A charge $Q$ moves parallel to a very long straight wire carrying a current $l$ as shown. The force on the charge is
AOpposite to $O X$
BAlong $O X$
COpposite to $OY$
DAlong $OY$
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AOpposite to $O X$
a (a)
$F=q(\vec{V} \times \vec{B})$
Using right hand thumb rule $F$ will be opposite to $O X$.
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