A plane EM wave is propagating along $x$ direction. It has a wavelength of $4 \mathrm{~mm}$. If electric field is in $y$ direction with the maximum magnitude of $60 \mathrm{~Vm}^{-1}$, the equation for magnetic field is :

<p>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.</p> <p>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 $) of the wave using the relationship:</p> <p>$ c = \lambda f $</p> <p>$ f = \frac{c}{\lambda} = \frac{3 \times 10^8}{4 \times 10^{-3}} = 75 \times 10^9 \mathrm{~Hz} $</p> <p>The angular frequency ($ \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} $.</p> <p>Knowing that the electric field ($ \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} $).</p> <p>$ B = \frac{E}{c} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \mathrm{~T} $</p> <p>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 $-direction), is:</p> <p>$ \mathbf{B} = B \sin(kx - \omega t) \hat{\mathbf{k}} $</p> <p>$ \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} $</p> <p>This is represented by Option B:</p> <p>$ \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} $</p> <p>Therefore, the correct answer is Option B.</p>