Question

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

• A. The expression of electric field is
  $\vec{E} = 120\,sin\left(\frac{\pi}{3}x - 100\pi\times10^{6} t\right)\hat{j}$
• B. The expression of electric field is
  $\vec{E} = 60\,sin\left(\frac{\pi}{3}x - 100\pi\times10^{6} t\right)\hat{j}$
• C. The expression of magnetic field is
  $\vec{B} = 40 \times 10^{-8}\,sin\left(\frac{\pi}{3}x - \pi\times10^{8} t\right)\hat{k}$
• D. Both (a) and (c)

Answer 1

Answer: (D)

Since the wave is propagating along x-direction,
$\therefore \quad$ Electric field $\vec{E}$ is along y - direction and magnetic field $\vec{B}$ along z-direction.
$\vec{E} = E_{0}\,sin\left(kx - \omega t\right) \hat{j}$
$= 120\,sin\left(\frac{\pi}{3}x - 100\pi\times10^{6} t\right)\hat{j}$
$\left(\because k = \frac{2\pi}{\lambda} = \frac{2\pi\upsilon}{c} = \frac{2\pi \times50\times 10^{6}}{3 \times 10^{8}} = \frac{\pi}{3}\right)$
$B_{0} = \frac{E_{0}}{c} = \frac{120}{3 \times 10^{8}} = 40 \times 10^{-8}\,T$
$\vec{B} = B_{0}\,sin\left(kx - \omega t\right) \hat{k}$
$= 40 \times 10^{-8}\,sin\left(\frac{\pi }{3}x - \pi \times 10^{8}t\right)\hat{k}$