An electromagnetic wave with frequency 65.0 Hz travels in aninsulating magnetic material that has dielectric constant 3.64 andrelative permeability 5.18 at this frequency. The electric fieldhas amplitude 7.20 \times 10^{ - 3} {\rm V}/{\rm 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?d.what is the intensity of the wave?

Respuesta :

To solve this problem, it is necessary to apply the concepts related to the wavelength, frequency, speed and intensity of an electromagnetic wave.

The expression for the speed of the electromagnetic wave is described as

[tex]\upsilon = \frac{1}{\mu_T \mu_0 (k\epsilon_0)}[/tex]

Where,

[tex]\mu_T[/tex]= Magnetic permeability

[tex]\mu_0[/tex]= Permeability of free space

[tex]\epsilon_0[/tex]= Permittivity of free space

The expression for the wavelength is

[tex]\lambda = \frac{v}{f}[/tex]

Where,

v = Velocity of wave

f = Frequency

The expression for the amplitude of the magnetic field is

[tex]B_{max} = \frac{E_{max}}{v}[/tex]

Where,

v = Speed

[tex]E_{max}[/tex] = Amplitude of the maximum electric field

The expression for the intensity of the wave is as follows:

[tex]I=\frac{1}{2} \epsilon v(E_{max})^2[/tex]

PART A ) Our values are given as

[tex]\mu_T = 5.18[/tex]

[tex]\mu_0 = 4\pi*10^{-7}H/m[/tex]

[tex]k = 3.64[/tex]

[tex]\epsilon_0 = 8.85*10^{-12}F/m[/tex]

Replacing at our first equation we have,

[tex]\upsilon = \frac{1}{\mu_T \mu_0 (k\epsilon_0)}[/tex]

[tex]\upsilon = \frac{1}{(5.18)(4\pi*10^{-7})(3.64)(8.85*10^{-12})}[/tex]

[tex]\upsilon = 4.768*10^{15}m/s[/tex]

Therefore the magnitude of the speed of the electromagnetic wave is equal to [tex]4.768*10^{15}m/s[/tex]

PART B) The value of our frequency is 65Hz and the velocity was previously found, then

[tex]\lambda = \frac{\upsilon}{f}[/tex]

[tex]\lambda = \frac{(4.768*10^{15})}{65}[/tex]

[tex]\lambda = 7.336*10^{13}m[/tex]

PART C) For the amplitude of the maximum magnetic field we have that

[tex]E_{max} = 7.2*10^{-3}V/m[/tex]

[tex]\upsilon = 4.768*10^{15}m/s[/tex]

Replacing we have that

[tex]B_{max} = \frac{7.2*10^{-3}}{4.768*10^{15}}[/tex]

[tex]B_{max} = 1.51*10^{-18}T[/tex]

Therefore the amplitude of the maximum magnetic field is equal to [tex]1.04*10^{-18}T[/tex]

PART D) Finally the intensity is given as

[tex]I = \frac{1}{2}k\epsilon_0 \upsilon(E_{max})^2[/tex]

[tex]I = \frac{1}{2}(3.64)(8.85*10^{-12})(6.9*10^7)(7.2*10^{-3})^2[/tex]

[tex]I = 5.76*10^{-8}W/m^2[/tex]

Therefore the intensity of the wave is equal [tex]5.76*10^{-8}W/m^2[/tex]

Answer:

a) the speed of propagation of the wave is 6.909x10⁷m/s

b) the wavelength of the wave is 1.063x10⁶m

c) the amplitude of the magnetic field is 1.042x10⁻¹⁰T

d) the intensity of the wave is 5.769x10⁻⁸W/m²

Explanation:

the solution is in the attached Word file

Ver imagen lcoley8