Question

A neutral particle is at rest in a uniform magnetic field $\overrightarrow{B}$. At $t=0$, particle explodes into two particles, each of mass $m$ and one of them have charge $q$. Both of these move off in separate paths lying in plane perpendicular to $\overrightarrow{B}$. After time $t$, both particles collide, then $t$ will be:

• A. $\frac{2\pi m}{qB}$
• B. $\frac{\pi m}{qB}$
• C. $\frac{\pi m}{4qB}$
• D. $\frac{\pi m}{6qB}$

Answer 1

The correct option is B $\frac{\pi m}{qB}$

Initially net charge on the particle is zero, thus as per conservation of charge after the explosion the charge on the counterpart will be $-q$.

Initial momentum of system, $p_i=0$

Explosion includes internal forces on system hence using conservation of linear momentum.

$p_i=p_f\Rightarrow 0=m{V}_A-m{V}_B$

$\therefore |V_A|=|V_B|$


As shown in the figure the force $F$ is directed towards centre $\text{C}$ of the circular path.

Since both particles have same mass and magnitude of charge, and lying in same uniform magnetic field, so both particles will collide at the diametrically opposite position w.r.t starting point.

Time taken to collide;

$t=\frac{\pi R}{V}=\frac{T}{2}$

$t=\left(\frac{2\pi m}{qB}\right)/2=\frac{\pi m}{qB}$

Hence, option $\left(b\right)$ is correct.