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NEET 2022
Exam
NEET
Year
2022
Subject
Physics
Chapter
Electromagnetic Waves

NEET 2022 · Physics · Electromagnetic Waves

The magnetic field of a plane electromagnetic wave is given by B =3...

The magnetic field of a plane electromagnetic wave is given by B→=3×10−8cos⁡(1.6×103x+\overrightarrow{\mathrm{B}}=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x+\right. 48×101048 \times 10^{10} t) j^\hat{j}, then the associated electric field will be:

Options

  1. A3×10−8cos⁡(1.6×103x+48×1010t)i^/m3 \times 10^{-8} \cos \left(1.6 \times 10^3 x+48 \times 10^{10} t\right) \hat{i} / \mathrm{m}
  2. B3×10−8sin⁡(1.6×103x+48×1010t)i^/m3 \times 10^{-8} \sin \left(1.6 \times 10^3 x+48 \times 10^{10} t\right) \hat{i} / \mathrm{m}
  3. C9sin⁡(1.6×103x−48×1010t)k^ V/m9 \sin \left(1.6 \times 10^3 x-48 \times 10^{10} t\right) \hat{k} \mathrm{~V} / \mathrm{m}
  4. D9cos⁡(1.6×103x+48×1010t)k^ V/m9 \cos \left(1.6 \times 10^3 x+48 \times 10^{10} t\right) \hat{k} \mathrm{~V} / \mathrm{m}

Correct answer

D. 9cos⁡(1.6×103x+48×1010t)k^ V/m9 \cos \left(1.6 \times 10^3 x+48 \times 10^{10} t\right) \hat{k} \mathrm{~V} / \mathrm{m}

Step-by-step solution

Given, B=3×10−8cos⁡(1.6×103x\mathrm{B}=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x\right. \left.+48 \times 10^{10} t\right)On comparing equations (1) & (2), we get;B0=3×10−8c=ωk=48×10101.6×103=3×108\begin{aligned} \mathrm{B}_0 & =3 \times 10^{-8} \\ c & =\frac{\omega}{k}=\frac{48 \times 10^{10}}{1.6 \times 10^3} \\ & =3 \times 10^8 \end{aligned}Andalso,And also,c=\frac{\mathrm{E}_0}{\mathrm{~B}_0}Hence,Hence,E_0=B_0 \times c=3×10−8×3×108=9\begin{aligned} & =3 \times 10^{-8} \times 3 \times 10^8 \\ & =9 \end{aligned}So,therequiredequationis:So, the required equation is:\mathrm{E}=9 \cos \left(1.6 \times 10^3 x+48 \times 10^{10} t\right)$
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