\(\overrightarrow{\mathrm{E}}=\mathrm{E}_{0}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right) \cos \pi\)
\(=-\mathrm{E}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\)
as \(\quad \overrightarrow{E} \times \overrightarrow{B}=\overrightarrow{\mathrm{c}}\)
\(+\mathrm{E}_{0}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right) \times \overrightarrow{\mathrm{B}}=\mathrm{c} \hat{\mathrm{k}}\)
\(\Rightarrow \overrightarrow{\mathrm{B}}=-\left(\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}}\right) \frac{\mathrm{E}_{0}}{\mathrm{c}}\)
\(\overline{\mathrm{F}}=\mathrm{q}\left(-\mathrm{E}_{0} \frac{(\hat{\mathrm{i}}+\hat{\mathrm{j}})}{\sqrt{2}}-\frac{\mathrm{v}_{0} \hat{\mathrm{k}}}{\mathrm{c}} \times(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \mathrm{E}_{\mathrm{o}}\right)\)
since \(\frac{v_{0}}{c} < < 1\)
\(\Rightarrow \mathrm{F}\) is antiparallel to \(\frac{\mathrm{i}+\mathrm{j}}{\sqrt{2}}\)
${V}$ કદ ધરાવતા નળાકારમાં રહેલી ઉર્જા $5.5 \times 10^{-12} \, {J}$ છે. તો ${V}$ નું મૂલ્ય $......{cm}^{3}$ હશે.
$\left(\right.$ given $\left.\in_{0}=8.8 \times 10^{-12} \,{C}^{2} {N}^{-1} {m}^{-2}\right)$