For a positively charged particle moving in a `x-y` plane initially along the `x-axis` , there is a sudden change in its path due to the presence of electric and//or magnetic fields beyond `p` . The curved path is shown in the ` x- y `plane and is found to be non - circular. Which one of the following combinations is possible ?

Options

A. `vec(E)=0,vec(B)=bhat(i)+chat(k)`
B. `vec(E)=ai,vec(B)=chat(k)+ahat(i)`
C. `vec(E)=0,vec(B)=chat(j)+bhat(k)`
D. `vec(E)=ai,vec(B)=chat(k)+bhat(j)`

Answer

2

Explanation

Electric field can deviate the path of the particle in the shown direction only when it is along negaive `y-` direction . In the given options `vec(E)` is either zero or along `x-` direction. Hence it is magnetic field which is really responsible for its curved path. Options `(a)` and `(c )` can't be accepted as the path will be helix in that case `(` when the velocity vector makes an angle other than `0^(@),180^(@) or 90^(@)` with the magnetic field, path is a helix `)` option `(d)` is wrong because in that case component of net force on the particle also comes in `k` deirection which is not acceptable as the particle is moving in `x-y` plane. Only in option `(b)` the particle can move in `x-y` plane.
In option `(d) : vec(F)_(n et)qvec(E)+q(vec(v)xxvec(B))`
Initial velocity is along `x-` direction. So let
`vec(v)=vhat(i),vec(F)_(n et)=qahat(i)+q[(vhat(i))xx(chat(k)+bhat(j))]`
`=qahat(i)-qvchat(j)=qvbhat(k)`
In option `(b)`
`vec(F)_(n et)=q(ahat(i))+q[(vhat(i))xx(chat(k)+ahat(i))=qahat(i)-qvchat(j)`