Physics · 2026
JEE Main · 28 January 2026, Shift 2 · Q40
A plane electromagnetic wave is moving in free space with velocity c=3 × 10^8 m / s and its electric field is given as E =54 sin (k z-ω t) j V / m,…
A plane electromagnetic wave is moving in free space with velocity $\displaystyle c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ and its electric field is given as $\displaystyle \vec{E}=54 \sin (k z-\omega t) \hat{j} \mathrm{~V} / \mathrm{m}$, where $\displaystyle \hat{j}$ is the unit vector along $\displaystyle y$-axis. The magnetic field vector $\displaystyle \vec{B}$ of the wave is:
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle -1.8 \times 10^{-7} \sin (k z-\omega t) \hat{i} \mathrm{~T}$
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.