A particle of mass m and charge q has an initial velocity v → = v 0 j ^ . If an electric field E → = E 0 i ^ and magnetic field B → = B 0 i ^ act on the particle, its speed will double after a time

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Acceleration produced by electric field,
$a = \frac{q E_{0}}{m} \hat{i}$

After time $t$ , velocity of the particle,
$v = u + a t$

$\Rightarrow v = v_{0} \hat{j} + (\frac{q E_{0} t}{m}) \hat{i} [∵ u = v = v_{0} \hat{j}]$

So, speed of the particle,
$|v| = \sqrt{v_{0}^{2} + {(\frac{q E_{0} t}{m})}^{2}}$

Now, given $|v| = 2 v_{0}$ after time $t$ , so
$2 v_{0} = \sqrt{v_{0}^{2} + {(\frac{q E_{0} t}{m})}^{2}}$

$\Rightarrow 4 v_{0}^{2} = v_{0}^{2} + {(\frac{q E_{0} t}{m})}^{2}$

$\Rightarrow 3 v_{0}^{2} = {(\frac{q E_{0} t}{m})}^{2}$

So, time interval, $t = \frac{\sqrt{3} m v_{0}}{q E_{0}}$

Here, we must note that no change in magnitude of velocity is caused by a perpendicular magnetic field. So, we are not taking effect of magnetic field while calculating change in speed.

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