Q64 / 90 · JEE Main 2024 (09 Apr Shift 1) · Physics

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 :

  • Correct answer
    A
    $\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2} \times 10^3\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}$
  • B
    $\mathrm{B}_z=60 \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}$
  • C
    $\mathrm{B}_x=60 \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{i} ~ T}$
  • D
    $\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}$

Solution

$\mathrm{E}=\mathrm{BC} \Rightarrow 60=\mathrm{B} \times 3 \times 10^8$ $\Rightarrow \mathrm{B}=2 \times 10^{-7}$ Also $\quad \mathrm{C}=\mathrm{f} \lambda$ $\begin{aligned} & \Rightarrow 3 \times 10^8=\mathrm{f} \times 4 \times 10^{-3} \\ & \Rightarrow \mathrm{f}=\frac{3}{4} \times 10^{11} \\ & \Rightarrow \omega=2 \pi \mathrm{f}=\frac{3}{4} \times 2 \pi \times 10^{11} \\ & \Rightarrow \omega=\frac{\pi}{2} \times 10^3 \mathrm{C}\end{aligned}$ $\Rightarrow \quad$ Electric field $\Rightarrow y$ direction Propagation $\Rightarrow \mathrm{x}$ direction Magnetic field $\Rightarrow \mathrm{z}$-direction

Hence, the correct answer is option: 'A'.

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