The electric field associated with an em wave in vacuum is given by E = 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

Answer: (a). From E = 40cos (kz - ω t): ω = 6× 10⁸ rad/s. In vacuum c = (ω)/(k), so k = (ω)/(c) = 6× 10⁸3× 10⁸. k = 2 m⁻¹. So the answer is (a).

The electric field associated with an em wave in vacuum is given by E… | NEET 2012 (AIPMT, Prelims)

Q83NEET 2012 · AIPMT · PRELIMSEasy

Nature of EM Waves & Wave Equation

The electric field associated with an em wave in vacuum is given by

E = $\hat{i}$ 40 cos (kz $-$ 6 $\times$ 108 t) where E, z and t are in volt/m, meter and seconds respectively. The value of wave vector k is

Show answer and solution

Answer: (a)

From $E = 40\cos(kz - \omega t)$ : $\omega = 6\times 10^{8}\ \text{rad/s}$ .
In vacuum $c = \frac{\omega}{k}$ , so $k = \frac{\omega}{c} = \frac{6\times 10^{8}}{3\times 10^{8}}$ .
$k = 2\ \text{m}^{-1}$ .
So the answer is (a).