Match List-I with List-II (For the parameters (in SI units) of an electromagnetic wave whose magnetic field is \( B_y = 20\sin(-500\,x + 20000\,t)\,\mu T \) )
| List-I | List-II |
| Parameter | Value |
| (A) Wavelength | (I) 3.18 × 103 |
| (B) Frequency | (II) 0.0126 |
| (C) Speed of propagation | (III) 6 × 103 |
| (D) Amplitude of electric field | (IV) 40 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
Compare with \( B_y = B_0\sin(\omega t - kx) \): the angular wave number k = 500 rad m−1, the angular frequency ω = 20000 rad s−1 and \( B_0 = 20\,\mu\mathrm{T} \).
(A) Wavelength — this matches (II) 0.0126.
\( \displaystyle \lambda = \frac{2\pi}{k} = \frac{2\pi}{500\,\mathrm{rad\,m^{-1}}} \approx 0.0126\,\mathrm{m} \)
(B) Frequency — this matches (I) 3.18 × 103.
\( \displaystyle \nu = \frac{\omega}{2\pi} = \frac{20000\,\mathrm{rad\,s^{-1}}}{2\pi} \approx 3.18\times10^{3}\,\mathrm{Hz} \)
(C) Speed of propagation — this matches (IV) 40.
\( \displaystyle v = \frac{\omega}{k} = \frac{20000}{500}\,\mathrm{m\,s^{-1}} = 40\,\mathrm{m\,s^{-1}} \)
(D) Amplitude of electric field — this matches (III) 6 × 103, using \( E_0 = cB_0 \).
\( \displaystyle E_0 = cB_0 = 3\times10^{8}\,\mathrm{m\,s^{-1}} \times 20\times10^{-6}\,\mathrm{T} = 6\times10^{3}\,\mathrm{V\,m^{-1}} \)
Note: It seems that there is a mistake in the original NTA paper, because the data of the wave do not agree with each other: the equation gives a speed of 40 m s−1, while 6 × 103 needs \( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \). The question can still be solved, because (A)–(II) and (B)–(I) follow directly from k and ω, and Option (2) is the only option containing both.
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Nature of Electromagnetic Waves – …
