Question

A neutral particle is at rest in a uniform magnetic field $\overline{B}$. At $t=0$, particle decays into two particles each of mass ${}^{\prime }{m}^{\prime }$ and one of them having charge ${}^{\prime }{q}^{\prime }$. Both of these move off in separate paths lying in plane perpendicular to $\overline{B}$. At later time, the particles collide. If this time of collision is $x\pi m/qB$ then $x$ is (neglecting the interaction force).

Answer 1

given that the parent particle is neutral so after explosion if the charge of one of the particles is q then the charge of another particle has to be-q.the charged particle will move in a circular path of radius r and will collide at a point diametrically opposite to the where the explosion took place

$qvB=\frac{m{v}^2}{r}\Rightarrow v\frac{rqB}{m}\Rightarrow \omega$ = $\frac{v}{r}=\frac{qB}{m}$

$T=\frac{2\pi }{\omega }=\frac{2\pi m}{qB}$

then on comparing

x=2