Question

If a charged particle of charge to mass ratio$\frac{q}{m}=\alpha$ is entering in a uniform magnetic field of strength $B$ which is extended up to $4d$ as shown in figure at a speed $v=\left(2\alpha dB\right)$
If the radius of the circular path described in the magnetic field is $nd$, find value of $n$.

Answer 1

Electromagnetic force will provide the necessary centripetal force,

$qBv=\frac{m{v}^2}{r}$
$\Rightarrow r=\frac{mv}{qB}=\frac{v}{B\alpha }=\frac{\left(2\alpha dB\right)}{\left(B\alpha \right)}=2d$
i.e. the electron will move out after travelling on a semicircular path of radius $r=2d$.