Question

An EM wave propagating in $x$-direction has a wavelength of $8mm$. The electric field vibrating $y-$ direction has maximum magnitude of $60V{m}^{-1}$. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum :

• A. $E_y=60\sin \left[\frac{\pi }{4}\times 10^3\left(x-3\times 10^{8}t\right)\right]\hat{j}V{m}^{-1}$
  $B_z=2\sin \left[\frac{\pi }{4}\times 10^3\left(x-3\times 10^{8}t\right)\right]\hat{k}T$
• B. $E_y=60\sin \left[\frac{\pi }{4}\times 10^3\left(x-3\times 10^{8}t\right)\right]\hat{j}V{m}^{-1}$
  $B_z=2\times 10^{-7}\sin \left[\frac{\pi }{4}\times 10^3\left(x-3\times 10^{8}t\right)\right]\hat{k}T$
• C. $E_y=2\times 10^{-7}\sin \left[\frac{\pi }{4}\times 10^3\left(x-3\times 10^{8}t\right)\right]\hat{j}V{m}^{-1}$
  $B_z=60\sin \left[\frac{\pi }{4}\times 10^3\left(x-3\times 10^{8}t\right)\right]\hat{k}T$
• D. $E_y=2\times 10^{-7}\sin \left[\frac{\pi }{4}\times 10^4\left(x-4\times 10^{8}t\right)\right]\hat{j}V{m}^{-1}$
  $B_z=60\sin \left[\frac{\pi }{4}\times 10^4\left(x-4\times 10^{8}t\right)\right]\hat{k}T$

Answer 1

Answer: (B)

$B_0=\frac{E_0}{c}=\frac{60}{3\times 10^8}=2\times 10^{-7}T$
$E\times B$ must be direction of propagation.
So, B $\to$ z-axis
$k=\frac{2\pi }{\lambda }=\frac{\pi }{4}\times 10^3{m}^{-1}$
$E_y=60\sin \left[\frac{\pi }{4}\times 10^3\left(x-3\times 10^{8}t\right)\right]\hat{j}V{m}^{-1}$
$B_z=2\times 10^{-7}\sin \left[\frac{\pi }{4}\times 10^3\left(x-3\times 10^{8}t\right)\right]kT$