Question

Your friend decides to produce electrical power by turning a coil of \(1.00 \cdot 10^{5}\) circular loops of wire around an axis perpendicular to 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 \(150 . \mathrm{Hz}\), what peak current will flow in a resistor, \(R=1.50 \mathrm{k} \Omega\), 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?

Short answer

Answer: The peak current flowing through the resistor is approximately 0.011783 A, and the average power obtained from this device is approximately 0.1039 W.

Step-by-step solution

Calculate the magnetic flux through a single loop of the coil

First, we need to find the magnetic flux (Φ) through a single loop of the coil. The formula for magnetic flux is given by: Φ = B * A * cos(θ), where B is the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop. Since the magnetic field is perpendicular to the loop, the angle (θ) is 0, and therefore cos(θ) = 1. Now we can calculate the area, A = π * r^2, and find the magnetic flux. B = 0.300 G = 0.300 * 10^{-4} T (converting from Gauss to Tesla) Radius (r) = 25.0 cm = 0.25 m A = π * (0.25)^2 = 0.19635 m^2 Calculating the magnetic flux (Φ): Φ = B * A * cos(θ) = 0.300 * 10^{-4} T * 0.19635 m^2 * 1 = 0.058905 * 10^{-4} Wb

Find the induced electromotive force (EMF) in the coil

Now we need to find the induced electromotive force (EMF) in the coil. Using Faraday's law of electromagnetic induction, the EMF formula is given by: EMF = - N * (ΔΦ/Δt), where N is the number of loops in the coil, and (ΔΦ/Δt) is the rate of change of the magnetic flux. Since the coil is turning at a frequency of 150 Hz, we can calculate the time period (T) of one rotation as T = 1/frequency = 1/150 s. The change in magnetic flux (ΔΦ) during one rotation is equal to 2 * Φ (as during the rotation, the magnetic field goes from the maximum value to the minimum value and back to the maximum value). Calculating the rate of change of magnetic flux (ΔΦ/Δt): ΔΦ/Δt = (2 * Φ) / T = (2 * 0.058905 * 10^{-4} Wb) / (1/150 s) = 1.7674 * 10^{-4} Wb/s Now, we can find the induced EMF in the coil: EMF = - N * (ΔΦ/Δt) = -1.00 * 10^5 * 1.7674 * 10^{-4} Wb/s = -17.674 V The negative sign indicates that the electromotive force opposes the change in magnetic flux, but since we are only interested in the magnitude, we can consider EMF = 17.674 V.

Calculate the peak current through the resistor

To find the peak current through the resistor (I), we can use Ohm's law: I = EMF / R Where R is the resistance of the resistor, which is given as 1.50 kΩ = 1500 Ω. Calculating the peak current (I): I = EMF / R = 17.674 V / 1500 Ω = 0.011783 A

Calculate the average current and power

Now, it's given that the average current is 0.7071 times the peak current. So, we can find the average current (I_avg) as: I_avg = 0.7071 * I = 0.7071 * 0.011783 A = 0.008326 A Finally, we can calculate the average power (P) obtained from this device using the formula: P = I_avg^2 * R Calculating the average power (P): P = (0.008326 A)^2 * 1500 Ω = 0.1039 W In conclusion, the peak current flowing through the resistor is approximately 0.011783 A, and the average power obtained from this device is approximately 0.1039 W.