Question

A deuteron particle (the nucleus of an isotope of hydrogen consisting of one proton and one neutron and having a mass of 3.34×10−27 kg ) moving horizontally enters a uniform, vertical, 0.620 T magnetic field and follows a circular arc of radius 58.5 cm .

A.) How fast was this deuteron moving just before it entered the magnetic field ? V₁ = _________ m/s

B.) How fast was this deuteron moving just after it came out of the field? V₂= _________ m/s

C.) What would be the radius of the arc followed by a proton that entered the field with the same velocity as the deuteron? R = ______ cm

Answer 1

According to the concept of the moving charges in magnetic field

F=Bqv

mv^2/r=BqV

mv=Bqr

V=Bqr/m

Given that

mass m=3.34*10^-27kg

magnetic field B=0.62 T

radius r=0.58.5 cm

now we find the speed

speed V=1.6*10^-19*0.62*0.585/3.34*10^-27

=0.174*10^8 m/s

now we find the speed deuteron leaving out to the field

According to the concept of lerontez force magnetic field and speed is always perpendicular to each other

so the velocity does not change only direction is changes

now we find the radius of the arc

radius r=mv/Bq=1.67*10^-27*0.174*10^8/1.6*10^-19*0.62=0.293 m=29.3 cm