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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2025 · 23 January · Morning Shift · Physics · Electromagnetic Waves

The electric field of an electromagnetic wave in free space is E...

The electric field of an electromagnetic wave in free space is E→=57cos⁡[7.5×106t−5×10−3(3x+4y)](4i^−3j^)N/C\overrightarrow{\mathrm{E}}=57 \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](4 \hat{i}-3 \hat{j}) N / C. The associated magnetic field in Tesla is

Options

  1. AB⇀=573×108cos⁡[7.5×106t−5×10−3(3x+4y)](k^)\stackrel{\rightharpoonup}{\mathrm{B}}=\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](\hat{k})
  2. BB→=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](k^)\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](\hat{k})
  3. CB→=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})
  4. DB→=573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\overrightarrow{\mathrm{B}}=\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})

Correct answer

C. B→=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})

Step-by-step solution

K→=3i^+4j^K^=3i^+4j^5E^=4i^−3j^5 B^=K^×E^B^=−Z^B0=E0C=573×108\begin{aligned} & \overrightarrow{\mathrm{K}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \\ & \hat{\mathrm{K}}=\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}}{5} \\ & \hat{\mathrm{E}}=\frac{4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}}{5} \\ & \hat{\mathrm{~B}}=\hat{\mathrm{K}} \times \hat{\mathrm{E}} \\ & \hat{\mathrm{B}}=-\hat{\mathrm{Z}} \\ & \mathrm{B}_0=\frac{\mathrm{E}_0}{\mathrm{C}}=\frac{57}{3 \times 10^8}\end{aligned}
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