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

Question

Verified by Toppr

Answer 1

$E = E_{0} \hat{i} ∥ B = B_{0} \hat{i}$

$v = v_{0} \hat{j} ⊥ E$

$v ⊥ B$

Hence, the path of the particle is a helix with speed remaining constant for the circular path in the $y z$ plane.

Magnitude of velocity of particle at any time $t :$ $v = v_{x}^{2} + v_{y}^{2} + v_{z}^{2} = (a_{x} t)^{2} + v_{0}^{2}$ where $v_{y}^{2} + v_{z}^{2} = v_{0}^{2}$

Given $v = 2 v_{0}$

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

$\Rightarrow (v_{x})^{2} = 3 (v_{0})^{2}$

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

$\Rightarrow a_{x} t = 3 v_{0}$

$\Rightarrow (\frac{q E}{m}) t = 3 v_{0}$

$\Rightarrow t = \frac{3 v _{0} m}{q E}$

Answer 2

E = E_{0} \hat{i} ∥ B = B_{0} \hat{i}

v = v_{0} \hat{j} ⊥ E

v ⊥ B

Hence, the path of the particle is a helix with speed remaining constant for the circular path in the

y z

plane.

Magnitude of velocity of particle at any time

t :

v = v_{x}^{2} + v_{y}^{2} + v_{z}^{2} = (a_{x} t)^{2} + v_{0}^{2}

where

v_{y}^{2} + v_{z}^{2} = v_{0}^{2}

Given

v = 2 v_{0}

\Rightarrow (2 v_{0})^{2} = v_{x}^{2} + v_{0}^{2}

\Rightarrow (v_{x})^{2} = 3 (v_{0})^{2}

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

\Rightarrow a_{x} t = 3 v_{0}

\Rightarrow (\frac{q E}{m}) t = 3 v_{0}

\Rightarrow t = \frac{3 v _{0} m}{q E}