A charged particle with charge `q` enters a region of constant, uniform and mututally orthogonal fields `vec(E) and vec(B)` with a velocity `vec(v)` perpendicular to both `vec(E) and vec(B)`, and comes out without any change in magnitude or direction of `vec(v)`. Then

Options

A. `vecv=vecB xx vecE//E^2`
B. `vecv=vecE xx vecB//B^2`
C. `vecv=vecB xx vecE//B^2`
D. `vecv=vecE xx vecB//E^2`

Answer

B

Explanation

When `vecE` and `vecB` are perpendicular and velocity has no change then `qE=qvB" i.e., "v=(E)/(B)`. The two forces oppose each other so, v is along `vecE xx vecB` i.e., `vecv=(vecExx vecB)/(B^(2))`
As `vecE and vecB` are perpendicular to each other
`(|vecE xx vecB|)/(B^(2))=(EB sin 90^(@))/(B^(2))=(E)/(B)`