An ion with a charge q, mass m, and speed v enters a magnetic field B and is deflected into a path with a radius of curvature R. If an ion with charge q, mass 2m, and speed 2v enters the same magnetic field, what will be the radius of curvature of the path it will be deflected into

R' = 4R

Explanation:

You have that the radius of curvature for a ion of mass m, charge q and speed v, is R.The ion is moving in a zone with a magnetic field. You take into account the following formula for the radius of the trajectory of the ion:

       (1)

m: mass of the ion

q: charge of the ion

B: magnitude of the magnetic field

v: speed of the ion

If another ion of mass 2m, charge q and speed 2v enters to the same magnetic field, you use the equation (1) to obtain the new radius of curvature:

Then, the new radius of curvature is four times the radius of curvature of the ion with mass m

i don't understand. explain it clearly