JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

The electric field a plane electromagnetic wave is given by :

$E_{y} = 69 sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t]$ V/m.

The expression for magnetic field associated with this electromagnetic wave is____ T.

A
$B_{z} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t]$
B
$B_{z} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x + 1.8 \times 1 0^{11} t]$
C
$B_{z} = 69 sin [0.6 \times 1 0^{3} x + 1.8 \times 1 0^{11} t]$
D
$B_{y} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t]$

Solution

The wave propagates along $+ \overset{x}{^}$ (from the form $k x - ω t$ ) and the electric field oscillates along $\overset{y}{^}$ . Since $\hat{E} \times \hat{B} = \overset{c}{^}$ , $\hat{B} = \overset{c}{^} \times \hat{E} = \overset{x}{^} \times \overset{y}{^} = \overset{z}{^}$ . So $B$ points along $\overset{z}{^}$ .

Magnitude: $∣ B ∣ = ∣ E ∣/ c$ , where $c = ω / k = (1.8 \times 1 0^{11}) / (0.6 \times 1 0^{3}) = 3 \times 1 0^{8}$ m/s. So $∣ B ∣ = 69/ (3 \times 1 0^{8}) = 2.3 \times 1 0^{- 7}$ T.

$B$ is in phase with $E$ , so $B_{z} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t]$ .

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