JEE MAIN - Physics (2024 - 1st February Evening Shift - No. 15)

If frequency of electromagnetic wave is $60 \mathrm{~MHz}$ and it travels in air along $z$ direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave (in $\mathrm{m}$ ) is :
2.5
5
10
2

Explanation

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

$$ c = f \lambda $$

Here, we are given that the frequency $$ f $$ of the electromagnetic wave is 60 MHz (megahertz), which can be converted to hertz (Hz) by multiplying by $$ 10^6 $$ (because 1 MHz = $$ 10^6 $$ Hz).

$$ f = 60 \times 10^6 \text{ Hz} $$

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

$$ \lambda = \frac{c}{f} $$

Substitute the given values of $$ c $$ and $$ f $$ into the equation to find the wavelength:

$$ \lambda = \frac{3 \times 10^8 \text{ m/s}}{60 \times 10^6 \text{ Hz}} $$

Now, simplify the equation by dividing the numbers:

$$ \lambda = \frac{3}{60} \times 10^{8-6} \text{ m} $$

$$ \lambda = \frac{1}{20} \times 10^2 \text{ m} $$

$$ \lambda = 5 \text{ m} $$

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 $$ z $$ direction.

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