1 \begin{problem*}{31.23}
2 Figure~P31.23 shows a top view of a bar that can slide on two
3 frictionless rails. The resistor is $R=6.00\U{\Ohm}$, and a
4 $2.50\U{T}$ magnetic field is directed perpendicularly downward, into
5 the paper. Let $l=1.20\U{m}$. \Part{a} Calculate the applied force
6 required to move the bar to the right at a constant speed of
7 $2.00\U{m/s}$. \Part{b} At what rate is energy delivered to the
23 MultiTerminal R = resistor(dir=90, "$R$", draw=false);
24 real rlen = R.terminal[1].y - R.terminal[0].y;
25 Vector B = BField(phi=-90);
26 vector_field(R.center + (0.75*u, 0), width=1.5*u, height=rlen, v=B);
28 real yt = R.terminal[1].y;
29 real yb = R.terminal[0].y;
30 wire((0, yt), (1.5*u, yt));
31 wire((0, yb), (1.5*u, yb));
32 pair p = (u,(yt+yb)/2);
33 Vector F = Force(p, Label("$F_\text{app}$", position=EndPoint)); F.draw();
34 Block block = Block(p, width=w, height=rlen); block.draw();
35 Distance Dl = Distance((u/2, yt), (u/2, yb), Label("$l$", embed=Shift));
43 Moving the bar to the right increases the magnetic flux directed into
44 the page by increasing the area of magnetic field enclosed by the
45 loop. Increasing the flux induces a counter-clockwise current to
49 |\EMF| &= |-\deriv{t}{\Phi_B}| = lB \deriv{t}{x} = lBv \\
50 0 &= \sum_\text{loop} V_i = \EMF - IR \\
51 I &= \frac{\EMF}{R} = \frac{lBv}{R} \;.
53 The counter-clockwise current is moving up through the sliding bar, so
54 the magnetic force on the bar is directed to the left, with a
57 F_B = IlB\sin(90\dg) = IlB = \frac{l^2B^2v}{R} \;.
60 For the bar to move to the right at a constant speed, the net force in
61 the $\ihat$ direction should be zero. We'll have to apply an external
62 $F_\text{app}$ to the right to counter $F_B$.
64 0 &= \sum_i F_{i,x} = F_\text{app} - F_B \\
65 F_\text{app} &= F_B = \frac{l^2B^2v}{R}
66 = \frac{(1.20\U{m}\cdot2.50\U{T})^2\cdot2.00\U{m/s}}{6.00\U{\Ohm}}
71 The power delivered to a resister is
73 P_R = IV = I^2R = \p({\frac{lBv}{R}})^2 R = \frac{(lBv)^2}{R}
74 = \frac{(1.20\U{m}\cdot2.50\U{T}\cdot2.00\U{m/s})^2}{6.00\U{\Ohm}}
78 If you want to look at the problem in terms of energy conservation,
79 the external force is putting energy into the system at a rate of
81 P_\text{in} = \deriv{t}{F_\text{app} x} = F_\text{app}\deriv{t}{x}
82 = F_\text{app}v = \frac{l^2B^2v}{R} \cdot v = \frac{(lBv)^2}{R} \;.
84 All this energy has to go somewhere, and the only place for it to go
85 is into resistor heat, so it's no surprise that we get the same
86 formula for $P_\text{in}$ that we got for $P_R$.