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Exam
NEET
Subject
Physics
Chapter
Electromagnetic Waves

NEET · Physics · Electromagnetic Waves

The magnetic field of an electromagnetic wave travelling in a...

The magnetic field of an electromagnetic wave travelling in a lossless, non-magnetic dielectric medium (dielectric constant ϵr=4\epsilon_r = 4) is given by:
Bz=2×10−8sin⁡(1.5π×108t+πx)B_z = 2 \times 10^{-8} \sin(1.5\pi \times 10^8 t + \pi x) T.

Match the wave parameters in List-I with their corresponding values in List-II.

List-IList-II
(A) Speed of the wave in the medium (in m/s)(I) 33
(B) Wavelength of the wave (in m)(II) +x+x
(C) Amplitude of the electric field (in V/m)(III) 1.5×1081.5 \times 10^8
(D) Direction of propagation(IV) 22
(V) −x-x


Choose the correct answer from the options given below:

Options

  1. A(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  2. B(A)-(III), (B)-(I), (C)-(IV), (D)-(V)
  3. C(A)-(III), (B)-(IV), (C)-(I), (D)-(V)
  4. D(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Correct answer

C. (A)-(III), (B)-(IV), (C)-(I), (D)-(V)

Step-by-step solution

The speed of the wave in the medium is v=cϵr=3×1084=1.5×108v = \dfrac{c}{\sqrt{\epsilon_r}} = \dfrac{3 \times 10^8}{\sqrt{4}} = 1.5 \times 10^8 m/s. (Matches A-III)

From the given equation, the wave vector k=πk = \pi rad/m. The wavelength is λ=2πk=2ππ=2\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{\pi} = 2 m. (Matches B-IV)

The amplitude of the electric field is E0=vB0=(1.5×108)×(2×10−8)=3E_0 = vB_0 = (1.5 \times 10^8) \times (2 \times 10^{-8}) = 3 V/m. (Matches C-I)

The phase of the wave is (1.5π×108t+πx)(1.5\pi \times 10^8 t + \pi x). The positive sign between the time and space terms indicates that the wave is propagating in the −x-x direction. (Matches D-V)

Thus, the correct matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(V).

Answer: (A)-(III), (B)-(IV), (C)-(I), (D)-(V)
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