Exercise 27.31 Part A Singly ionized (one electron removed) atoms are accelerate

Question

Exercise 27.31 Part A Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 151 V/m and the magnetic field is 3.18x10 T. The ions next enter a uniform magnetic field of magnitude 1.72x10-2 T that is oriented perpendicular to their velocity. How fast are the ions moving when they emerge from the velocity selector? m/S Submit My Answers Give Up Part B If the radius of the path of the ions in the second magnetic field is 17.8 cm , what is their mass? Tn = Submit My Answers Give Up

Explanation / Answer

Given,

E = 151 V/m ; B = 3.18 x 10^-2 T ; B' = 1.72 x 10^-2 T

A)The electric force is balanced by the magnetic force, so

Fe = F(B)

q E = q v B => v = E/B

v = 151/3.18 x 10^-2 = 4.75 x 10^3 m/s

Hence, v = 4.75 x 10^3 m/s

B)Now the magnetic force balances the centripital force,

F(B) = Fc

q v B = m v^2/R

m = q B R/v = 1.6 x 10^-19 x 1.72 x 10^-2 x 0.178/(4.75 x 10^3) = 1.88 x 10^-25 kg

Hence, m = 1.03 x 10^-25 kg

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

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