Question

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 $R_2$, respectively. The ratio of the masses of $A$ and $B$ is

• A. $\sqrt{R_1/R_2}$
• B. $R_1/R_2$
• C. ${\left(R_1/R_2\right)}^2$
• D. ${\left(R_2/R_1\right)}^2$

Answer 1

Answer: (C)

Radius of circular path followed by charged particle is given by
$R=\frac{mv}{qB}=\frac{\sqrt{2mK}}{qB}$
$[\because p=mv=\sqrt{2mK}]$
where, $K$ is kinetic energy of particle. Charged particle $q$ is accelerated through some potential difference $V$, such that kinetic energy of particle is
$K=qV$
$\therefore R=\frac{\sqrt{2mqV}}{qB}$
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, $\frac{R_1}{R_2}=\sqrt{\frac{m_A}{m_B}}$
$\Rightarrow \frac{m_A}{m_B}={\left(\frac{R_1}{R_2}\right)}^2$