Problem 24.30 - Enhanced with Feedback If the magnetic field inside the cyclotro

Question

Problem 24.30 - Enhanced with Feedback If the magnetic field inside the cyclotron is 1.25 T, what is the diameter of the deuterons' largest orbit, just before they exit? 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 × 10-27 kg. The deuterons exit the cyclotron with a kinetic energy of 5.90MeV. Express your answer with the appropriate units Id= 1 Value Units You may want to review (Lm-pages 778-783) Submit My Answers Give Up Incorrect Try Again, 4 attempts remaining Part C If the beam current is 410 A how many deuterons strike the target each second? deuetrons Submit Mv Answers Give Up

Explanation / Answer

Ek= (1/2) mv^2

isolate for velocity:

v = sqrt(2Ek/m)

where Ek = 5.90 MeV x 1.602 x -19 J

v = sqrt[(2 x 5.90 MeV x 1.602 x 10^-19) / (3.34 x 10^-27)]

v = 2.38 x 10^7 m/s

b)

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

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

rearrange for r:

r = mv / qB

r = (3.34 x 10^-27 x 2.38 x 10^7) / (1.6 x 10^-19 x 1.25)

r = 0.397 m

d = 2r

d = 0.795 m

c)

I = q/t

since t =1s, I=q.

Therefore the total charge that strikes the target each second is 410 microcoulomb.

so divide that total charge by each charge of a deuteron,

Q = (410 x 10^-6 ) / (1.6 x 10^-19)

Q = 2.5625 x 10^15 deuetrons.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.