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Two particles, $A$ and $B$, having equal charges, after being accelerated through the same potential difference enter into a region of uniform magnetic field and the particles describe circular paths of radii $R_{1}$ and $R_{2}$ resporctively. The ratio of the masses of $A$ and $B$ is
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Verified Answer
The correct answer is:
$\left(R_{1} / R_{2}\right)^{2}$
Radius of circular path followed by charged particle is given by
$$
R=\frac{m v}{q B}=\frac{\sqrt{2 m K}}{q B} \quad[\because p=m v=\sqrt{2 m K}]
$$
where, $K$ is kinetic energy of particle. Charged particle $q$ is accelerated through some potential difference $V$, such that kinetic energy of particle is
$\therefore$
$$
\begin{aligned}
K &=q V \\
R &=\frac{\sqrt{2 m q V}}{q B}
\end{aligned}
$$
As the two charged particles of same magnitude and being accelerated through same potential, enters into a uniform magnetic field region, then $R \propto \sqrt{m}$
So,
$$
\frac{R_{1}}{R_{2}}=\sqrt{\frac{m_{A}}{m_{B}}} \Rightarrow \frac{m_{A}}{m_{B}}=\left(\frac{R_{1}}{R_{2}}\right)^{2}
$$
$$
R=\frac{m v}{q B}=\frac{\sqrt{2 m K}}{q B} \quad[\because p=m v=\sqrt{2 m K}]
$$
where, $K$ is kinetic energy of particle. Charged particle $q$ is accelerated through some potential difference $V$, such that kinetic energy of particle is
$\therefore$
$$
\begin{aligned}
K &=q V \\
R &=\frac{\sqrt{2 m q V}}{q B}
\end{aligned}
$$
As the two charged particles of same magnitude and being accelerated through same potential, enters into a uniform magnetic field region, then $R \propto \sqrt{m}$
So,
$$
\frac{R_{1}}{R_{2}}=\sqrt{\frac{m_{A}}{m_{B}}} \Rightarrow \frac{m_{A}}{m_{B}}=\left(\frac{R_{1}}{R_{2}}\right)^{2}
$$
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