QJEEM24Ph5fbe: JEE Main 2024 Physics - Class 12 JEEM24Ph5fbe Solution: A plane EM wave is propagating along direction. It has a…

To find the correct equation for the magnetic field of the plane electromagnetic wave given its parameters, we can use a couple of known relationships from electromagnetism.

Firstly, the wavelength ( $\lambda$ ) of the wave is given as $4 \mathrm{~mm} = 4 \times 10^{-3} \mathrm{~m}$ . The speed of light (and all electromagnetic waves in vacuum) is $c = 3 \times 10^8 \mathrm{~m/s}$ . Using these values, we can find the frequency ( $f$ ) of the wave using the relationship:

$c = \lambda f$

$f = \frac{c}{\lambda} = \frac{3 \times 10^8}{4 \times 10^{-3}} = 75 \times 10^9 \mathrm{~Hz}$

The angular frequency ( $\omega$ ) which appears in wave equations is related to the frequency by $\omega = 2\pi f$ . However, in this context, what we need is the wave vector ( $k$ ), which defines how the phase of the wave changes with space. The wave vector $k = \frac{2\pi}{\lambda}$ . So, for this wave, $k = \frac{2\pi}{4 \times 10^{-3}} = \frac{\pi}{2} \times 10^3 \, \mathrm{m}^{-1}$ .

Knowing that the electric field ( $\mathbf{E}$ ) and magnetic field ( $\mathbf{B}$ ) are related as $E = cB$ in a vacuum, where $E$ and $B$ are the magnitudes of the electric and magnetic fields, respectively, we can calculate the magnitude of the magnetic field using the provided maximum electric field magnitude ( $E = 60 \, \mathrm{Vm}^{-1}$ ).

$B = \frac{E}{c} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \mathrm{~T}$

Therefore, the wave equation for the magnetic field, considering it propagates in the $x$ -direction and oscillates in a direction perpendicular to both the $x$ -direction and the direction of the electric field (thus, in the $z$ -direction if $\mathbf{E}$ is in the $y$ -direction), is:

$\mathbf{B} = B \sin(kx - \omega t) \hat{\mathbf{k}}$

$\mathbf{B} = 2 \times 10^{-7} \sin \left(\frac{\pi}{2} \times 10^3(x - 3 \times 10^8t)\right) \hat{\mathbf{k}} \mathrm{T}$

This is represented by Option B:

$\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2} \times 10^3\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}$

Therefore, the correct answer is Option B.