Question

In the mass spectrometer shown above, particles having a net charge +Q are accelerated from rest through a potential difference in Region I. They then move in a straight line through Region II, which contains a magnetic field B and an electric field E. Finally, the particles enter Region III, which contains only a magnetic field B, and move in a semicircular path of radius R before striking the detector. The magnetic fields in Regions II and III are uniform, have the same magnitude B, and are directed out of the page as shown. 

a. In the figure above, indicate the direction of the electric field necessary for the particles to move in a straight line through Region II. 

In terms of any or all the quantities Q, B, E, and R, determine expressions for 

b. the speed v of the charged particles as they enter Region III; 

c. the mass m of the charged particles; d. the accelerating potential V in Region I; 

e. the acceleration a of the particles in Region III; 

f. the time required for the particles to move along the semicircular path in Region III.

Answer 1

(a) Based on the RHR, the magnetic force on the + charge is down, so the electric force should point up. For + charges, and E field upwards would be needed to make a force up.

(b) The speed on region III is equal the whole time and is the same as the speed of the particles in region II. For region II we have … F_e = F_b … Eq = qvB … v = E/B

(c) Using region III … F_net(C) = mv²/r … qvB = mv² / r … m = QBR / v (sub in v) … = QB²R / E

(d) In between the plates, W = K … Vq = ½ mv² … V = mv²/2Q … (sub in v and m) … = RE / 2

(e) In region three, the acceleration is the centripetal acceleration. a_c = v² / R … (sub in v) … E² / RB²

(f) Time of travel can be found with v = d / t with the distance as half the circumference (2πR/2) then sub in v giving … t = πRB / E