First slide

Motion of a charged particle

Question

A particle of charge q and mass m starts moving from the origin under the action of an electric field $\vec{E} = E_{0} \hat{i}$ and magnetic field $\vec{B} = B_{0} \hat{i}$ with a velocity $\vec{v} = v_{0} \hat{j}$ . The speed of the particle will become $2 v_{0}$ after a time.

Difficult

A

$t = \frac{2 m v_{0}}{q E_{0}}$
B

$t = \frac{2 B q}{m v_{0}}$
C

$t = \frac{\sqrt{3} B q}{m v_{0}}$
D

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

Solution

1. Magnetic field can’t change speed of charged particle, it will only change the direction of velocity $v_{0} \hat{j}$ such that it always lie in y-z plane.
2. Since the electric field is in x-direction, hence it would not affect the velocity in the y-z plane.
After time t, the electric field will inject some velocity $(v_{x})$ in the x- direction such that the magnitude of velocity changes to $2 v_{0}$ .
$∴ \sqrt{v_{0}^{2} + v_{x}^{2}} = 2 v_{0}$
$\Rightarrow v_{x}^{2} = 4 v_{0}^{2} - v_{0}^{2} = 3 v_{0}^{2}$
$\Rightarrow v_{x} = \sqrt{3} v_{0}$
3. Electric force $F = q E_{0}$
4. Acceleration $= \frac{F}{m}$
5. The time taken by x-component velocity to become from zero to $\sqrt{3} v_{0}$ due to uniform electric field $= \frac{{(v_{x})}_{final} - {(v_{x})}_{initial}}{acceleration}$
$= \frac{\sqrt{3} v_{0} - 0}{a}$ here $a = \frac{E_{0} q}{m}$
$t = \frac{\sqrt{3} v_{0}}{(\frac{E_{0} q}{m})} = \frac{\sqrt{3} v_{0} m}{E_{0} q}$
Therefore, the correct answer is $(D)$

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