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STANDARDPhysics-Electromagnetic Waves

STANDARD Physics Wave Propagation MCQ Question

Type: MCQ-conceptual-Medium-Class 12

The magnetic field of an electromagnetic wave is given by B=1.6×10−6⋅cos⁡(2×107z+6×1013t)(2i+j)Wbm2\mathbf{B} = 1.6 \times 10^{-6} \cdot \cos(2 \times 10^7 z + 6 \times 10^{13} t) (2\mathbf{i} + \mathbf{j}) \frac{\text{Wb}}{\text{m}^2}. The associated electric field will be

A

E = 4.8 ×\times 10210^2 cos⁡\cos(2 ×\times 10710^7 z + 6 ×\times 101310^{13} t) (-\mathbf{i} + 2\mathbf{j}) $$\textV/m{V/m}

B

E = 4.8 ×\times 10210^2 cos⁡\cos(2 ×\times 10710^7 z - 6 ×\times 101310^{13} t) (-2\mathbf{j} + \mathbf{i}) $$\textV/m{V/m}

C

E = 4.8 ×\times 10210^2 cos⁡\cos(2 ×\times 10710^7 z + 6 ×\times 101310^{13} t) (\mathbf{i} - 2\mathbf{j}) $$\textV/m{V/m}

D

E = 4.8 ×\times 10210^2 cos⁡\cos(2 ×\times 10710^7 z - 6 ×\times 101310^{13} t) (2\mathbf{i} + \mathbf{j}) $$\textV/m{V/m}

Correct Answer

Option C

Detailed Explanation

The electric field E\mathbf{E} is perpendicular to the magnetic field B\mathbf{B} and the direction of propagation. Given B=1.6×10−6⋅cos⁡(2×107z+6×1013t)(2i+j)\mathbf{B} = 1.6 \times 10^{-6} \cdot \cos(2 \times 10^7 z + 6 \times 10^{13} t) (2\mathbf{i} + \mathbf{j}), the electric field E\mathbf{E} must be perpendicular to (2i+j)(2\mathbf{i} + \mathbf{j}), leading to (i−2j)(\mathbf{i} - 2\mathbf{j}).

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