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The electric field of a plane electromagnetic wave is given by $\overrightarrow{ E }= E _{0}(\hat{ x }+\hat{ y }) \sin ( kz -\omega t )$ Its magnetic field will be given by
$\frac{ E _{0}}{ c }(\hat{ x }-\hat{ y }) \cos ( kz -\omega t )$
$\frac{ E _{0}}{ c }(-\hat{ x }+\hat{ y }) \sin ( kz -\omega t )$
$\frac{ E _{0}}{ c }(\hat{ x }-\hat{ y }) \sin ( kz -\omega t )$
$\frac{ E _{0}}{ c }(\hat{ x }+\hat{ y }) \sin ( kz -\omega t )$
Solution
$\overrightarrow{ E }= E _{0}(\hat{ x }+\hat{ y }) \sin ( kz -\omega t )$
direction of propagation $=+\hat{ k }$
$\hat{ E }=\frac{\hat{ i }+\hat{ j }}{\sqrt{2}}$
$\hat{ k }=\hat{ E } \times \hat{ B }$
$\hat{ k }=\left(\frac{\hat{ i }+\hat{ j }}{\sqrt{2}}\right) \times \hat{ B } \quad \Rightarrow \hat{ B }=\frac{-\hat{ i }+\hat{ j }}{\sqrt{2}}$
$\therefore \overrightarrow B =\frac{ E _{0}}{ C }(-\hat{ x }+\hat{ y }) \sin ( kz -\omega t )$
