the magnetic field $B_1$ created by
$I_1$ at the location of $I_2$?
\Part{b} What is the force per unit length exerted by $I_1$ on $I_2$?
\Part{c} What is the magnitude of the magnetic field $B_2$ created by
$I_2$ at the location of $I_1$?
\Part{d} What is the force per unit length exerted by $I_2$ on $I_1$?
\end{problem*} % problem 22.34

\begin{solution}
\Part{a}
From Ampere's law, the $B$ field generated by a long, thin current is
\begin{equation}
 B = \frac{\mu_0 I}{2 \pi r}
\end{equation}
Plugging in $I_1$, we have
\begin{equation}
 B_1 = \frac{\mu_0 I_1}{2 \pi r} = \ans{10.0\U{$\mu$T}} \label{eqn.34_b1}
\end{equation}
This $B$ field depends on your distance from $I_1$, but because the
wires are parallel, the $B$ field from $I_1$ is constant along $I_2$
We can use the right hand rule to determine that $\vect{B}_1$ is
perpendicular to both $I_1$ and $r$.

\Part{b}
From $F_B = q\vect{v}\times\vect{B}$ we have the force on a current
carrying wire in a uniform magnetic field as
\begin{equation}
 F_B = I\vect{l}\times\vect{B}
\end{equation}

Combining these two equations, we have the force per unit length of
$I_1$ on $I_2$ as
\begin{equation}
 F_{B12}/l = I_2 B_1 = \frac{\mu_0 I_1 I_2}{2 \pi r} = \ans{80.0\U{$\mu$N}} \label{eqn.34_fb12}
\end{equation}
where there is no $\sin\theta$ term in the cross product, because
$B_1$ is perpendicular to $I_2$.
By drawing the situation and doing some right hand rules, you can
convince yourself that this force is \emph{attractive}.

\Part{c}
Because the situation in \Part{c} is identical to \Part{a} with $I_1
\leftrightarrow I_2$, we simply relabel eqn.\ \ref{eqn.34_b1}.
\begin{equation}
 B_2 = \frac{\mu_0 I_2}{2 \pi r} = \ans{16.0\U{$\mu$T}}
\end{equation}

\Part{d}
Eqn.\ \ref{eqn.34_fb12} is identical under the relabeling, so we have
another attractive force at the same magnitude
\begin{equation}
 F_{B21}/l = \ans{80\U{$\mu$N}}
\end{equation}
as we would expect from Newton's third law (for every action there is
an equal and opposite reaction).
\end{solution}
