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 $\vec{B} = B_{0} \hat{i}$ with a velocity $\vec{v} = v_{0} \hat{j}$ . The speed of the particle will be become $3 v_{0}$ after a time

Difficult

A

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

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

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

$t = \frac{(2 \sqrt{2}) m v_{0}}{q E}$

Solution

$\vec{E}$ is parallel to $\vec{B}$ and $\vec{v}$ is perpendicular to both. Therefore, path of the particle is a helix with increasing pitch.
Speed of particle at any time t is $v = \sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}}$
Here, in yz plane speed will remain same. Therefore, $v_{y}^{2} + v_{z}^{2} = v_{0}^{2}$
Given : $v = 3 v_{0}$
$v = \sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}} \Rightarrow v^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2} \Rightarrow {(3 v_{o})}^{2} = v_{x}^{2} + {v_{o}}^{2} \Rightarrow v_{x}^{2} = 8 {v_{o}}^{2} \Rightarrow v_{x} = 2 \sqrt{2} v_{o}$
Equation of Kinematics, along x axis
$v_{x} = u_{x} + a_{x} t \Rightarrow 2 \sqrt{2} v_{0} = 0 + \frac{q E}{m} t \Rightarrow t = \frac{(2 \sqrt{2}) m v_{0}}{q E}$ .

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