1 \begin{problem*}{23.10}
2 A piece of insulated wire is shaped into a figure eight as shown in
3 Figure P23.10. The radius of the upper circle is $r_s = 5.00\U{cm}$
4 and that of the lower circle is $r_b = 9.00\U{cm}$. The wire has a
5 uniform resistance per unit length of $\lambda = 3.00\U{$\Omega$/m}$.
6 A uniform magnetic field is applied perpendicular to the plane of the
7 two circles, in the direction shown. The magnetic field is increasing
8 at a constant rate of $dB/dt = 2.00\U{T/s}$. Find the magnitude and
9 direction of the induced current in the wire.
10 \end{problem*} % problem 23.10
13 Pick a direction for the current to be counterclockwise in the bottom
14 loop (so clockwise in the top). Thus, the area vector of the top loop
15 is antiparallel to \vect{B} and that of the bottom loop is parallel to
16 \vect{B}. The magnetic flux is then
18 \Phi_B = \vect{A}\cdot\vect{B} = (\pi r_s^2 - \pi r_b^2)B \;.
23 \varepsilon = - \frac{d\Phi_B}{dt} = \pi(r_b^2 - r_s^2)\frac{dB}{dt} \;.
26 The resistance of the entire figure eight is
28 R = \lambda (2 \pi r_s + 2 \pi r_b) \;.
31 Plugging that into Ohm's law yields
33 \varepsilon &= V = I R \\
34 I &= \frac{(r_b^2 - r_s^2) \frac{dB}{dt}}{2 \lambda (r_s + r_b)}
35 = \ans{25.2\U{mA}} \;.