99 Percentile Qs Bank for JEE MainPhysicsElectromagnetic Waves
The electric field for an electromangetic wave in free space is E = i 30 (k z-5 10^8 t ) , where magnitude of E is in V / m . The magnitude of wave vector, k is (velocity of em wave in free space =3 10^8 ~m / s )
Options
- A0.46 rad m ⁻¹
- B3 rad m ⁻¹
- C1.66 rad m ⁻¹
- D0.83 rad m m ⁻¹
Correct answer
C. 1.66 rad m ⁻¹
Step-by-step solution
The given E = i 30 (k z-5 10^8 t ) We know that, E = iV (k z- t) Comparing the both equations, =5 10^8 rad / s But we know that also K= 2 and =2 where, is wavelength and v is frequency of the wave array ll & k = 2 v 2 / =v =C & C= k or & k= C = 5 10^8 3 10^8 & k= 5 3 =1.66 rad / m array