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 $\frac{m}{qB}=\frac{4}{5}$

Length of the electric field region is

• A. $\frac{3}{5}(2+\frac{1}{\sqrt{2}})m$
• B. $\frac{4}{5}(1+\frac{1}{\sqrt{2}})m$
• C. $\frac{4}{5}(1-\frac{3}{\sqrt{2}})m$
• D. None of these.

Answer 1

The correct option is B $\frac{4}{5}(1+\frac{1}{\sqrt{2}})m$

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
$t=\frac{T}{8}$
$T\to {360}^0$
$t\to \frac{{360}^0}{8}={45}^0$
$⟹x=rcos{45}^0$
$⟹x=\frac{r}{\sqrt{2}}$

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

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

Length of electric field region = $l-x=l-rcos\theta$
$⟹l-x=\frac{4\sqrt{2}}{5}(\sqrt{2}+1)-\frac{4}{5}(1+\frac{1}{\sqrt{2}})$
$⟹l-x=\frac{4}{5}(1+\frac{1}{\sqrt{2}})$