Question

A magnetic dipole antenna is used to detect an electromagnetic wave. The antenna is a coil of 50 turns with radius \(5.0 \mathrm{cm} .\) The EM wave has frequency \(870 \mathrm{kHz}\) electric field amplitude $0.50 \mathrm{V} / \mathrm{m},\( and magnetic field amplitude \)1.7 \times 10^{-9} \mathrm{T} .$ (a) For best results, should the axis of the coil be aligned with the electric field of the wave, or with the magnetic field, or with the direction of propagation of the wave? (b) Assuming it is aligned correctly, what is the amplitude of the induced emf in the coil? (Since the wavelength of this wave is much larger than \(5.0 \mathrm{cm},\) it can be assumed that at any instant the fields are uniform within the coil.) (c) What is the amplitude of the emf induced in an electric dipole antenna of length \(5.0 \mathrm{cm}\) aligned with the electric field of the wave?

Short answer

Answer: The amplitude of the induced emf in the coil is approximately 3.66 μV.

Step-by-step solution

Determine the best alignment of the coil axis

To obtain the best results, the coil should be aligned so that its axis is parallel with the magnetic field of the wave. This will cause the greatest amount of change in magnetic flux through the coil and, thus, the largest induced emf.

Calculate the induced emf in the coil

We will now calculate the amplitude of the induced emf when the coil is aligned correctly using Faraday's Law. First, we need to find the magnetic flux through the coil. The magnetic flux (\(\Phi_{B}\)) is given by the formula: \(\Phi_{B} = B \times A \times \cos{\theta}\) where B is the magnetic field amplitude, A is the area of the coil and θ is the angle between B and the normal vector of A. Since the coil is aligned to its axis, \(\theta = 0\). The area of the coil can be calculated using the formula: \(A = \pi r^{2}\) So, \(A = \pi (0.05 \thinspace m)^{2} \approx 7.854 \thinspace m^{2}\) Now we can calculate the magnetic flux through the coil: \(\Phi_{B} = (1.7\times10^{-9} \thinspace \mathrm{T}) \times (7.854 \thinspace m^{2}) \times \cos{0} = 1.3378\times10^{-11} \thinspace \mathrm{Wb}\) The maximum amplitude of the EMF induced in the coil can be calculated using Faraday's Law: \(\epsilon_{max} = -N \times \frac{d\Phi_{B}}{dt}\) As we were given the frequency of the wave, we can calculate the angular frequency (\(\omega\)) as follows: \(\omega = 2 \pi f\) \(\omega = 2 \pi (870 \times 10^{3} \thinspace \mathrm{Hz}) \approx 5.469 \times 10^{6} \thinspace \mathrm{rad/s}\) Now, the maximum rate of change of the magnetic flux (\(\frac{d\Phi_{B}}{dt}\)) can be determined: \(\frac{d\Phi_{B}}{dt} = \omega \times \Phi_{B}\) \(\frac{d\Phi_{B}}{dt} = (5.469 \times 10^{6} \thinspace \mathrm{rad/s}) \times (1.3378\times10^{-11} \thinspace \mathrm{Wb}) \approx 7.320\times10^{-5} \thinspace \mathrm{Wb/s}\) Finally, we can calculate the induced emf: \(\epsilon_{max} = |N \times \frac{d\Phi_{B}}{dt}|\) \(\epsilon_{max} = |50 \times 7.320\times10^{-5} \thinspace \mathrm{Wb/s}| \approx 3.66 \thinspace \mathrm{\mu V}\)

Calculate the emf induced in the electric dipole antenna

Now, we will calculate the amplitude of the emf induced in an electric dipole antenna of length 5.0 cm aligned with the electric field of the wave using the formula: Emf = Electric Field \(\times\) Antenna Length Emf = \(0.50 \thinspace \mathrm{V/m} \times 0.05 \thinspace \mathrm{m} \approx 0.025 \thinspace \mathrm{V}\) So, the amplitude of the emf induced in an electric dipole antenna of length 5.0 cm aligned with the electric field of the wave is approximately 0.025 V.