| ⇦ | 
| ⇨ |
The electric field associated with an e.m. wave in vacuum is given by E = i 40 cos (kz – 6 x 10⁸ t), where E, z and t are in volt/m, meter and seconds respectively. The value of wave vector k is:
Options
(a) 2 m⁻¹
(b) 0.5 m⁻¹
(c) 6 m⁻¹
(d) 3 m⁻¹
Correct Answer:
2 m⁻¹
Explanation:
On comparing the given equation to
E = a₀i cos (ωt – kz) ω = 6 x 10⁸ ,
k = 2π / r = ω / c
k = ω / c = 6 x 10⁸ / 3 x 10⁸ = 2 m⁻¹
Related Questions: - The potential difference applied to an X-ray tube is 5 kV and the current through it
- A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily
- A radioactive substance decays to 1/16th of its initial activity in 40 days.
- A body has 80 microcoulomb of charge. Number of additional electrons in it will be
- Rutherford’s α-scattering experiment concludes that
Topics: Electromagnetic Waves
(28)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The potential difference applied to an X-ray tube is 5 kV and the current through it
- A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily
- A radioactive substance decays to 1/16th of its initial activity in 40 days.
- A body has 80 microcoulomb of charge. Number of additional electrons in it will be
- Rutherford’s α-scattering experiment concludes that
Topics: Electromagnetic Waves (28)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply