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

Question

In a certain region, uniform electric field E and magnetic field B are present in the opposite direction. A particle of mass m and of charge q enters in this region with a velocity v at an angle θ from the magnetic field. The time after which the speed of the particle would be minimum is equal to (A) (2 π m/B q) (B) (m v sin θ /q E) (C) (m v cos θ /q E) (D) (m v/q E)

Tardigrade Admin · Accepted Answer

The charges experiences a retarding force

F = qE

along x-axis
Retardation

= \frac{qE}{m}

.

It is clear that the speed of the particle will be minimum when its component of velocity is along the direction of electric field.
Then, using the equation of motion,

v = u + a t

\Rightarrow 0 = v cos θ - \frac{qE}{m} t

\Rightarrow t = \frac{m v cos θ}{qE}

$\frac{2 πm}{Bq}$

$\frac{m v s in θ}{qE}$

$\frac{m v cos θ}{qE}$

$\frac{m v}{qE}$

Bq2πm​

qEmvsinθ​

qEmvcosθ​

qEmv​

The charges experiences a retarding force F=qE along x-axis Retardation =mqE​ . It is clear that the speed of the particle will be minimum when its component of velocity is along the direction of electric field. Then, using the equation of motion, v=u+at ⇒0=vcosθ−mqE​t ⇒t=qEmvcosθ​

$\frac{2 πm}{Bq}$

$\frac{m v s in θ}{qE}$

$\frac{m v cos θ}{qE}$

$\frac{m v}{qE}$