Part A Problem 20.15 How fast was this deuteron moving just before it entered th

Question

Part A Problem 20.15 How fast was this deuteron moving just before it entered the magnetic field? A deuteron particle (the nucleus of an isotope of hydrogen consisting of one proton and one neutron and having a mass of 3.34x10-27 kg ) moving horizontally enters a uniform, vertical, 0.660 T magnetic field and follows a circular arc of radius 57.5 cm m/s Submit My Answers Give Up Part B How fast was this deuteron moving just after it came out of the field? m/s Submit My Answers Give Up Part C What would be the radius of the arc followed by a proton that entered the field with the same velocity as the deuteron? cm Submit My Answers Give Up

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 = 1.6*10^-19 coul * 0.575m * 0.660T / 3.34*10^-27 Kg = 1.81*10^7 m/s, or roughly 5% the speed of light

b) it would be going the same speed (1.81*10^7 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 1.67*10^-27 kg, so if


q*B=m*V/r
r = m*V/(q*B) = 0.286 m, or 28.6 cm (one half of the other radius)

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