Question

Please explain clearly how they got the answer, thank you!

Please explain clearly how they got the answer, thank you! A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60-T magnetic field. What is the radius of the resulting path?

Answer 1

Here ,

the potential difference , V = 2.5 kV

V = 2500 V

magnetic field , B =0.60 T

let the radius is r = sqrt(2 * m * V *q)/(B*q)

r = sqrt(2 * 1.67 *10^-27 * 2500 * 1.602 *10^-19)/(1.602 *10^-19 * 0.6)

r = 0.012 m

r = 12 mm

the radius of path of proton is 12 mm