Question

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?

Answer 1

r = 1,224 10⁻² m

Step-by-step 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 =
`(m \ v)/(q \ B)`

let's calculate

r =
`(1.7 \ 10^(-27) \ 5.760063 \ 10^7 )/(1.6 \ 10^(-19) \ 0.5 )`

r = 1,224 10⁻² m