Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Answer:
The wavelength of the wave is [tex]1.06\times10^6 m[/tex]
Explanation:
Lets calculate
We know an electromagnetic wave is propagating through an insulating magnetic material of dielectric constant K and relative permeability [tex]K_m[/tex] ,then the speed of the wave in this dielectric medium is [tex]\nu[/tex] is less than the speed of the light c and is given by a relation
[tex]\nu=\frac{c}{\sqrt{KK_m} }[/tex] --------- 1
In case the electromagnetic wave propagating through the insulating magnetic material , the amplitudes of electric and magnetic fields are related as -
[tex]E_m_a_x= \nu B_m_a_x[/tex]
The magnitude of the 'time averaged value' of the pointing vector is called the intensity of the wave and is given by a relation
[tex]I = S_a_v[/tex]
[tex]\frac{E_m_a_xB_m_a_x}{2K_m\mu0}[/tex]----------- 3
now , we will find the speed of the propagation of an electromagnetic wave by using equation 1
[tex]\nu=\frac{c}{\sqrt{KK_m} }[/tex]
Putting the values ,
=[tex]\nu= \frac{3.00\times10^8}{\sqrt{(3.64)(5.18)} }[/tex]
=[tex]0.6908\times10^8m/s[/tex]
= [tex]6.91\times10^7m/s[/tex]
Now , using this above solution , we will find the wavelength of the wave -
[tex]\lambda=\frac{\nu}{f}[/tex]
Putting the values from above equations -
[tex]\frac{6.91\times10^7m/s}{65.0Hz}[/tex]
[tex]\lambda= 1.06\times10^6 m[/tex]
Hence , the answer is [tex]\lambda= 1.06\times10^6 m[/tex]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
