Answer
a. 0.15 A.
b. 1.5 mW.
Work Step by Step
a. The current in the coil is the induced emf, divided by the coil’s resistance. Use equation 18-3 t find the resistance of the coil.
$$I=\frac{\epsilon}{R}=\frac{NA_{coil}\frac{dB}{dt}}{\rho L/A_{wire}}$$
$$I=\frac{NA_{coil} A_{wire}}{\rho L }\frac{dB}{dt}$$
$$I=\frac{30\pi (0.110m)^2 \pi (0.0013m)^2}{(1.68\times10^{-8}\Omega \cdot m) 30 (2 \pi (0.110m))}(8.65\times10^{-3}T/s)=0.15A$$
b. The rate at which thermal energy is produced in the wire is the dissipated power.
$$P=I^2R=I^2\rho L/A_{wire}=(0.1504A)^2\frac{(1.68\times10^{-8}\Omega \cdot m) 30 (2 \pi (0.110m))}{ \pi (0.0013m)^2}=1.5 mW$$
