Question

Suppose that the electric field amplitude of an electromagnetic wave propagating along x-djrection is $E_0=120N{C}^{-1}$ and that its frequency is $\upsilon =50.0MHz$.

• A. The expression of electric field is
  $\vec{E}=120sin\left(\frac{\pi }{3}x-100\pi \times 10^{6}t\right)\hat{j}$
• B. The expression of electric field is
  $\vec{E}=60sin\left(\frac{\pi }{3}x-100\pi \times 10^{6}t\right)\hat{j}$
• C. The expression of magnetic field is
  $\vec{B}=40\times 10^{-8}sin\left(\frac{\pi }{3}x-\pi \times 10^{8}t\right)\hat{k}$
• D. Both (a) and (c)

Answer 1

Answer: (D)

Since the wave is propagating along x-direction,
$\therefore$ Electric field $\vec{E}$ is along y - direction and magnetic field $\vec{B}$ along z-direction.
$\vec{E}=E_{0}sin\left(kx-\omega t\right)\hat{j}$
$=120sin\left(\frac{\pi }{3}x-100\pi \times 10^{6}t\right)\hat{j}$
$\left(\because k=\frac{2\pi }{\lambda }=\frac{2\pi \upsilon }{c}=\frac{2\pi \times 50\times 10^6}{3\times 10^8}=\frac{\pi }{3}\right)$
$B_0=\frac{E_0}{c}=\frac{120}{3\times 10^8}=40\times 10^{-8}T$
$\vec{B}=B_{0}sin\left(kx-\omega t\right)\hat{k}$
$=40\times 10^{-8}sin\left(\frac{\pi }{3}x-\pi \times 10^{8}t\right)\hat{k}$