Question

Protons are accelerated in a vacuum through a potential difference of 610 V, then enter a vacuum chamber with a ?"⃗ field of strength 0.050 T. What is the smallest sized chamber which can contain the protons? (A vacuum chamber is a chamber with no air in it.) (mp=1.67 x 10-27 kg, qp=e=1.6 x 10-19 C).

(Please provide visual representation so I can understand the solution better if possible, thank you!)

Answer 1

given

delta_V = 610 V
mass of proton, m = 1.67*10^-27 kg
charge of proton, q = 1.6*10^-19 C
B = 0.05 T

let v is the speed of the proton when it enters the chamber.

use, Work-energy theorem,

Wnet = gain in kinetic energy

q*delta_V = (1/2)*m*v^2

==> v = sqrt(2*q*delta_V/m)

= sqrt(2*1.6*10^-19*610/(1.67*10^-27))

= 3.419*10^5 m/s

let r is the radius of the orbit followed by proton.

use, magnetic force acting on proton,

F = q*v*B*sin(90)

m*a = q*v*B

m*v^2/r = q*v*B

==> r = m*v/(B*q)

= 1.67*10^-27*3.419*10^5/(0.05*1.6*10^-19)

= 0.0714 m

so, size of the chamber = diameter

= 2*r

= 2*0.0714

= 0.1428 m (or) 14.28 cm <<<<<<--------------Answer