Question

The magnetic field in a plane electromagnetic wave is given by: $$ \vec{B} = (3 \times 10^{-7}) \sin(3 \times 10^4 x + 9 \times 10^{12} t) \hat{j} $$ What is the corresponding electric field $ \vec{E} $ of the wave?

• A. \( 90 \sin(3 \times 10^4 x + 9 \times 10^{12} t)\, \hat{i} \, \text{Vm}^{-1} \)
• B. \( 90 \sin(3 \times 10^4 x + 9 \times 10^{12} t)\, \hat{k} \, \text{Vm}^{-1} \)
• C. \( 45 \sin(3 \times 10^4 x + 9 \times 10^{12} t)\, \hat{i} \, \text{Vm}^{-1} \)
• D. \( 45 \sin(3 \times 10^4 x + 9 \times 10^{12} t)\, \hat{k} \, \text{Vm}^{-1} \)

Hint:

In EM waves: \( E = cB \), and \( \vec{E}, \vec{B}, \vec{k} \) form a right-handed coordinate system.

Answer 1

The Correct Option is B

In an electromagnetic wave: \[ E = cB, \quad \text{and } \vec{E} \perp \vec{B},\ \vec{E},\vec{B} \perp \text{direction of propagation} \] Given: - \( B_0 = 3 \times 10^{-7}\, \text{T} \) - \( E_0 = c \cdot B_0 = 3 \times 10^8 \cdot 3 \times 10^{-7} = 90\, \text{Vm}^{-1} \) Since \( \vec{B} \parallel \hat{j} \), and propagation is along \( \hat{i} \), then \( \vec{E} \parallel \hat{k} \) \[ \vec{E} = 90 \sin(3 \times 10^4 x + 9 \times 10^{12} t)\, \hat{k} \]