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=vecExx(vecB)/(B^(2))`
B. `vecv=vecBxx(vecE)/(B^(2))`
C. `vecv=vecExx(vecB)/(E^(2))`
D. `vecv=vecBxx(vecE)/(E^(2))`

Answer

A

Explanation

As v of charged particle is remaining constant, it means force acting on charged particle is zero:
`f=qE+q(vxx B)=0, qE-(vxxB)=0`
So `q(vxxB)=qEimpliesvxxB=Eimpliesv=(ExxB)/(B^(2))`