Respuesta :
Answer:
The resulting magnetic field component of the wave
[tex]B_{max}=2.166\times10^{-12}\ T[/tex]
[tex]k =25645.65\ m^{-1}[/tex]
[tex]\omega =4.274\times10^{12}[/tex]
Explanation:
Given that,
Wavelength = 245 μm
Electric field [tex]E= 6.50\times10^{-3}\ V/m[/tex]
We need to calculate the wave number and angular frequency
Using formula of wave number
[tex]k=\dfrac{2\pi}{\lambda}[/tex]
[tex]k=\dfrac{2\pi}{245\times10^{-6}}[/tex]
[tex]k=25645.65\ m^{-1}[/tex]
The angular frequency is
[tex]\omega_{0}=\dfrac{2\pi}{T}[/tex]
[tex]\omega_{0}=\dfrac{2\pi}{\dfrac{\lambda}{c}}[/tex]
[tex]\omega_{0}=25645.65\times3\times10^{8}[/tex]
[tex]\omega_{0}=7.693695\times10^{12}\ rad/sec[/tex]
If the wave period increases by a factor of 1.80 times
[tex]\omega=\dfrac{2\pi}{1.80T}[/tex]
[tex]\omega=\dfrac{\omega_{0}}{1.80}[/tex]
[tex]\omega=\dfrac{7.693695\times10^{12}}{1.80}[/tex]
[tex]\omega=4.274\times10^{12}\ rad/sec[/tex]
The maximum [tex]\vec{E}[/tex] and [tex]\vec{B}[/tex] related by the equation
[tex]B_{max}=\dfrac{E_{max}}{c}[/tex]
Put the value in the equation
[tex]B_{max}=\dfrac{6.50\times10^{-3}}{3\times10^{8}}[/tex]
[tex]B_{max}=2.166\times10^{-12}\ T[/tex]
The magnetic field is perpendicular to the electric field.
So, the equation is
[tex]B=[2.166\times10^{-12}]exp(25645.65\ x-4.274\times10^{12}\ t). \hat{Z}[/tex]
Hence, The resulting magnetic field component of the wave
[tex]B_{max}=2.166\times10^{-12}\ T[/tex]
[tex]k =25645.65\ m^{-1}[/tex]
[tex]\omega =4.274\times10^{12}[/tex]
