Physics · Electromagnetic Waves

JEE Main 2024 — 9 April, Shift 1 — Question 33

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

Option A:
$B_{z}=60 \sin \left[\frac{\pi}{2}\left(x-3 \times 10^{8} t\right)\right] \hat{k} T$
Option B:
$\mathrm{B}_{\mathrm{z}}=2 \times 10^{-7} \sin \left[\frac{\pi}{2} \times 10^{3}\left(\mathrm{x}-3 \times 10^{8} \mathrm{t}\right)\right] \hat{\mathrm{k} T}$
Option C:
$\mathrm{B}_{\mathrm{x}}=60 \sin \left[\frac{\pi}{2}\left(\mathrm{x}-3 \times 10^{8} \mathrm{t}\right)\right] \hat{\mathrm{i} T}$
Option D:
$\mathrm{B}_{\mathrm{z}}=2 \times 10^{-7} \sin \left[\frac{\pi}{2}\left(\mathrm{x}-3 \times 10^{8} \mathrm{t}\right)\right] \hat{\mathrm{k} T}$
Correct

Answer: D

Step-by-step solution

$\mathrm{E}=\mathrm{BC} \Rightarrow 60=\mathrm{B} \times 3 \times 10^{8}$

$\Rightarrow \mathrm{B}=2 \times 10^{-7}$

Also $\mathrm{C}=\mathrm{f} \lambda$

$\Rightarrow 3 \times 10^{8}=\mathrm{f} \times 4 \times 10^{-3}$

$\Rightarrow \mathrm{f}=\frac{3}{4} \times 10^{11}$

$\Rightarrow \omega=2 \pi \mathrm{f}=\frac{3}{4} \times 2 \pi \times 10^{11}$

$\Rightarrow \omega=\frac{\pi}{2} \times 10^{3} \mathrm{C}$

$\Rightarrow \quad$

Electric field $\Rightarrow \mathrm{y}$ direction

Propagation $\Rightarrow \mathrm{x}$ direction

Magnetic field $\Rightarrow$ z-direction

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Physics
Chapter: Electromagnetic Waves
Topic: Displacement Current and Equation of EM Waves