Question

An AM radio station broadcasts at \(570 \mathrm{kHz}\). (a) What is the wavelength of the radio wave in air? (b) If a radio is tuned to this station and the inductance in the tuning circuit is \(0.20 \mathrm{mH},\) what is the capacitance in the tuning circuit? (c) In the vicinity of the radio, the amplitude of the electric field is \(0.80 \mathrm{V} / \mathrm{m} .\) The radio uses a coil antenna of radius \(1.6 \mathrm{cm}\) with 50 turns. What is the maximum emf induced in the antenna, assuming it is oriented for best reception? Assume that the fields are sinusoidal functions of time.

Short answer

Question: Calculate the wavelength of a 570 kHz AM radio wave, the capacitance of the tuning circuit with a 0.20 mH inductance, and the maximum emf induced in a coil antenna with a radius of 1.6 cm, and 50 turns when exposed to an electric field of amplitude 0.80 V/m. Answer: The wavelength of the 570 kHz AM radio wave is approximately 526.32 meters. The capacitance of the tuning circuit with a 0.20 mH inductance is approximately \(4.87×10^{-11} \mathrm{F}\). The maximum emf induced in the coil antenna is approximately 0.115 V.

Step-by-step solution

(a) Finding the Wavelength of the Radio Wave

To find the wavelength of the AM radio wave, we can use the equation: speed of light = frequency * wavelength The speed of light, c, is approximately \(3.0 × 10^8 \mathrm{m/s}\). The frequency, f, is given in kilohertz, so we need to convert it to hertz: \(570 \mathrm{kHz} = 570 × 10^3 \mathrm{Hz}\) Now we can plug the values into the equation: \(3.0 × 10^8 \mathrm{m/s} = 570 × 10^3 \mathrm{Hz} * \lambda\) Solving for the wavelength, \(\lambda\): \(\lambda= \frac{3.0 × 10^8 \mathrm{m/s}}{570 × 10^3 \mathrm{Hz}}\) \(\lambda= 526.32 \mathrm{m}\) So, the wavelength of the radio wave in the air is 526.32 meters.

(b) Finding the Capacitance in the Tuning Circuit

To find the capacitance in the tuning circuit, we can use the resonant frequency formula, which is given by: \( f_{res} = \frac{1}{2\pi\sqrt{LC}}\) We are given the value of the inductance, L = \(0.20 \mathrm{mH}\). We need to find the capacitance, C, while the resonant frequency, \(f_{res}\) = \(570 \mathrm{kHz}\) First, let's convert the given inductance value to henry: \(L = 0.20 \mathrm{mH} = 0.20 × 10^{-3} \mathrm{H}\) Now we can rewrite the resonant frequency formula to find the capacitance: \(C = \frac{1}{4\pi^2f_{res}^2L}\) Plugging in the values: \(C = \frac{1}{4\pi^2(570 × 10^3 \mathrm{Hz})^2 × 0.20 × 10^{-3} \mathrm{H}}\) \(C = 4.87 × 10^{-11} \mathrm{F}\) Therefore, the capacitance in the tuning circuit is \(4.87×10^{-11} \mathrm{F}\).

(c) Finding the Maximum EMF Induced in the Antenna

To find the maximum emf induced in the coil antenna, we can use Faraday's Law of electromagnetic induction: \(\epsilon = N \frac{\Delta \phi}{\Delta t}\) We are given the amplitude of the electric field, E = \(0.80 \mathrm{V/m}\), the radius of the coil antenna, r = \(1.6 \mathrm{cm}\) and the number of turns, N = 50 turns. First, let's convert r to meters: \(r = 1.6 \mathrm{cm} = 0.016 \mathrm{m}\) Next, we can find the area of the coil: \(A = \pi r^2 \approx 8.042 × 10^{-4} \mathrm{m^2}\) Since the fields are sinusoidal, the maximum field variation is \(\Delta \phi = E * A\): \(\Delta \phi = 0.80 \mathrm{V/m} * 8.042 × 10^{-4} \mathrm{m^2} = 6.434 × 10^{-4} \mathrm{Wb}\) Now, we can use Faraday's Law to find the maximum emf induced: \(\epsilon = N \frac{\Delta \phi}{\Delta t}\) Considering the field as sinusoidal, we can find the maximum emf by differentiating it with respect to time. The frequency is given as 570 kHz. Then, we can find the angular frequency, \(\omega\): \(\omega = 2\pi f = 2\pi(570 × 10^3 \mathrm{Hz})\approx 3.585 × 10^6 \mathrm{s^{-1}}\) Now, we can plug in the values: \(\epsilon = 50 * 6.434×10^{-4}\mathrm{W b} * 3.585 × 10^6 \mathrm{s^{-1}}\) \(\epsilon = 0.115 \mathrm{V}\) So, the maximum emf induced in the antenna is approximately 0.115 V.