For E0 = 120 N/C, v = 50 MHz = 50 × 106 Hz
i. λ = \(\frac{c}{v}=\frac{3\times10^8}{50\times10^6}\) = 6 m
B0 = \(\frac{E_0}{v}=\frac{120}{3\times10^8}\) = 4 × 10-7 T = 400 nT
k = \(\frac{2\pi}{λ}=\frac{2\pi}{6}\) = 1.0472 rad/m
ω = 2πv = 2π × 50 × 106
= 3.14 × 108 rad/s.
ii. Assuming motion of em wave along X-axis, expression for electric field vector may lie along Y-axis,
∴ \(\overset\rightarrow{E}\) = E0 sin (kx – ωt)
= 120 sin (1.0472 × – 3.14 × 108 t) \(\hat{j}\) N/C.
Also, magnetic field vector will lie along Z-axis, expression for magnetic field vector,
∴ \(\overset\rightarrow{E}\) = B0 sin (kx – ωt)
= 4 × 10-7 sin (1.0472 × – 3.14 × 108 t) \(\hat{k}\) T.