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MCQMediumJEE Main 2025 · 23 January, Shift 1Characteristics of EM Waves

Physics Question from JEE Main 2025 · 23 January, Shift 1

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

  • A

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

  • B

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

  • C

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

  • D

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

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