Question

A neutral particle at rest in a magnetic field decays into two charged particles of different masses but have equal amount of charges. The energy released goes into their kinetic energies. Then, what can be the path(s) of the two particles.

Neglect any interaction between the two charges.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B


When neutral particle decays, using the conservation of charges,

$q_1+q_2=0\Rightarrow q_1=-q_2$

As there is no external force, then the momentum of the system is conserved.

$\therefore m_1{v}_1+m_2{v}_2=0\Rightarrow m_1{v}_1=-m_2{v}_2...\left(1\right)$

$\Rightarrow$ Direction of $v_1$ is opposite to $v_2$.

Using the right-hand thumb rule, we can see that, both the particles experiences magnetic force in the downward direction, which provides the necessary centripetal force for their circular motion.

As we know, the radius of path of a charged particle in a uniform magnetic field when $\overrightarrow{v}\perp \overrightarrow{B}$ is given by,

$r=\frac{mv}{qB}$

$\Rightarrow \left|\begin{array}{c}{r}_1\end{array}\right|=\left|\begin{array}{c}{r}_2\end{array}\right|\left[\text{From}\right(1\left)\right]$

Also, angular velocity of the particle is given by,

$\omega =\frac{qB}{m}$

$\Rightarrow \omega$ is not equal as $m_1\ne m_2$.

So, collision does not occur at the diametrically opposite points.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(B\right)$ is the correct answer.