JEE Main 2026 — Jan 28, Shift 2 — Physics solution

JEE Main2026Jan 28, Shift 2Physics

A plane electromagnetic wave is moving in free space with velocity $c = 3 \times 1 0^{8}$ m/s and its electric field is given as $E = 54 sin (k z - ω t) \hat{j}$ V/m, where $\hat{j}$ is the unit vector along $y$ -axis. The magnetic field vector $B$ of the wave is :

A
$- 1.8 \times 1 0^{- 7} sin (k z - ω t) \hat{i}$ T
B
$1.4 \times 1 0^{- 7} sin (k z - ω t) \hat{k}$ T
C
$1.4 \times 1 0^{- 7} sin (k z - ω t) \hat{i}$ T
D
$+ 1.8 \times 1 0^{- 7} sin (k z - ω t) \hat{i}$ T

Solution

Direction of propagation is $\hat{k}$ (along $+ z$ ). For an EM wave, $\hat{B} = \hat{C} \times \hat{E} = \hat{k} \times \hat{j} = - \hat{i}$ .

Magnitude: $B_{0} = \frac{E _{0}}{c} = \frac{54}{3 \times 1 0 ^{8}} = 1.8 \times 1 0^{- 7}$ T.

$∴ B = - 1.8 \times 1 0^{- 7} sin (k z - ω t) \hat{i}$ T.

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