Your friend decides to produce electrical power by turning a coil of \(1.00 \cdot 10^{5}\) circular loops of wire around an axis parallel to a diameter in the Earth's magnetic field, which has a local magnitude of \(0.300 \mathrm{G}\). The loops have a radius of \(25.0 \mathrm{~cm} .\) a) If your friend turns the coil at a frequency of \(150.0 \mathrm{~Hz}\) what peak current will flow in a resistor, \(R=1500 . \Omega\) connected to the coil? b) The average current flowing in the coil will be 0.7071 times the peak current. What will be the average power obtained from this device?

To find the peak magnetic flux, we need to calculate the maximum possible magnetic flux through each loop. The magnetic flux (Φ_B) is given by: Φ_B = B * A * cos(θ) where: B - the magnetic field magnitude, A - the area of the loop, θ - the angle between the magnetic field and the normal to the loop surface. Since the loop is rotated parallel to the Earth's magnetic field diameter, at the maximum possible moment, the normal to the loop surface will be perpendicular to the magnetic field, and the angle θ will be 0 degrees. Therefore, cos(θ) is 1 at this moment. The area A of the loop can be calculated as: A = π * r^2 Using the given magnetic field magnitude (B = 0.3 G = 0.3 * 10^{-4} T) and the radius of the loops (r = 25 cm = 0.25 m), we get: Φ_B_peak = (B * π * r^2)

Next, we will use Faraday's Law of electromagnetic induction to find the peak electromotive force (EMF) induced in the coil. Faraday's Law states that the induced EMF (ε) in a coil is equal to the negative rate of change of magnetic flux (Φ_B) with respect to time (t): ε = -N * d(Φ_B) / dt where N is the number of loops in the coil. Since the coil is turning at a frequency f, the angular velocity (ω) can be found as: ω = 2 * π * f The change of magnetic flux with respect to time can be determined as the product of the peak flux and the derivative of cos(ωt): d(Φ_B) / dt = Φ_B_peak * (-ω * sin(ωt)) The induced EMF at any time t is: ε(t) = N * (Φ_B_peak * ω * sin(ωt)) The peak EMF (ε_peak) occurs when sin(ωt) = 1: ε_peak = N * Φ_B_peak * ω Using the given values for N (1.00 * 10^{5}), f (150 Hz), and our previous calculation for Φ_B_peak, find the peak EMF value: ε_peak = N * Φ_B_peak * ω

Now we will use Ohm's Law to find the peak current (I_peak) flowing in the resistor: I_peak = ε_peak / R Where R is the resistance (1500 Ω). Use the peak EMF value calculated in Step 2 to calculate the peak current.

The average current flowing in the coil (I_avg) is given as: I_avg = 0.7071 * I_peak The average power (P_avg) can be calculated as: P_avg = I_avg^2 * R Using the calculated values for the average current and resistance, find the average power obtained from this device.