A beam of protons is accelerated through a potential difference of 0.725kV and then enters a uniform magnetic field traveling perpendicular to the field.

A) What magnitude of field is needed to bend these protons in a circular arc of diameter 1.72m?

B) What magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the protons?

A) For the proton field, B = 4.53 * 10^-3 T

B) For the electron field, B = - 2.47 * 10^-6 T

Explanation:

V = 0.725kV = 725V

KE gained by proton (charge + e) = eV (J)

KE = (1.60 * 10^-19 C) (725.0 J/C) = 1.16 * 10^-16 J

KE = ½ mv²

v² = 2. KE/mp ... (mp = mass of proton)

v = √[2. (1.16 * 10^-16J) / (1.67 * 10^-27 kg) ] = 3.73 * 10^5 m/s

A) For the curved path equate centripetal and magnetic force

mv²/R = Bqv

B = mv/Rq

Diameter of the arc = 1.72m

Radius of the arc = 1.72/2 = 0.86m

Mass of the proton = 1.67 * 10^-27 kg

q = 1.60 * 10^-19 C

B = (1.67 * 10^-27 kg) (3.73 * 10^5 m/s) / (0.86m) (1.60 * 10^-19 C)

B = 4.53 * 10^-3 T

B) For the electrons

R = 0.86m

Mass of the electron = 9.10 * 10^-31 kg

q = - 1.60 * 10^-19 C

B = mv/Rq = (9.10^-31 kg) (3.73^5 m/s) / (0.86m) (-1.60 * 10^-19 C)

B = - 2.47 * 10^-6 T (opposite direction to proton field)

(a) magnitude of field is needed to bend these protons is 421.5 N/C

(b) magnitude of the field will be the same as in proton.

Explanation:

Given;

potential difference of the proton, V = 0.725kV = 725 V

distance of the circular arc = 1.72 m

Part (a) magnitude of field is needed to bend these protons in a circular arc of diameter 1.72m.

V = E*d

E = V/d = 725/1.72

E = 421.5 N/C

Part (b) magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the proton.

If the electron have the same speed with proton, and move on the same path of 1.72m. Thus, the magnitude of the field will be same as in proton.