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If $\vec{E} = 60 \sin (8 x 10^{10} t - ky) \hat{k}$ represents the electric field associated with an electromagnetic wave in vacuum, then its magnetic field component can be represented by (Here E, t and y are in V/m, second and meter respectively).

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$\vec{B} = 12 \sin (8 x 10^{10} - kx) \hat{j}$

$\vec{B} = 20 x 10^{- 8} \sin (8 x 10^{10} t - ky) \hat{i}$

$\vec{B} = 8 x 10^{- 8} \sin (8 x 10^{10} t - ky) \hat{k}$

$\vec{B} = 6 x 10^{- 6} \sin (8 x 10^{10} + ky) \hat{i}$

answer is C.

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Detailed Solution

$\frac{E_{m}}{B_{m}} = c \Rightarrow B_{m} = \frac{60}{3 x 10^{8}} T = 20 x 10^{- 8} T$

Direction of propagation is determined by the direction of $\vec{E} x \vec{B}$ .

The wave is propagating along y-axis. Since, $\hat{k} x \hat{i} = \hat{j}, \vec{B}$ must be along x-axis.

Also $\vec{E} and \vec{B}$ are always in the same phase.

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