MCQ
Two charged particles of mass $m$ and charge $q$ each are projected from origin simultaneously with same speed $V$ in transverse magnetic field. If ${\vec r_1}$ and ${\vec r_2}$ are the position vectors of particles (with respect to origin) at $t = \frac{{\pi m}}{{qB}}$ then the value of  ${\vec r_1}.{\vec r_2}$ at that time is
  • A
    ${\left( {\frac{{mv}}{{qB}}} \right)^2}$
  • B
    $\frac{1}{2}{\left( {\frac{{mv}}{{qB}}} \right)^2}$
  • $2{\left( {\frac{{mv}}{{qB}}} \right)^2}$
  • D
    $4{\left( {\frac{{mv}}{{qB}}} \right)^2}$

Answer

Correct option: C.
$2{\left( {\frac{{mv}}{{qB}}} \right)^2}$
c
$\overrightarrow{\mathrm{r}}_{1} \cdot \overrightarrow{\mathrm{r}}_{2}=\left(\mathrm{r} \times \mathrm{r} \times \cos \left(60^{\circ}\right)\right.$

$=\frac{1}{2} r^{2}=\frac{1}{2}(2 R)^{2}=2 R^{2}=2\left(\frac{m v}{q B}\right)$

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