- Home
- Standard 12
- Physics
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 velocity ${\rm{\vec v}} = {{\rm{v}}_0}\hat j$. The speed of the particle will become $2v_0$ after a time
$t = \frac{{2m{{\rm{v}}_0}}}{{qE}}$
$t = \frac{{2Bq}}{{m{{\rm{v}}_0}}}$
$t =\frac{{\sqrt 3 \,Bq}}{{m{{\rm{v}}_0}}}$
$t =\frac{{\sqrt 3 \,m{{\rm{v}}_0}}}{{qE}}$
Solution
$\vec{E}=E_{0} i \| \vec{B}=B_{0} \hat{i}$
$\vec{v}=v_{0} \hat{j} \perp \vec{E}$
$\vec{v} \perp \vec{B}$
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 $t: v=\sqrt{v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}=\sqrt{\left(a_{x} t\right)^{2}+v_{0}^{2}}$
where $v_{y}^{2}+v_{z}^{2}=v_{0}^{2}$
Given $v=2 v_{0}$
$\Rightarrow\left(2 v_{0}\right)^{2}=v_{x}^{2}+v_{0}^{2}$
$\Rightarrow\left(v_{x}\right)^{2}=3\left(v_{0}\right)^{2}$
$\Rightarrow v_{x}=\sqrt{3} v_{0}$
$\Rightarrow a_{x} t=\sqrt{3} v_{0}$
$\Rightarrow\left(\frac{q E}{m}\right) t=\sqrt{3} v_{0}$
$\Rightarrow t=\frac{\sqrt{3} v_{0} m}{q E}$
