Answer
See answers.
Work Step by Step
a. Calculate the magnitude of the emf induced in the ring. The change in magnetic flux is the change in magnetic field (0.68 T) multiplied by the ring’s area, $A=1.767\times10^{-4}m^2$.
$$emf=\frac{\Delta \Phi}{\Delta t}=\frac{(\Delta B)(A)}{\Delta t}$$
$$=\frac{(0.68 T)( 1.767\times10^{-4}m^2)}{0.045s}=2.67\times10^{-3}V$$
The average power dissipated in the ring is $I^2R$, where the current is emf/R. Finally, the thermal energy is the power multiplied by the time.
$$Q=P\Delta t=I^2R \Delta t = \frac{emf^2}{R}\Delta t$$
$$Q= \frac{0.00267V^2}{(55\times10^{-6}\Omega)}0.045s\approx 5.8mJ$$
b. Now find the temperature change from the thermal energy, the mass of the ring, and the specific heat of gold.
$$\delta T = \frac{Q}{mc}\frac{5.833\times10^{-3}J}{(0.015kg)(129J/(kg \cdot C^{\circ}))}=3.0\times10^{-3} C^{\circ}$$
