Question

An electromagnetic wave with frequency 65.0 Hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency. The electric field has amplitude $\mathbf{7}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{V}\mathbf{/}\mathbf{m}$. (a) What is the speed of propagation of the wave? (b) What is the wavelength of the wave? (c) What is the amplitude of the magnetic field?

Short answer

1. The speed of propagation of the wave is $6.91\times {10}^7\mathrm{m}/\mathrm{s}$.
2. The wavelength of the wave is $1.06\times {10}^6\mathrm{m}$.
3. The amplitude of the magnetic field is$1.04\times {10}^{-10}\mathrm{T}$ .

Step-by-step solution

Step 1: Define frequency and write formulas.

The number of waves that moves a fixed point in unit time is known as frequency.

$f=\frac{1}{T}or\frac{\omega }{2\pi }$

Where,T =Time period in seconds and $\omega$is angular frequency in .

The frequency, wavelength, and speed of light of any wave are related by the wavelength-frequency relationship.

The wavelength-frequency relationship is:

$c=\lambda f$

Where, is the speed of light which is equal to $3\times {10}^8\mathrm{m}/\mathrm{s}$,$\lambda$is the wavelength and f is the frequency.

The wavelength is inversely proportional to the frequency.

$\mathrm{\lambda }=\frac{\mathrm{c}}{\mathrm{f}}\mathrm{or}\frac{\mathrm{v}}{\mathrm{f}}$

Where, v is velocity of wave in m/s .

The relation between wave number and wavelength is:

$k=\frac{2\pi }{\lambda }$

Where, k is wavenumber in rad/m.

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

$$\begin{gathered} E_{max}=\sqrt{\frac{2l}{{\epsilon }_{0}c}} \\ {B}_{max}=\frac{E_{max}}{c} \end{gathered}$$

Determine the speed of propagation of wave.

Given that,

$$\begin{gathered} K=3.64 \\ {K}_m=5.18 \end{gathered}$$

The speed of the propagation is:

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{c}}{\sqrt{{\mathrm{KK}}_{\mathrm{m}}}} \\ =\frac{3\times {10}^8}{\sqrt{\left(3.64\right)\left(5.18\right)}} \\ =6.91\times {10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, speed of propagation of the wave is $6.91\times {10}^7\mathrm{m}/\mathrm{s}$.

Determine the wavelength of wave.

Given that, $f=$Hz

The wavelength of wave is:

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{v}}{\mathrm{f}} \\ =\frac{6.91\times {10}^7}{65} \\ =1.06\times {10}^6\mathrm{m} \end{gathered}$$

Hence the wavelength of the wave is$1.06\times {10}^6\mathrm{m}$ .

Determine the amplitudes of magnetic fields.

Given that, ${\mathrm{E}}_{\max }=7.2\times {10}^{-3}\mathrm{V}/\mathrm{m}$

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

$B_{max}=\frac{{\mathrm{E}}_{\max }}{c}$

Substitute the values of variables

$$\begin{gathered} {\mathrm{B}}_{\max }=\frac{7.2\times {10}^{-3}}{6.91\times {10}^3} \\ =1.04\times {10}^{-10}\mathrm{T} \end{gathered}$$

Hence, the amplitude of the magnetic fields of the wave$1.04\times {10}^{-10}\mathrm{T}$ is respectively.