Question

A proton is first accelerated from rest through a potential difference V and then enters a uniform 0.750-T magnetic field oriented perpendicular to its path. In this field, the proton follows a circular arc having a radius of curvature of 1.84 cm. What was the potential difference V

Answer 1

The magnetic force acting on a charged particle moving perpendicular to the field is:

$F_{b}$ = qvB

$F_{b}$ is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.

The centripetal force acting on a particle moving in a circular path is:

$F_{c}$ = mv²/r

$F_{c}$ is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.

If the magnetic force is acting as the centripetal force, set $F_{b}$ equal to $F_{c}$ and solve for v:

qvB = mv²/r

v = qBr/m

Due to the work-energy theorem, the work done on the proton by the potential difference V becomes the proton's kinetic energy:

W = KE

W is work, KE is kinetic energy

W = Vq

KE = 0.5mv²

Therefore:

Vq = 0.5mv²

Substitute v = qBr/m and solve for V:

V = 0.5qB²r²/m

Given values:

m = 1.67×10⁻²⁷kg (proton mass)

B = 0.750T

q = 1.60×10⁻¹⁹C (proton charge)

r = 1.84×10⁻²m

Plug in the values and solve for V:

V = (0.5)(1.60×10⁻¹⁹)(0.750)²(1.84×10⁻²)²/1.67×10⁻²⁷

V = 9120V