A proton (with charge of 1.6 x 10^-19 C and mass of 1.7*10^-27 kg) traveling at a speed of 57,600,630 m/s in the + x-direction enters a region of space where there is a magnetic field of strength 0.5 T in the - z-direction. What would be the radius of the circular motion that the proton would go into if it is "trapped" in this magnetic field region?

r = 1,224 10⁻² m

Explanation:

For this exercise let's use Newton's second law

          F = m a

the force is magnetic

         F = q v x B

The bold letters indicate vectors, the module of this expesion is

          F = q v B

The direction of the force is found by the right hand rule

thumb points in the direction of the velicad + x

fingers extended in the direction of B -z

the palm is in the direction of the force + and

the acceleration of the proton is cenripetal

          a = v² / r

we substitute

          q v B = m v² / r

          r =

let's calculate

         r =

         r = 1,224 10⁻² m

another example of kinetic energy would be,

kinetic energy   >     moving energy

answer   >   letter choice (c),   a hammer falling off a table.

hope that !                                         : )