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QJEEM24Phc227: JEE Main 2024 Physics - Class 12 JEEM24Phc227 Solution: If frequency of electromagnetic wave is and it travels in air…

If frequency of electromagnetic wave is 60 MHz60 \mathrm{~MHz} and it travels in air along zz direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave (in m\mathrm{m} ) is :

Multiple Choice Options

  • A 2.5
  • B 5
  • C 10
  • D 2

Step-by-Step Verified Solution

The speed of electromagnetic waves in air (and in a vacuum) is approximately the speed of light, which we denote as cc . The speed of light cc is 3×1083 \times 10^8 meters per second. The relationship between the speed of light cc , the frequency ff , and the wavelength λ\lambda of an electromagnetic wave is given by the equation:

c=fλc = f \lambda

Here, we are given that the frequency ff of the electromagnetic wave is 60 MHz (megahertz), which can be converted to hertz (Hz) by multiplying by 10610^6 (because 1 MHz = 10610^6 Hz).

f=60×106 Hzf = 60 \times 10^6 \text{ Hz}

To find the wavelength λ\lambda in meters, we can rearrange the wave equation to solve for λ\lambda :

λ=cf\lambda = \frac{c}{f}

Substitute the given values of cc and ff into the equation to find the wavelength:

λ=3×108 m/s60×106 Hz\lambda = \frac{3 \times 10^8 \text{ m/s}}{60 \times 10^6 \text{ Hz}}

Now, simplify the equation by dividing the numbers:

λ=360×108−6 m\lambda = \frac{3}{60} \times 10^{8-6} \text{ m}

λ=120×102 m\lambda = \frac{1}{20} \times 10^2 \text{ m}

λ=5 m\lambda = 5 \text{ m}

Therefore, the wavelength of the wave is 5 meters, corresponding to option B. Additionally, in an electromagnetic wave, the electric and magnetic field vectors are indeed mutually perpendicular to each other and to the direction of propagation, which in this case is the zz direction.

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