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Electromagnetic Waves question

2025 · 23 Jan · Shift 1 · Q64
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  5. /2025 · 23 Jan · Shift 1 · Q64

Electromagnetic Waves question

2025 · 23 Jan · Shift 1 · Q64

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
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 / CE=57cos[7.5×106t−5×10−3(3x+4y)](4i^−3j^​)N/C. The associated magnetic field in Tesla is
  1. A
    B→=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})B=3×10857​cos[7.5×106t−5×10−3(3x+4y)](k^)
  2. B
    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})B=3×10857​cos[7.5×106t−5×10−3(3x+4y)](5k^)
  3. C
    B→=−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})B=−3×10857​cos[7.5×106t−5×10−3(3x+4y)](k^)
  4. D
    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})B=−3×10857​cos[7.5×106t−5×10−3(3x+4y)](5k^)
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