Correct Answer - A::B::D
A charged particle will move with a constant velocity in a region if the force on it due to electron field as well as magnetic field is zero or force on charged particle due to electric field is equal and opposite to the force on it due to magnetic field.
Force on charged particle due to electric field, `vecF_E=qvecE`
Force on charged particle due to magnetic field, `vecF_m=q(vecvxxvecB)` or `F_m=qvBsintheta`.
Now, `F_E=0` if `E=0` and `F_m=0` if `sintheta=0` or `theta=0^@` or `180^@`
A charged particle would continue to move with a constant velocity in a region if the force due to electric field and magnetic field is zero, i.e.,
(i) `E=0` and particle is moving along the direction of magnetic field (i.e., `theta=0^@` or `180^@`). Hence, `B!=0`. Option (a) is correct.
(ii) `E=0` and `B=0`. Option (d) is correct.
(iii) The resultant force, `qvecE+q(vecvxxvecB)=0`. In this case `E!=0` and `B!=0`, i.e., Option (b) is correct.
(iv) If `B=0` but `E!=0`, then the charged particle will be accelerated by electric field. Option (c) is wrong.
