The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional to

An electron is moving along the positive $X$-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$-axis. This can be done by applying the magnetic field along

When an electron placed in a uniform magnetic field is accelerated from rest through a potential difference $V_1$,it experiences a force $F$. If the potential difference is changed to $V_2$,the force experienced by the electron in the same magnetic field is $2F$. Then,the ratio of potential differences $\frac{V_2}{V_1}$ is:

An ion beam of specific charge $5 \times 10^7 \ C/kg$ enters a uniform magnetic field of $4 \times 10^{-2} \ T$ with a velocity $2 \times 10^5 \ m/s$ perpendicularly. The radius of the circular path of the ions in meters will be:

When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be:
$(1)$ a straight line
$(2)$ a circle
$(3)$ a helix

$A$ singly ionized magnesium atom $(A=24)$ ion is accelerated to kinetic energy $5\,keV$ and is projected perpendicularly into a magnetic field $B$ of magnitude $0.5\,T$. The radius of the path formed will be . . . . . . $cm$.