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Two particles of masses $m_{a}$ and $m_{b}$ and same charge are projected in a perpendicular magnetic field. They travel along circular paths of radius $r_{a}$ and $r_{b}$ such that $r_{a}>r_{b}$. Then which is true?
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Verified Answer
The correct answer is:
$m_{a} v_{a}>m_{b} v_{b}$
Radius of circular path $r_{a}=\frac{m_{a} v_{a}}{q B}$
and
$$
r_{b}=\frac{m_{b} v_{b}}{q B}
$$
According to question
$$
r_{a}>r_{b}
$$
$\therefore$
$$
\frac{m_{a} v_{a}}{q B}>\frac{m_{b} v_{b}}{q B}
$$
or $\quad m_{a} v_{a}>m_{b} v_{b}$
and
$$
r_{b}=\frac{m_{b} v_{b}}{q B}
$$
According to question
$$
r_{a}>r_{b}
$$
$\therefore$
$$
\frac{m_{a} v_{a}}{q B}>\frac{m_{b} v_{b}}{q B}
$$
or $\quad m_{a} v_{a}>m_{b} v_{b}$
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