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=vecBxxvecE//E^2`
B. `vecv=vecExxvecB//B^2`
C. `vecv=vecBxxvecE//B^2`
D. `vecv=vecExxvecB//E^2`

Answer

B

Explanation

When charged particle goes undefected, then
`qE=qvB` or `v=E//B`
The two forces acting on moving charged particle due to electric and magnetic fields oppose each other if `vecv` is along `vecExxvecB`
Now, `v=(EB)/(B^2)=(EB sin 90^@)/(B^2)=(|vecExxvecB|)/(B^2)`
`:. vecv=((vecExxvecB))/(B^2)`