Answer :
Answer with Explanation:
We are given that
[tex]E=-(375V/m)sin(5.97\times 10^{15}(rad/s)t+(1.99\times 1067(rad/m)x)[/tex]
a.General equation of electric field wave
[tex]E=E_0sin(\omega t+kx)[/tex]
Where [tex]E_0[/tex]=Amplitude of electric field wave
By comparing
[tex]\omega=5.97\times 10^{15}rad/s[/tex]
[tex]k=1.99\times 10^7rad/m[/tex]
a.Amplitude of electric field wave=[tex]E_0=375V/m[/tex]
b.Amplitude of magnetic field wave,[tex]B_0=\frac{E_0}{c}[/tex]
Where [tex]c=3\times 10^8 m/s[/tex]
Amplitude of magnetic field wave=[tex]B_0=\frac{375}{3\times 10^8}=125\times 10^{-8} T[/tex]
c.Frequency of wave,[tex]f=\frac{\omega}{2\pi}=\frac{5.97\times 10^{15}}{2\pi}=0.95\times 10^{15}Hz[/tex]
d.Wavelength,[tex]\lambda=\frac{2\pi}{k}[/tex]
[tex]\lambda=\frac{2\pi}{1.99\times 10^7}=3.16\times 10^{-7} m[/tex]
e.Period of wave,[tex]T=\frac{1}{f}=\frac{1}{0.95\times 10^{15}}=1.05\times 10^{-15} s[/tex]
f.Speed of wave,[tex]v=f\lambda=0.95\times 10^{15}\times 3.16\times 10^{-7}=3.00\times 10^8 m/s[/tex]