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. The associated magnetic field in Tesla is (a) B = 573× 10⁸cos [7.5× 10⁶t - 5× 10⁻³(3x + 4y)](k) (b) B = 573× 10⁸cos [7.5× 10⁶t - 5× 10⁻³(3x + 4y)](5k) (c) B = -573× 10⁸cos [7.5× 10⁶t - 5× 10⁻³(3x + 4y)](k) (d) B = -573× 10⁸cos [7.5× 10⁶t - 5× 10⁻³(3x + 4y)](5k)

Answer: (d). k = 5×10⁻³(3 i + 4 j), | k| = 2.5×10⁻², k = (3 i+4 j)/(5); ω/| k| = 3×10⁸ m/s. B = ( k× E)/(c), same phase. ((3 i+4 j))/(5)×(4 i - 3 j) = (-9-16)/(5) k = -5 k B = -573×10⁸cos [](5 k) So the answer is (d).

The electric field of an electromagnetic wave in free space is E =… | JEE Main 2025 (23 Jan, Shift 1)

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Nature of EM Waves & Wave Equation

The electric field of an electromagnetic wave in free space is

$\vec{E} = 57\cos\left[7.5\times 10^{6}t - 5\times 10^{-3}(3x + 4y)\right](4\hat{i} - 3\hat{j})\text{ N/C}$ . The associated magnetic field in Tesla is

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Answer: (d)

$\vec k = 5\times10^{-3}(3\hat i + 4\hat j)$ , $|\vec k| = 2.5\times10^{-2}$ , $\hat k = \tfrac{3\hat i+4\hat j}{5}$ ; $\omega/|\vec k| = 3\times10^{8}$ m/s.
$\vec B = \dfrac{\hat k\times\vec E}{c}$ , same phase.
$\tfrac{(3\hat i+4\hat j)}{5}\times(4\hat i - 3\hat j) = \tfrac{-9-16}{5}\hat k = -5\hat k$
$\vec B = -\dfrac{57}{3\times10^{8}}\cos[\dots](5\hat k)$
So the answer is (d).

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