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The electric field in an electromagnetic wave is given by $\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 40 \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right) N \mathrm{NC}^{-1}$. The magnetic field induction of this wave is (in SI unit):
$\overrightarrow{\mathrm{B}}=\hat{\mathrm{i}} \frac{40}{\mathrm{c}} \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)$
$\vec{B}=\hat{j} 40 \cos \omega\left(t-\frac{z}{c}\right)$
$\overrightarrow{\mathrm{B}}=\hat{\mathrm{k}} \frac{40}{\mathrm{c}} \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)$
$\vec{B}=\hat{j} \frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right)$
Solution
$\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 40 \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)$
$\overrightarrow{\mathrm{E}}$ is along $+\mathrm{x}$ direction
$\overrightarrow{\mathrm{v}}$ is along $+\mathrm{z}$ direction
So direction of $\vec{B}$ will be along +y and magnitude of $B$ will be $\frac{E}{C}$
So answer is $\frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right) \hat{j}$
