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 ADetermine the current in the loop.

To determine the current in the loop, we first need to find the induced emf (electromotive force) in the coil using Faraday's law of electromagnetic induction. The law states that the induced emf in a coil is equal to the rate of change of magnetic flux through the coil.

Step 1: Calculate the area of the coil The diameter of the coil is 21.0 cm, so the radius r is 21.0 cm / 2 = 10.5 cm = 0.105 m. The area A of the coil is given by the formula for the area of a circle, A = πr^2 = π*(0.105 m)^2 = 0.0347 m^2.

Step 2: Calculate the change in magnetic flux The magnetic flux Φ through the coil is given by Φ = BA, where B is the magnetic field and A is the area of the coil. The rate of change of magnetic flux is given by ΔΦ/Δt = BΔA/Δt + AΔB/Δt. Since the area of the coil is constant, ΔA/Δt = 0, so ΔΦ/Δt = A*ΔB/Δt = 0.0347 m^2 * 9.30×10^−3 T/s = 3.23×10^−4 Wb/s.

Step 3: Calculate the induced emf According to Faraday's law, the induced emf E is equal to the negative rate of change of magnetic flux, E = -N*ΔΦ/Δt, where N is the number of turns in the coil. The negative sign indicates that the induced emf opposes the change in magnetic flux, but since we're only interested in the magnitude of the emf, we can ignore the negative sign. So, E = 50 turns * 3.23×10^−4 Wb/s = 0.0162 V.

Step 4: Calculate the resistance of the wire The resistance R of the wire is given by R = ρL/A, where ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire. The length of the wire is the circumference of the coil times the number of turns, L = 2πrN = 2π0.105 m50 turns = 33.0 m. The cross-sectional area of the wire is A = π*(d/2)^2 = π*(2.0 mm / 2)^2 = 3.14 mm^2 = 3.14×10^−6 m^2. So, R = 1.68×10^−8 Ω*m * 33.0 m / 3.14×10^−6 m^2 = 0.177 Ω.

Step 5: Calculate the current in the loop The current I in the loop is given by Ohm's law, I = E/R = 0.0162 V / 0.177 Ω = 0.0915 A. So, the current in the loop is approximately 0.0915 A.