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In a certain region static electric and magnetic fields exist. The magnetic field is given by $\vec B = {B_0}\left( {\hat i + 2\hat j - 4\hat k} \right)$. If a test charge moving with a velocity $\vec v = {v_0}\left( {3\hat i - \hat j + 2\hat k} \right)$ experiences no force in that region, then the electric field in the region, in $SI\, units$, is
$\vec E = - {v_0}{B_0}\left( {3\hat i - 2\hat j - 4\hat k} \right)$
$\vec E = - {v_0}{B_0}\left( {\hat i + \hat j + 7\hat k} \right)$
$\vec E = {v_0}{B_0}\left( {14\hat j + 7\hat k} \right)$
$\vec E = - {v_0}{B_0}\left( {14\hat j + 7\hat k} \right)$
Solution
According to question, as the test charge experiences no net force in that region i.e., sum of electric force $\left( {{{\text{F}}_{\text{e}}} = {\text{q}}\overrightarrow {\text{E}} } \right)\,$ and magnetic forces $[{{\text{F}}_{\text{m}}} = {\text{q}}(\overline {\text{v}} \times \overline {\text{B}} ]$ will be zero.
Hence, $F_{e}+F_{m}=0$
$\mathrm{F}_{\mathrm{e}}=-\mathrm{q}(\overline{\mathrm{v}} \times \overline{\mathrm{B}})$
$=-\mathrm{B}_{0} \mathrm{v}_{0}[(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \times(\mathrm{i}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})]$
$=-B_{0} v_{0}(14 \hat{j}+7 \hat{k})$
