## Guest

deepak
84 Points
5 years ago
consider an setup like $\overrightarrow{E}$ along the +ve x direction and $\overrightarrow{B}$ −y axis and the velocity $\overrightarrow{v}$ along the positive direction

force due to $\overrightarrow{E}$ on the particle will be in the +ve x direction =Eq

force due to $\overrightarrow{B}$ on the particle will be in the -ve x direction =qvB sin(90)

if there is no alteration in the path of the particle then Fdue to B= Fdue to E    .....they can cancle as they are in the opposite directions

=> qvB=Eq
=> v = E/B = EB/B2
we also know that the velocity is in the direction of $\overrightarrow{E}$ × $\overrightarrow{B}$ (vectorially by right hand rule)

hence $\overrightarrow{v}$ = $[\overrightarrow{E} * \overrightarrow{B}] / B^{2}$

hence option d is correct