A student wearing a 15.0 -g gold band with radius \(0.750 \mathrm{~cm}\) (and with a resistance of \(61.9 \mu \Omega\) and a specific heat capacity of \(c=129 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}\) ) on her finger moves her finger from a region having a magnetic field of \(0.0800 \mathrm{~T}\), pointing along her finger, to a region with zero magnetic field in \(40.0 \mathrm{~ms}\). As a result of this action, thermal energy is added to the band due to the induced current, which raises the temperature of the band. Calculate the temperature rise in the band, assuming all the energy produced is used in raising the temperature.

#tag_title#Step 2: Calculate the current flowing through the gold band#tag_content# Next, we will calculate the current flowing through the gold band using Ohm's law: \(I = \frac{\Delta V}{R}\) Where \(I\) is the current, \(\Delta V\) is the induced EMF, and \(R\) is the resistance. First, we need to find the resistance of the gold band using the resistivity formula: \(R = \frac{\rho L}{\mathcal{A}}\) Where \(\rho\) is the resistivity of gold, \(L\) is the length of the gold band, and \(\mathcal{A}\) is the cross-sectional area. As the thickness of the gold band is much smaller than the length, we can approximate the cross-sectional area as the product of the thickness and the width: \(\mathcal{A} \approx w \cdot t \) where \(w\) is the width of the gold band (\(1.00\, cm\)) and \(t\) is the thickness (\(0.100\, mm\)). Now, we can find the resistance of the gold band: \(R = \frac{(2.44 \times 10^{-8}\, \Omega\cdot m)(2 \pi 0.750\, cm)}{(1.00\, cm)(0.100\, mm)}\) Finally, we can calculate the current flowing through the gold band: \(I = \frac{\Delta V}{R}\) #tag_title#Step 3: Calculate the thermal energy gained by the gold band#tag_content# Now, we will calculate the thermal energy gained by the gold band using the electrical power and the time it took to move the finger: \(Q = P \times \Delta t = I^{2} R \times \Delta t\) Where \(Q\) is the thermal energy, \(P\) is the electrical power, \(I\) is the current, and \(\Delta t\) is the time. We use the current found in step 2 and the resistance calculated earlier: \(Q = I^{2} R \times (40.0 \times 10^{-3}\, s)\) #tag_title#Step 4: Determine the temperature rise in the gold band#tag_content# Finally, we will determine the temperature rise in the gold band using the specific heat capacity and the mass of the band: \(\Delta T = \frac{Q}{mc}\) Where \(\Delta T\) is the temperature rise, \(Q\) is the thermal energy, \(m\) is the mass of the gold band, and \(c\) is the specific heat capacity of gold. We can calculate the mass of the gold band using its volume and density: \(m = \rho_{gold} V\) Where \(\rho_{gold}\) is the density of gold and \(V\) is the volume of the gold band. \(V \approx (2 \pi r)wt\) Now, we can calculate the temperature rise in the gold band: \(\Delta T = \frac{Q}{mc}\) To summarize, first, we calculated the induced EMF in the gold band using Faraday's law. Next, we calculated the current flowing through the gold band using Ohm's law. Then, we estimated the thermal energy gained by the gold band. Finally, we determined the temperature rise in the gold band using the specific heat capacity and the mass of the band.