The electric field a plane electromagnetic wave is given by : $\mathrm{E}_{\mathrm{y}}=69 \sin \left[0.6 \times 10^{3} \mathrm{x}-1.8 \times 10^{11} \mathrm{t}\right] \mathrm{V} / \mathrm{m}$. The expression for magnetic field associated with this electromagnetic wave is \_\_\_\_ T.

(A) $\mathrm{B}_{\mathrm{z}}=2.3 \times 10^{-7} \sin \left[0.6 \times 10^{3} \mathrm{x}-1.8 \times 10^{11} \mathrm{t}\right]$

(B) $\mathrm{B}_{\mathrm{z}}=2.3 \times 10^{-7} \sin \left[0.6 \times 10^{3} \mathrm{x}+1.8 \times 10^{11} \mathrm{t}\right]$

(C) $\mathrm{B}_{\mathrm{y}}=69 \sin \left[0.6 \times 10^{3} \mathrm{x}+1.8 \times 10^{11} \mathrm{t}\right]$

(D) $\mathrm{B}_{\mathrm{y}}=2.3 \times 10^{-7} \sin \left[0.6 \times 10^{3} \mathrm{x}-1.8 \times 10^{11} \mathrm{t}\right]$

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · EM Waves

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) $B_{z} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t]$ ✓ Correct
(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) $B_{y} = 69 sin [0.6 \times 1 0^{3} x + 1.8 \times 1 0^{11} t]$
(D) $B_{y} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t]$

Answer: (A)

The wave travels along

+ x

with its electric field along

y

, so the magnetic field must lie along

z

, since the propagation direction satisfies

\overset{c}{^} = \hat{E} \times \hat{B}

with

\hat{i} = \hat{j} \times \hat{k}

. Its amplitude is

B_{0} = \frac{E _{0}}{c} = \frac{69}{3 \times 1 0 ^{8}} = 2.3 \times 1 0^{- 7} T

, and it oscillates in phase with

E

. Hence

B_{z} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t] T

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option