A charged particle of charge ' \(q\) ' is accelerated by a potential difference ' \(V\) ' enters a region of uniform magnetic field ' \(B\) ' at right angles to the direction of field. The charged particle completes semicircle of radius ' \(r\) ' inside magnetic field. The mass of the charged particle is

When charged particle enters perpendicular to a uniform magnetic field, then it follows a circular path.
Its radius is given by,
\(R=\frac{m v}{q B}=\frac{\sqrt{2 m E}}{q B}\)
As the charged particle is accelerated through potential difference of V ,
\(\mathrm{K} \cdot \mathrm{E}=\mathrm{qV}\)
\(\begin{array}{ll}
& \text { Substituting in (i), } \\
& R=\frac{\sqrt{2 m q V}}{q B}=\sqrt{\frac{2 m V}{q}} \times \frac{1}{B} \\
\therefore \quad & R^2=\frac{2 m V}{q} \times \frac{1}{B^2} \\
\therefore \quad & m=\frac{R^2 q^2}{2 V}=\frac{r^2 q B^2}{2 V}
\end{array}\)
\(\ldots \text { (given, } \mathrm{R}=\mathrm{r} \text { ) }\)