Question

A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields \[E\] and \[\overrightarrow{B}\] with a velocity v perpendicular to both and comes out without any change in its magnitude or direction. Then
(A) \[v=\frac{\overrightarrow{B}\times \overrightarrow{E}}{{{E}^{2}}}\]
(B) \[v=\frac{\overrightarrow{E}\times \overrightarrow{B}}{{{B}^{2}}}\]
(C) \[v=\frac{\overrightarrow{B}\times \overrightarrow{E}}{{{B}^{2}}}\]
(D) \[v=\frac{\overrightarrow{E}\times \overrightarrow{B}}{{{E}^{2}}}\]

Answer 1

Hint: here given a charged particle the magnitude of charge is q, so, this is a positive charge. It enters into a region where both the electric field and magnetic field are present. Electric field will exert electric force and since charge is in motion magnetic force will also be there. We can calculate the magnitude of the forces and then can add them to find resultant force.

Complete step by step answer:Magnitude of charge=q
Fields are orthogonal and the velocity is perpendicular to both it means the angle between electric field vector and magnetic field vector is \[{{90}^{0}}\]and the velocity vector makes an angle of \[{{90}^{0}}\] with both the fields.
The charge comes out without any change in its velocity neither in magnitude nor in direction. This can only happen when the electric force and magnetic force are equal and cancel out each other.
\[\begin{align}
  & {{F}_{E}}={{F}_{M}} \\
 & qE=qvB \\
 & v=\frac{E}{B} \\
\end{align}\]
The two forces oppose each other so, v is along \[\overrightarrow{E}\times \overrightarrow{B}\]
\[\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{{{B}^{2}}}\]
So the correct option is (B)

Additional Information: A velocity selector is a region in which there are uniform electric field and a uniform magnetic field. The fields are perpendicular to one another, and perpendicular to the velocity of the charged particle.

Note:we have to take into account directions of forces of both electric force and magnetic force and then we have to add the two forces using vector law of addition. This is the standard way of solving the question.