Question

In a certain region uniform electric field $\overrightarrow{E}$ and uniform magnetic field $\overrightarrow{B}$ are present in opposite directions. A particle of mass $m$ and of charge $q$ enters in this region with a velocity $v$ at an angle $\theta$ from the magnetic field. The time after which the speed of the particle would be minimum is equal to :

• A. $\frac{2\pi m}{qB}$
• B. $\frac{mv\sin \theta }{qE}$
• C. $\frac{mv\cos \theta }{qE}$
• D. $\frac{mv}{qE}$

Answer 1

The correct option is C $\frac{mv\cos \theta }{qE}$

The charge experiences a retarding force $\overrightarrow{F}=q\overrightarrow{E}$ along $x-$ axis. Retardation $=\frac{qE}{m}$
It is clear that the speed of the particle will be minimum when its component of velocity along the direction of electric field is zero.
Thus, using the equation of motion.
$v=u+at$
$0=v\cos \theta -\frac{qE}{m}t$
or $t=\frac{mv\cos \theta }{qE}$