Wave Equation Parameters
The electric field of an electromagnetic wave is described by E = 60\sin(0.5 \times 10^{3}x - 1.5 \times 10^{11}t) in SI units. What is the wavelength of this wave?
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About This Question
- Subject
- physics
- Chapter
- electromagnetic waves
- Topic
- wave equation parameters
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.26 cm
A travelling wave written as E = E_0\sin(kx - \omega t) encodes the angular wave number k as the coefficient of x and the angular frequency \omega as the coefficient of t. Here k = 0.5 \times 10^{3} = 500\ \text{rad/m}. The wavelength follows from \lambda = 2\pi/k = 2\pi/500 = 1.26 \times 10^{-2}\ \text{m}, which equals 1.26 cm. The 2.50 cm option mistakenly uses \lambda = 1/k without the factor 2\pi scaled correctly. The 0.63 cm value halves the true wavelength, perhaps by confusing k with 2k. The 12.6 cm answer misplaces the decimal by a factor of ten. Reading wave parameters directly from the argument of the sine function is a routine JEE skill rooted in the NCERT treatment of progressive waves, where every travelling wave shares the same canonical form regardless of whether it is mechanical or electromagnetic. The positive x and negative t signs together tell us the wave moves in the +x direction, another piece of information hidden in the argument. As a consistency check, the wave speed is \omega/k = (1.5 \times 10^{11})/500 = 3.0 \times 10^{8}\ \text{m/s}, exactly the speed of light, confirming this is a genuine electromagnetic wave in vacuum and that the extracted wave number is correct.
This medium difficulty physics question is from the chapter electromagnetic waves, covering the topic of wave equation parameters. It appeared in the 2025 exam.
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