A particle of mass m and charge q has an initial velocity v-v0- If an electric field E-E0- and magnetic field B-B0- act on the particle- its speed will double after a time -

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Electric Field

A particle of...

Question

A particle of mass $m$ and charge $q$ has an initial velocity $\to v = v_{0}^j$ . If an electric field $\to E = E_{0}^i$ and magnetic field $\to B = B_{0}^i$ act on the particle, its speed will double after a time :

\frac{2 m v_{0}}{q E_{0}}

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\frac{3 m v_{0}}{q E_{0}}

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\frac{\sqrt{3} m v_{0}}{q E_{0}}

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\frac{\sqrt{2} m v_{0}}{q E_{0}}

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Solution

The correct option is C $\frac{\sqrt{3} m v_{0}}{q E_{0}}$

As $B$ is perpendicular to $v$ , force due to magnetic field changes only the direction and does not change the speed of the particle.

Due to electric field, in the $x -$ direction,

$F_{x} = q E$

$\Rightarrow m a_{x} = q E_{0} (∵ \to E = E_{0}^i)$

$\Rightarrow a_{x} = \frac{q E_{0}}{m}$

Given that, $v_{y} = v_{0}$

For speed to be doubled,

$v_{x}^{2} + v_{y}^{2} = (2 v_{0})^{2}$

$v_{x}^{2} + v_{0}^{2} = (2 v_{0})^{2}$

$\Rightarrow v_{x} = \sqrt{3} v_{0}$

Using equation of motion for uniformly accelerated motion in $x -$ direction,

$v_{x} = u_{x} + a_{x} t$

$\Rightarrow \sqrt{3} v_{0} = 0 + \frac{q E_{0} t}{m}$
$\Rightarrow t = \frac{\sqrt{3} v_{0} m}{q E_{0}}$

Hence, option $(C)$ is correct.

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