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(C) The Lorentz force on a charged particle is given by $\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{v} 	imes \overrightarrow{B})$.At

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antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

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(C) The Lorentz force on a charged particle is given by $\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{v} 	imes \overrightarrow{B})$.At $t=0$ and $z = \frac{\pi}{k}$,the electric field is $\overrightarrow{E} = E_{0} \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}ight) \cos(\pi) = -E_{0} \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}ight)$.For an electromagnetic wave,the direction of propagation is $\hat{k}_{prop} = \frac{\overrightarrow{E} 	imes \overrightarrow{B}}{|E||B|}$. Here,the wave propagates in the $-z$ direction (since the argument is $kz + \omega t$),so $\hat{k}_{prop} = -\hat{k}$.Thus,$\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}ight) 	imes \overrightarrow{B} = -\hat{k} \cdot \frac{E_{0}}{c}$. Solving this,we get $\overrightarrow{B} = -\frac{E_{0}}{c} \left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}ight)$.The magnetic force is $\overrightarrow{F}_{m} = q(\overrightarrow{v} 	imes \overrightarrow{B}) = q(v_{0}\hat{k} 	imes [-\frac{E_{0}}{c} \frac{\hat{i}-\hat{j}}{\sqrt{2}}]) = -q \frac{v_{0}E_{0}}{c} \left(\frac{\hat{j}+\hat{i}}{\sqrt{2}}ight)$.Since $\frac{v_{0}}{c} \ll 1$,the magnetic force is negligible compared to the electric force.Therefore,$\overrightarrow{F} \approx q\overrightarrow{E} = -q E_{0} \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}ight)$,which is antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$.

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