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 $E = E_{o} \hat{i} a n d B = B_{o} \hat{i}$ with velocity $\vec{V} = V_{o} \hat{j}$ . The speed of the particle will become $2 V_{0}$ after a time

Difficult

A

$t = \frac{2 m V_{o}}{E q}$
B

$t = \frac{2 B q}{m V_{o}}$
C

$t = \frac{3 B q}{m V_{o}}$
D

$t = \frac{\sqrt{3} m V_{o}}{E q}$

Solution

$G i v e n, E = E_{o} \hat{i}, B = B_{o} \hat{i} a n d V = V_{o} \hat{j}$
$∴ \vec{V} ⊥ \vec{E} a n d \vec{V} ⊥ \vec{B}$
Hence, the path is helix with speed remaining constant for the circular path in the YZ plane and uniform accelerated motion $a_{x} = \frac{E q}{m}$ along x axis.
$∴ {V_{y}}^{2} + {V_{z}}^{2} = {V_{o}}^{2} ............. (1) N o w a t a n y g i v e n t i m e s p e e d i s g i v e n b y, V = \sqrt{{V_{x}}^{2} + {V_{y}}^{2} + {V_{z}}^{2}} \Rightarrow V^{2} = {V_{x}}^{2} + {V_{o}}^{2} [using (1)]$
Given, $V = 2 V_{0}$ ; and here $V_{x} = a_{x} t$
$4 {V_{o}}^{2} = {(a_{x} t)}^{2} + {V_{o}}^{2} \Rightarrow {(a_{x} t)}^{2} = 3 {V_{o}}^{2} \Rightarrow t = \frac{\sqrt{3} V_{o}}{a_{x}} = \sqrt{3} \frac{V_{o} m}{E q} (∵ a_{x} = \frac{E q}{m})$

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