\begin{problem7}{22.21}
A rectangular coil consists of $N=100$ closely wrapped turns and has
dimensions $a = 0.400\U{m}$ and $b = 0.300\U{m}$.  The coil is hinged
along the $y$ axis, and its plane makes an angle $\theta = 30.0\dg$
with the $x$ axis (Fig.~P22.21).  What is the magnitude of the torque
exerted on the coil by a uniform magnetic field $B = 0.800\U{T}$
directed along the x axis whwn the current is $I=1.20\U{A}$ in the
direction shown?  What is the expected direction of motion of the
coil?
\begin{center}
\begin{empfile}[7]
\begin{emp}(0,0)
  pair p[];
  numeric dirz, dirl, Dx, Dy, Dz, Dlxz, Dly, ddy;
  dirz := -160;
  dirl := -20;
  Dx := 2cm;
  Dy := 2cm;
  Dz := 1cm;
  Dlxz := 1.5cm;
  Dly := 1.5cm;
  ddy := 5pt;
  % draw axes
  drawarrow origin--(Dx,0) withcolor black withpen pencircle scaled 0pt;
  label.bot(btex x etex, (Dx,0));
  drawarrow origin--(0,Dy) withcolor black withpen pencircle scaled 0pt;
  label.lft(btex y etex, (0,Dy));
  drawarrow origin--(Dz*dir(dirz)) withcolor black withpen pencircle scaled 0pt;
  label.lft(btex z etex, (Dz*dir(dirz)));
  % draw box
  p[0] := origin;
  p[1] := Dlxz*dir(dirl);
  p[2] := p[1]+(0,Dly);
  p[3] := p[0]+(0,Dly);
  draw p[0]--p[1]--p[2]--p[3]--cycle withcolor (1,.2,0) withpen pencircle scaled 2pt; 
  label.lft(btex $a = 0.400\mbox{ m}$ etex, draw_length(p[3],p[0],3pt));
  label.bot(btex $b = 0.300\mbox{ m}$ etex, draw_length(p[0],p[1],3pt));
  label.rt(btex $30^\circ$ etex, draw_langle((Dx,0),p[0],p[1],Dlxz/2.2));
  drawarrow (p[3]+0.2(p[2]-p[3])+(0,ddy))--(p[2]-0.2(p[2]-p[3])+(0,ddy)) withcolor (1,.2,1) withpen pencircle scaled 1pt;
  label.urt(btex $I = 1.20\mbox{ A}$ etex, (p[2]+p[3])/2+(0,ddy));
\end{emp}
\end{empfile}
\end{center}
\end{problem*} % problem 22.21

\begin{solution}
Using our formula for force on a wire segment $\vect{F} =
I\vect{l}\times\vect{B}$, and recalling that torque is defined $\tau =
\vect{r}\times\vect{F}$, we can use the right hand rule to find the
direction of motion.

The torque from the portion of the coil lying on the $y$ axis is zero,
because the lever arm is zero ($r$ in the torque equation).  The
torque from the top portion is also zero, because the force is in the
\jhat\ direction and the coil is not free to rotate in that direction.
Similarly the torque from the bottom portion is zero, because the
force is in the $-\jhat$ direction.  All the torque comes from the
force on the outer leg, giving a force in the \khat\ direction.  So
\ans{we expect the angle $\theta$ to increase}.

To find the magnitude of the torque, we simply plug in
\begin{equation}
 \tau = \vect{r}\times\vect{F} = bF\cos\theta = b\cos\theta\cdot(NIaB)
 = IabB\cos\theta
      = \ans{9.98\U{J}}
\end{equation}
Where we multiplied the force from a single wire by $N$ because there
are $N$ wraps, and took $\cos\theta$ to get the perpendicular force
because $\theta$ is the complement of the angle between $r$ and $F$.
\end{solution}
