# Q. The electric field of a plane polarized electromagnetic wave in free space at time t= 0 is given by an expression $\vec{E} (x,y) = 10 \hat{j} \cos [(6x + 8z)]$ The magnetic field $\vec{B} (x, z, t)$ is given by : (c is the velocity of light)

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Solution:

## $\vec{E} =10\hat{j} \cos \left[\left(6\hat{i} +8\hat{k}\right) . \left(x\hat{i} +z\hat{k}\right)\right]$ $=10\hat{j} \cos\left[\vec{K} .\vec{r}\right]$ $\therefore \vec{K} = 6\hat{i} +8 \hat{k} ;$ direction of waves travel. i.e. direction of 'c'. $\therefore$ Direction of $\hat{B}$ will be along $\hat{C} \times\hat{E} = \frac{-4 \hat{i} +3\hat{k}}{5}$ Mag. of $\vec{B}$ will be along $\hat{C} \times \hat{E} = \frac{-4 \hat{i} +3\hat{k}}{5}$ Mag. of $\vec{B} = \frac{E}{C} = \frac{10}{C}$ $\therefore \vec{B} = \frac{10}{C} \left( \frac{-4\hat{i} +3\hat{k}}{5}\right) = \frac{\left(-8\hat{i}+6\hat{k}\right)}{C}$

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