The magnetic field in a plane electromagnetic wave is B y = (3.5 × 10 -7 ) sin (1.5 × 10 3 x +0.5 . .× 10 11 t ) T . The corresponding electric field will be

E z =105 sin (1.5 × 10 3 x +0.5 × 10 11 t ) Vm -1

E y =1.17 sin (1.5 × 10 3 x +0.5 × 10 11 t ) Vm -1

E z =1.17 sin (1.5 × 10 3 x +0.5 × 10 11 t ) Vm -1

E y =10.5 sin (1.5 × 10 3 x +0.5 × 10 11 t ) Vm -1

Physics · Electromagnetic Waves

JEE Main 2024 — 9 April, Shift 2 — Question 34

The magnetic field in a plane electromagnetic wave is $\mathrm{B}_{\mathrm{y}}=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^{3} \mathrm{x}+0.5\right.$ $\left.\times 10^{11} \mathrm{t}\right) \mathrm{T}$ .

The corresponding electric field will be

Option A:
$\mathrm{E}_{\mathrm{y}}=1.17 \sin \left(1.5 \times 10^{3} \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^{-1}$
Option B:
$\mathrm{E}_{\mathrm{z}}=105 \sin \left(1.5 \times 10^{3} \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^{-1}$
Correct
Option C:
$\mathrm{E}_{\mathrm{z}}=1.17 \sin \left(1.5 \times 10^{3} \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^{-1}$
Option D:
$\mathrm{E}_{\mathrm{y}}=10.5 \sin \left(1.5 \times 10^{3} \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^{-1}$

Answer: B

Step-by-step solution

${E}_{0}=\mathrm{B}_{0} \mathrm{C}$

$E_{0}=3 \times 10^{8} \times\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^{3} \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right)$

$\mathrm{E}_{0}=105 \sin \left(1.5 \times 10^{3} \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^{-1}$

Data inconsistent while calculating speed of wave. You can challenge for data.

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Electromagnetic Waves
Topic: Displacement Current and Equation of EM Waves