The electric field associated with an e. m. wave in vacuum is given by vec E = hat i 40 cos ( kz -6 × 108 t ), where E , z and t are in volt / m, meter and seconds respectively. The value of wave vector k is :

Question

The electric field associated with an e. m. wave in vacuum is given by vec E = hat i 40 cos ( kz -6 × 108 t ), where E , z and t are in volt / m, meter and seconds respectively. The value of wave vector k is : (A) 2 m-1 (B) 0.5 m-1 (C) 6 m-1 (D) 3 m-1

Tardigrade Admin · Accepted Answer

Compare the given equation with E=E0 cos (k z-ω t) we get, ω=6 × 10x s -1 Wave vector, k=(ω/c)=(6 × 108 s -1/3 × 108 m s -1)=2 m -1

Q. The electric field associated with an e. $m$ . wave in vacuum is given by $E = \hat{i} 40 cos (k z - 6 \times 1 0^{8} t)$ , where $E, z$ and $t$ are in volt $/ m$ , meter and seconds respectively. The value of wave vector $k$ is :

$2 m^{- 1}$

$0.5 m^{- 1}$

$6 m^{- 1}$

$3 m^{- 1}$

Compare the given equation with
$E = E_{0} cos (k z - ω t)$
we get, $ω = 6 \times 1 0^{x} s^{- 1}$
Wave vector, $k = \frac{ω}{c} = \frac{6 \times 1 0 ^{8} s ^{- 1}}{3 \times 1 0 ^{8} m s ^{- 1}} = 2 m^{- 1}$

Q. The electric field associated with an e. m. wave in vacuum is given by E=i^40cos(kz−6×108t), where E,z and t are in volt /m, meter and seconds respectively. The value of wave vector k is :

2m−1

0.5m−1

6m−1

3m−1