A plane electromagnetic wave is moving in free space with velocity c = 3× 10⁸ m/s and its electric field is given as E = 54sin (kz - ω t) j V/m, where j is the unit vector along y-axis. The magnetic field vector B of the wave is: (a) -1.8× 10⁻⁷sin (kz - ω t) i T (b) +1.8× 10⁻⁷sin (kz - ω t) i T (c) 1.4× 10⁻⁷sin (kz - ω t) k T (d) 1.4× 10⁻⁷sin (kz - ω t) i T

Answer: (a). sin (kz-ω t): wave moves along +z. B₀ = 543×10⁸ = 1.8×10⁻⁷ T Need j× B = k: j×(- i) = k, so B is along - i. So the answer is (a) -1.8×10⁻⁷sin (kz-ω t) i T.

A plane electromagnetic wave is moving in free space with velocity c… | JEE Main 2026 (28 Jan, Shift 2)

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Nature of EM Waves & Wave Equation

A plane electromagnetic wave is moving in free space with velocity $c = 3\times 10^{8}$ m/s and its electric field is given as $\vec{E} = 54\sin(kz - \omega t)\,\hat{j}$ V/m, where $\hat{j}$ is the unit vector along y-axis. The magnetic field vector $\vec{B}$ of the wave is:

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Answer: (a)

$\sin(kz-\omega t)$ : wave moves along $+z$ .
$B_0 = \dfrac{54}{3\times10^{8}} = 1.8\times10^{-7}$ T
Need $\hat j\times\hat B = \hat k$ : $\hat j\times(-\hat i) = \hat k$ , so $\vec B$ is along $-\hat i$ .
So the answer is (a) $-1.8\times10^{-7}\sin(kz-\omega t)\hat i$ T.

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