Radius of the circular orbit is inversely proportional to the magnetic field.

Suppose the value of the magnetic field is doubled, then the value of radius is halved.

Let the radius of new circular paths are

$r’_{U-238} = dfrac{r_{U-238}}{2}$ and $r’_{U-235} = dfrac{r_{U-235}}{2}$

The new separation between the isotopes is

$d’ = 2 r’_{U-238} – 2r’_{U-235} = cancel{2} dfrac{r_{U-238}}{cancel{2}} – cancel{2} dfrac{r_{U-235}}{cancel{2}} = r_{U-238} – r_{U-235}= dfrac{d}{2}$

Thus if the field is doubled the separation is also halved.

So if the magnetic field is increased the separation decreases.

The radius of proton’s circular path that moves with the speed $6.27times 10^5::m/s$, perpendicular to the magnetic field of $0.45:T$ is given by

$r = dfrac{m_p v}{e B} = dfrac{1.673 times 10^{-27} times 6.27 times 10^5}{1.6 times 10^{-19} times 0.45} = 14.6 times 10^{-3}:m = 14.6:mm$

The speed of the electron that moves in a circular orbit of radius $r=2.1:cm = 2.1 times 10^{-2}:m$ and perpendicular to the magnetic field of $B =0.0033:T$ is given by

$v=dfrac{|e| B r}{m_e} = dfrac{1.6 times 10^{-19} times 0.0033 times 2.1 times 10^{-2}}{9.11 times 10^{-31}} = 121.7 times 10^5:m/s$