1 Part A A cyclotron is used to produce a beam of highenergy

1.)

Part A.

A cyclotron is used to produce a beam of high-energy deuterons that then collide with a target to produce radioactive isotopes for a medical procedure. Deuterons are nuclei of deuterium, an isotope of hydrogen, consisting of one neutron and one proton, with total mass 3.34×1027kg. The deuterons exit the cyclotron with a kinetic energy of 4.60 MeV .

Q\'s:

What is the speed of the deuterons when they exit?

If the magnetic field inside the cyclotron is 1.25 T, what is the diameter of the deuterons\' largest orbit, just before they exit?

If the beam current is 450 A how many deuterons strike the target each second?

Part B

A rail gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basic mechanism of acceleration is relatively simple and can be illustrated in the following example. A metal rod of mass 10.0 g and electrical resistance 0.500 rests on parallel horizontal rails that have negligible electric resistance. The rails are a distance L = 9.00 cm apart. (Figure 1) The rails are also connected to a voltage source providing a voltage of V = 5.00 V .
The rod is placed in a vertical magnetic field. The rod begins to slide when the field reaches the value B = 5.44×10² T . Assume that the rod has a slightly flattened bottom so that it slides instead of rolling. Use 9.80 m/s2 for the magnitude of the acceleration due to gravity.

Q:

Find s, the coefficient of static friction between the rod and the rails.

Give your answer numerically.

Solution

Part A)

1) Ek= (1/2) mv^2,

isolate for velocity: v=sqrt(2Ek/m)

where Ek= 4.60 MeV * 1.6*10^(-19) J = 7.36*10^(-13) J

v = sqrt(2*7.36*10^(-13)/(3.34*10^(-27))) = 2.1*10^7 m/s

2) We know that the equation for centripital force is mv^2/r, we also know that the only force present in the system is the force caused by the magnetic field, qv x B, therefore

mv^2/r = qvB, where q is the charge of one proton

r = mv/qB

r = 3.34*10^(-27)*2.1*10^7/(1.6*10^(-19)*1.25) = 0.35 m

diameter = 0.7 m

3)I=q/t

since t=1s, I=q. Therefore the total charge that strikes the target each second is 450 microcoulomb.

However, we are trying to find the NUMBER of the deuterons, so divide that total charge by each charge of a deuteron (which is the same as the charge of a proton) and get

no of deuterons = 450*10^(-6)/(1.6*10^(-19)) = 2.81*10^15

Part B)

Current I in the rod = V/R = 5/0.5 = 10 A

Length through which B crosses = 0.09 m

Magnetic field B = 5.44*10^(-2) T

Force acting on the rod = ILB = 10*0.09*5.44*10^(-2) = 0.0489 N

Frictional force = u*mg

Net force = 0.0489 - u*mg = 0 when motion just begins.

u = 0.0489 / (0.01*9.8) = 0.499