1 \begin{problem*}{23.13}
2 Figure P23.12 shows a top view of a bar that can slide without
3 friction. The resistor is $R = 6.00\Omega$, and a $B = 2.50\U{T}$
4 magnetic field is directed perpendicularly downward, into the paper.
6 \Part{a} Calculate the applied force required to move the bar to the
7 right at a constant speed $v = 2.00\U{m/s}$.
8 \Part{b} At what rate is energy delivered to the resistor?
9 \end{problem*} % problem 23.13
13 Let $x$ be the width of the enclosed loop. The magnetic flux is then
19 \varepsilon = -\frac{d\Phi_B}{dt} = -lB \frac{dx}{dt} = -lvB
21 So the induced current is
23 I = \frac{\varepsilon}{R} = \frac{-lvB}{R}
25 The $-$ sign indicates the current is counterclockwise (out of the
26 page), so current flows upward through the bar, so the magnetic force
27 on the bar is to the left, so our applied force must be \ans{to the right}.
29 The work begin done by the applied force is
33 So the power input from the force is
35 P_F = \frac{W}{dt} = F \frac{dx}{dt} = Fv \;.
38 All of this power must be dissipated by the resistor, so the current is
41 I &= \sqrt{\frac{P}{R}} = \sqrt{\frac{Fv}{R}} \;.
44 We combine both current equations to yield
46 \frac{-lvB}{R} &= \sqrt{\frac{Fv}{R}} \\
48 F &= \frac{v(lB)^2}{R} = \ans{3.00\U{N}} \;.
52 Going back and plugging in $F$,
54 P_F = Fv = \ans{6.00\U{W}} \;.