The radius of curvature of the path of a 14.4 MeV proton in a 6.00 T magnetic field is 0.091 m.
The radius of curvature can be calculated with Lorentz force:

Since there is no electric field (E = 0) and the Lorentz force is equal to the centripetal force, we have:

Replacing
for
in the above equation, we get:

Where:
m: is the mass of the proton = 1.67x10⁻²⁷ kg
q: is the charge of the proton = 1.602x10⁻¹⁹ C
r: is the radius of curvature =?
v: is the <em>tangential </em>velocity
B: is the magnetic field = 6.00 T
The <em>magnetic field</em> is <u>perpendicular</u> to the charged <em>particles</em>, so:
![\frac{mv^{2}}{r} = q[vBsin(90)]](https://tex.z-dn.net/?f=%20%5Cfrac%7Bmv%5E%7B2%7D%7D%7Br%7D%20%3D%20q%5BvBsin%2890%29%5D%20)

The <em>tangential velocity</em> can be calculated from the <em>energy</em>:

Finally, the radius is:
Therefore, the radius of curvature of the path is 0.091 m.
Find more here:
brainly.com/question/13791875?referrer=searchResults
I hope it helps you!