Question

A particle of mass $m$ and charge $q$ enters with velocity $v_0$ perpendicular to a magnetic field B (coming out of the plane of the paper) as shown in the figure. It moves in the magnetic field for time $t=\frac{\pi m}{4qB}$, and then enters into a constant electric field region. The electric and magnetic fields are present only in a rectangular region of thickness $d$. The length of rectangular region is $l$. The particle enters parallel to and grazing the side $RQ$ and leaves the region at $P$.Given : $v_0=(\sqrt{2}+1)m{s}^{-1}$.
$\frac{E}{B}=8m{s}^{-1},l=\left|\frac{4\sqrt{2}}{5}{v}_0\right|$ metres and $\frac{m}{qB}=\frac{4}{5}$

Time taken by the particle to cross the electric field region is

• A. $\frac{4}{5}s$
• B. $\frac{3}{5}s$
• C. $\frac{1}{5}s$
• D. None of these

Answer 1

The correct option is A $\frac{4}{5}s$

The trajectory of particle before it enters the region of electric field can be shown as
Consider a point $C$ to be centre of the circular path followed by the particle.


Displacement of the particle in $x$ direction = $rcos\theta$
We know that $t=\frac{\pi m}{4qB}$-----(1)
and $\frac{m{v}^2}{r}=qvB$
$⟹m\omega =qB$
$⟹m\times \frac{2\pi }{T}=qB$
$⟹T=\frac{2\pi m}{qB}$------(2)

On comparing (1) and (2) we get
$t=\frac{T}{8}$
$T\to {360}^0$
$t\to \frac{{360}^0}{8}={45}^0$

The angle made by velocity when it exits the magnetic region with horizontal $={45}^0$

$⟹$ Horizontal component of velocity
$=vcos{45}^0$

To calculate time we can say
$t\times vcos45=l-x$
$⟹t=\frac{l-rcos{45}^0}{\frac{v_0}{\sqrt{2}}}$
$l=\frac{4}{5}\sqrt{2}{v}_0$-----(Given)

We know that $r=\frac{m{v}_0}{qB}$
and $x=rcos\theta$
$\therefore x=\frac{m{v}_0}{qB\sqrt{2}}$

Further it is given that $v_0=(\sqrt{2}+1)m{s}^{-1}$ and $\frac{m}{qB}=\frac{4}{5}$
$⟹x=\frac{4}{5}\times (\sqrt{2}+1)\times \frac{1}{\sqrt{2}}$
$\therefore x=\frac{4}{5}(1+\frac{1}{\sqrt{2}})m$
$rcos\theta =\frac{4}{5}(1+\frac{1}{\sqrt{2}})=\frac{4}{5}\frac{\sqrt{2}+1}{\sqrt{2}}=\frac{4}{5}\frac{v_0}{\sqrt{2}}$

Now, $t=\left(\frac{l-rcos\theta }{v_0}\right)\sqrt{2}$

$⟹t=\left[\frac{\frac{4}{5}\sqrt{2}{v}_0-\frac{4}{5}\frac{v_0}{\sqrt{2}}}{v_0}\right]\sqrt{2}$

$⟹t=\frac{4}{5}s$