The electric field in a plane electromagnetic wave is given by : E_y = 69 sin [0.6 × 10^3 x − 1.8 × 10^11 t] V/m. (A) B_z = 2.3 × 10^−7 sin [0.6 × 10^3 x − 1.8 × 10^11 t] (B) B_z = 2.3 × 10^−7 sin [0.6 × 10^3 x + 1.8 × 10^11 t] (C) B_y = 2.3 × 10^−7 sin [0.6 × 10^3 x − 1.8 × 10^11 t] (D) B_y = 69 sin [0.6 × 10^3 x + 1.8 × 10^11 t]

The electric field in a plane electromagnetic wave is given by Ey = 69 sin [0.6 × 10^3 x − 1.8 × 10^11 t] V/m

Q. The electric field in a plane electromagnetic wave is given by :

\(E_y = 69 \sin [0.6 \times 10^3 x - 1.8 \times 10^{11} t] \, \text{V/m}\)

(A) \(B_z = 2.3 \times 10^{-7} \sin [0.6 \times 10^3 x - 1.8 \times 10^{11} t]\)

(B) \(B_z = 2.3 \times 10^{-7} \sin [0.6 \times 10^3 x + 1.8 \times 10^{11} t]\)

(C) \(B_y = 2.3 \times 10^{-7} \sin [0.6 \times 10^3 x - 1.8 \times 10^{11} t]\)

(D) \(B_y = 69 \sin [0.6 \times 10^3 x + 1.8 \times 10^{11} t]\)

Correct Answer: A

Explanation

Given wave:

\[ E_y = 69 \sin(kx - \omega t) \]

So wave is travelling in +x direction.

For electromagnetic wave:

\[ E = cB \]

\[ B = \frac{E}{c} \]

\[ B_0 = \frac{69}{3 \times 10^8} \]

\[ B_0 = 2.3 \times 10^{-7} \, T \]

Direction rule:

\[ \vec{E} \times \vec{B} = \text{direction of propagation} \]

E is along y-axis, wave propagates along +x.

So B must be along z-axis.

Phase must remain same (kx − ωt).

Therefore option (A) is correct.

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