A charged particle of charge ' $q$ ' is accelerated by a potential difference ' $V$ ' enters a region of…

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
  1. $\frac{r^2 q B^2}{2 V}$
  2. $\frac{r^2 q^2 B^2}{\sqrt{2} V}$
  3. $\frac{q r B}{2 \mathrm{~V}}$
  4. $\frac{q^2 r^2 B^2}{V}$

Solution

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 , $\mathrm{K} \cdot \mathrm{E}=\mathrm{qV}$ $\begin{array}{ll} & \text { Substituting in (i), } \\ & R=\frac{\sqrt{2 m q V}}{q B}=\sqrt{\frac{2 m V}{q}} \times \frac{1}{B} \\ \therefore \quad & R^2=\frac{2 m V}{q} \times \frac{1}{B^2} \\ \therefore \quad & m=\frac{R^2 q^2}{2 V}=\frac{r^2 q B^2}{2 V} \end{array}$ $\ldots \text { (given, } \mathrm{R}=\mathrm{r} \text { ) }$ ^

Asked in: MHT CET 2024 (04 May Shift 1)

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