Suppose that the electric field amplitude of an electromagnetic wave i

1.	Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is v = 50.0 MHz.(i) Determine, B0, ω, k, and λ.(ii) Find expressions for vectors E and B.
Answer» 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.

Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is v = 50.0 MHz.(i) Determine, B0, ω, k, and λ.(ii) Find expressions for vectors E and B.

Answer»

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.