Two large metal sheets carry surface currents as depicted in the...

The magnetic field produced by a single large metal sheet carrying a surface current density $K$ is given by $B = \dfrac{\mu_0 K}{2}$.

For the top sheet, the current is directed out of the plane of the paper. According to the right-hand rule, the magnetic field produced by the top sheet at point Q (which lies below it) is directed towards the right, with magnitude $B_1 = \dfrac{\mu_0 K}{2}$.

For the bottom sheet, the current is directed into the plane of the paper. The magnetic field produced by the bottom sheet at point Q (which lies above it) is also directed towards the right, with magnitude $B_2 = \dfrac{\mu_0 K}{2}$.

The net magnetic field at point Q is the vector sum of the fields from both sheets. Since they point in the same direction, the net magnetic field is:

$B_{net} = B_1 + B_2 = \dfrac{\mu_0 K}{2} + \dfrac{\mu_0 K}{2} = \mu_0 K$

When a particle of charge $q$ and mass $m$ is projected with speed $v$ into a uniform magnetic field perpendicular to its velocity, it moves in a circular path. The magnetic force provides the necessary centripetal force:

$q v B_{net} = \dfrac{m v^2}{r}$

Solving for the speed $v$, we get:

$v = \dfrac{q B_{net} r}{m}$

Substituting the value of $B_{net}$:

$v = \dfrac{\mu_0 K q r}{m}$