A 21.0-cmcm-diameter coil consists of 50 turns of circular copper wire 2.0 mmmm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 9.30×10−3 T/sT/s . The resistivity of copper is 1.68×10−8Ω⋅m×10−8Ω⋅m.Part APart completeDetermine the current in the loop.Express your answer to two significant figures and include the appropriate units.I𝐼 =9.1×10−2 AA SubmitPrevious Answers CorrectCorrect answer is shown. Your answer 0.0915A0.0915A was either rounded differently or used a different number of significant figures than required for this part.Part BDetermine the rate at which thermal energy is produced.

To determine the rate at which thermal energy is produced, we need to use the formula for power, which is P = I^2 * R, where I is the current and R is the resistance.

First, we need to calculate the resistance of the wire. The resistance of a wire can be calculated using the formula R = ρ * (L/A), where ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire.

The length of the wire can be calculated by multiplying the circumference of the coil by the number of turns. The circumference of a circle is given by the formula C = π * D, where D is the diameter. So, the length of the wire is L = π * D * N = π * 0.21 m * 50 = 33.0 m.

The cross-sectional area of the wire can be calculated using the formula A = π * (d/2)^2, where d is the diameter of the wire. So, the cross-sectional area is A = π * (0.002 m / 2)^2 = 3.14 * 10^-6 m^2.

Now we can calculate the resistance: R = ρ * (L/A) = 1.68 * 10^-8 Ω*m * (33.0 m / 3.14 * 10^-6 m^2) = 0.177 Ω.

Finally, we can calculate the power: P = I^2 * R = (0.091 A)^2 * 0.177 Ω = 0.00147 W or 1.47 mW. This is the rate at which thermal energy is produced.