A deuteron particle (the nucleus of an isotope of hydrogen consisting of one pro
ID: 1790741 • Letter: A
Question
A deuteron particle (the nucleus of an isotope of hydrogen consisting of one proton and one neutron and having a mass of 3.34×1027 kg ) moving horizontally enters a uniform, vertical, 0.500 T magnetic field and follows a circular arc of radius 56.0 cm. A) How fast was this deuteron moving just before it entered the magnetic field? B) How fast was this deuteron moving just after it came out of the field? C) What would be the radius of the arc followed by a proton that entered the field with the same velocity as the deuteron?
Explanation / Answer
a) the force exerted on the deutereon is given by F=q*V*B
the centripetal force is F=M*V^2/r
this gives us q*V*B=m*V^2/r
q*B=m*V/r
q*r*B/m = 1.6E-19 coul * .56m * .50T / 3.34E-27 Kg = 1.3413 E7 m/s,
b) it would be going the same speed (1.3413 E7 m/s) as it leaves the field, because the lorentz force always acts perpendicular to the velocity: it can only change direction, not speed.
c) if it were a proton, not a deuteron, q would be the same, v would be the same, B would be the same, but m would be only 1.67E-27 kg, so if
q*B=m*V/r
r = m*V/(q*B) = 0.279 m, or 27.9 cm