\(\overrightarrow{\mathrm{E}}\) 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 \(\frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right) \hat{j}\)
$\left[\varepsilon_{0}=8.85 \times 10^{-12}\; C ^{2} N ^{-1} m ^{-2}, c =3 \times 10^{8}\; ms ^{-1}\right.$ લો.]