Hi! I saw the answer of problem 27 in the chapter 19. Theanswer is said that mas

Question

Hi! I saw the answer of problem 27 in the chapter 19. Theanswer is said that mass of the particle is 2.0x10kg. But thequestion is asking about magnetic field. the question isbelow, "A proton moving freely in a circular path perpendicular to aconstant magnetic field taks 1.00s to complete onerevolution. Determin the magnitude of the magnetic field." I just want to make sure this is the right anawer for thequestion. Tamiko Hi! I saw the answer of problem 27 in the chapter 19. Theanswer is said that mass of the particle is 2.0x10kg. But thequestion is asking about magnetic field. the question isbelow, "A proton moving freely in a circular path perpendicular to aconstant magnetic field taks 1.00s to complete onerevolution. Determin the magnitude of the magnetic field." I just want to make sure this is the right anawer for thequestion. Tamiko

Explanation / Answer

time taken to complete one revolution T = 1 s = 1* 10 ^-6 s i.e., time period T = 10 ^ -6 s angular frequency w = 2 / T = 6.28 * 10^ 6 rad / s In magnetic field Bvq = mr w ^ 2 from this magnetic field B = mrw^ 2 / v q                                        = mrw ^ 2 / rw q                                        = mw / q                                       = [ (1.67 * 10 ^ -27 kg ) ( 6.28 * 10 ^ 6 rad / s ) ] / [ 1.6 * 10^ -19 C ]                                       = 6.55 * 10 ^ -2 T

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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