1 \begin{problem7}{22.21}
2 A rectangular coil consists of $N=100$ closely wrapped turns and has
3 dimensions $a = 0.400\U{m}$ and $b = 0.300\U{m}$. The coil is hinged
4 along the $y$ axis, and its plane makes an angle $\theta = 30.0\dg$
5 with the $x$ axis (Fig.~P22.21). What is the magnitude of the torque
6 exerted on the coil by a uniform magnetic field $B = 0.800\U{T}$
7 directed along the x axis whwn the current is $I=1.20\U{A}$ in the
8 direction shown? What is the expected direction of motion of the
14 numeric dirz, dirl, Dx, Dy, Dz, Dlxz, Dly, ddy;
24 drawarrow origin--(Dx,0) withcolor black withpen pencircle scaled 0pt;
25 label.bot(btex x etex, (Dx,0));
26 drawarrow origin--(0,Dy) withcolor black withpen pencircle scaled 0pt;
27 label.lft(btex y etex, (0,Dy));
28 drawarrow origin--(Dz*dir(dirz)) withcolor black withpen pencircle scaled 0pt;
29 label.lft(btex z etex, (Dz*dir(dirz)));
32 p[1] := Dlxz*dir(dirl);
35 draw p[0]--p[1]--p[2]--p[3]--cycle withcolor (1,.2,0) withpen pencircle scaled 2pt;
36 label.lft(btex $a = 0.400\mbox{ m}$ etex, draw_length(p[3],p[0],3pt));
37 label.bot(btex $b = 0.300\mbox{ m}$ etex, draw_length(p[0],p[1],3pt));
38 label.rt(btex $30^\circ$ etex, draw_langle((Dx,0),p[0],p[1],Dlxz/2.2));
39 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;
40 label.urt(btex $I = 1.20\mbox{ A}$ etex, (p[2]+p[3])/2+(0,ddy));
44 \end{problem*} % problem 22.21
47 Using our formula for force on a wire segment $\vect{F} =
48 I\vect{l}\times\vect{B}$, and recalling that torque is defined $\tau =
49 \vect{r}\times\vect{F}$, we can use the right hand rule to find the
52 The torque from the portion of the coil lying on the $y$ axis is zero,
53 because the lever arm is zero ($r$ in the torque equation). The
54 torque from the top portion is also zero, because the force is in the
55 \jhat\ direction and the coil is not free to rotate in that direction.
56 Similarly the torque from the bottom portion is zero, because the
57 force is in the $-\jhat$ direction. All the torque comes from the
58 force on the outer leg, giving a force in the \khat\ direction. So
59 \ans{we expect the angle $\theta$ to increase}.
61 To find the magnitude of the torque, we simply plug in
63 \tau = \vect{r}\times\vect{F} = bF\cos\theta = b\cos\theta\cdot(NIaB)
67 Where we multiplied the force from a single wire by $N$ because there
68 are $N$ wraps, and took $\cos\theta$ to get the perpendicular force
69 because $\theta$ is the complement of the angle between $r$ and $F$.