In a mass spectrometer, a beam of protons enters a magnetic field. Some protons

Question

In a mass spectrometer, a beam of protons enters a magnetic field. Some protons make exactly a one-quarter circular arc of radius 0.35 m .

1) If the field is always perpendicular to the proton's velocity, what is the field's magnitude if exiting protons have a kinetic energy of 12 keV ? IN T

ANS: I got (to two sigs.) 1500000, BUT ITS WRONG! :(

2) How long does it take the proton to complete the quarter circle? IN SECONDS

3) Find the net force (magnitude) on a proton while it is in the field. IN NEWTONS

Explanation / Answer

(1/2) mv2 = 12*1.6*10-19

(1/2)*1.67*10-27*v2  = 12*1.6*10-19

v = 47952.07 m/s

3) F = mv2/r = 1.67*10-27 *(47952.07)2 / 0.35 = 1.097*10-17 N

(1) F = qvB

1.097*10-17 = 1.6*10-19 * 47952.07 * B

B = 1.429 * 10-3 T

(2) t = (R/2v) = 1.146*10-5 seconds

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

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