\begin{problem*}{29.40}
Consider the system pictured in Figure~P29.40.  A $15.0\U{cm}$
horizontal wire of mass $15.0\U{g}$ is placed between two thin,
vertical conductors, and a uniform magnetic field acts perpendicular
to the page.  The wire is free to move vertically without friction on
the two vertical conductors.  When a $5.00\U{A}$ curent is directed as
shown in the figure, the horizontal wire moves upward at a constant
velocity in the presence of gravity.  \Part{a} What forces act on the
horizontal wire, and \Part{b} under what condition is the wire able to
move upward at a constant velocity.  \Part{c} Find the magnitude and
direction of the minimum magnetic field required to move the wire at a
constant speed.  \Part{d} What happens if the magnetic field exceeds
this minimum value?
\begin{center}
% |     |  15.0 cm between verticals
% +--<--+
% |  5A |
% |     |
% v 5A  ^ 5A
\begin{asy}
import Mechanics;
import Circ;

real u = 1cm;

pair ul = (-u, u);    // upper left
pair ml = (-u, 0);    // middle left
pair ll = (-u, -2u);  // lower left
pair ur = (u, u);     // upper right
pair mr = (u, 0);
pair lr = (u, -2u);

MultiTerminal I = current(label=Label("$5.00\U{A}$"), draw=false);
I.centerto(lr, mr);  I.label.align = E;  I.draw();
wire(lr, I.terminal[0]);
wire(I.terminal[1], ur);
I.centerto(mr, ml);  I.label.align = S;  I.draw();
wire(mr, I.terminal[0]);
wire(I.terminal[1], ml);
I.centerto(ml, ll);  I.label.align = W;  I.draw();
wire(ul, I.terminal[0]);
wire(I.terminal[1], ll);
dot(mr);
dot(ml);

Distance D = Distance(ml, mr, offset=-u/2, Label("$15.0\U{cm}$", align=N));
D.draw();
\end{asy}
\end{center}
\end{problem*}

\begin{solution}
\Part{a}
The only forces acting on the horizontal wire are \ans{gravity and the
force of the magnetic field on the current carried by the wire}.

\Part{b}
If the wire is moving at a constant velocity,
\begin{align}
  0 &= \deriv{t}{\vect{v}} = \vect{a} = \frac{\vect{F}}{m} \\
  \ans{\vect{F}} &= \ans{0} \;.
\end{align}
In English, this means the net force on the wire must be zero.
Because there are only two forces acting on the wire, this means these
forces must be equal in magnitude and opposite in direction.

\Part{c}
The gravitational force on the wire will be $F_g=mg$ directed
downwards, and the magnetic force on the wire will be
$F_B=I\vect{L}\times\vect{B}$ directed upwards.  Because the magnetic
field is perpendicular to the page, and the current is directed along
the page, the angle between $\vect{L}$ and $\vect{B}$ is $90\dg$, so
$F_B=ILB\sin{90\dg}=ILB$.  Summing forces in the vertical direction,
\begin{align}
  0 &= \sum_i F_{i,y} = F_B - F_g = ILB - mg \\
  ILB &= mg \\
  B &= \frac{mg}{IL}
    = \frac{0.0150\U{kg} \cdot 9.80\U{m/s$^2$}}{5.00\U{A} \cdot 0.150\U{m}}
    = \ans{196\U{mT}} \;.
\end{align}
In order to generate an upwards force on a leftwards current, the
magnetic field must be directed \ans{out of the page}.

\Part{d}
If the magnetic field exceeds this value, \ans{the rod will accelerate
  upwards}, moving up faster and faster.  If the magnetic field is
less than this value, the rod will accelerate downwards, moving up
slower and slower, eventually pausing, and then moving back down
faster and faster.  At the critical magnetic field, where the forces
balance, the rod will keep moving forever at it's initial velocity (or
hang in place if the initial velocity was zero).
\end{solution}
