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

The Correct Option is C

Radius of circular path followed by charged particle is given by
$R=\frac{m v}{q B}=\frac{\sqrt{2 m K}}{q B}$
$[\because p=m v=\sqrt{2 m K}]$
where, $K$ is kinetic energy of particle. Charged particle $q$ is accelerated through some potential difference $V$, such that kinetic energy of particle is
$K =q V$
$\therefore R =\frac{\sqrt{2 m q V}}{q B}$
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, $\frac{R_{1}}{R_{2}}=\sqrt{\frac{m_{A}}{m_{B}}}$
$\Rightarrow \frac{m_{A}}{m_{B}}=\left(\frac{R_{1}}{R_{2}}\right)^{2}$