question_answer

A proton moving with a velocity v is subjected to an electric field E and magnetic field B. The proton will remain moving without any deflection, if:

A)  E and B both are parallel to v

B)  E is parallel to v but perpendicular to B

C)  E is perpendicular to B

D)  E, v and B are mutually perpendicular to each other and \[v=\frac{E}{B}\]

Correct Answer: D

Solution :

 Key Idea: No deflection is observed, if force due to electric field is equal to that due to magnetic field. When a proton enters an electric field E, the electric force acting on it, is \[F=qE\] ... (i) where q is charge and magnetic force when in magnetic field is \[F=qv\,B\sin \theta \] for \[\theta ={{90}^{o}}\] \[F=qvB\] ... (ii) For no deflection, \[qE=qBv\] \[\Rightarrow \] \[v=\frac{E}{B}\] Hence, v, E and B should be mutually perpendicular to each other.