A Circulating Charged Particle — Question 9

Section 28-5 · Physics pages 4568–4569

Question 9 Two helical particles can regain their common transverse phase after different turn counts Two positive particles have the same charge q and masses m and 3m/2. They start from the same point with the same nonzero perpendicular velocity and the same parallel velocity u in uniform B. Let T1 = 2πm/(qB).⃗ B p helix projected onto a plane Schematic; labels, not drawing lengths, specify the data. (a) Find the earliest positive time at which both have completed integer numbers of turns and returned to their initial transverse positions. (b) Find their common axial displacement then and compare their radii and pitches.
Show solutionHide solution
Question 9 - Solution Two helical particles can regain their common transverse phase after different turn counts⃗ B p helix projected onto a plane (a) Method and calculation T2 = 3 2T1, n1T1 = n2T2 ⇒2n1 = 3n2. n1 = 3, n2 = 2, treturn = 3T1. (b) Interpretation and further calculation ∆z = utreturn = 3uT1. r2/r1 = m2/m1 = 3/2, p2/p1 = T2/T1 = 3/2. Physical check Their transverse radii differ, but integer-turn returns remove that spatial difference at the specified instant.

Question 9

Two helical particles can regain their common transverse phase after different turn counts

Two positive particles have the same charge q and masses m and 3m/2. They start from the same point with the same nonzero perpendicular velocity and the same parallel velocity u in uniform B. Let T1=2πm/(qB)T_1=2\pi m/(qB).

Diagram shown in the original SVG worksheet.

  1. Find the earliest positive time at which both have completed integer numbers of turns and returned to their initial transverse positions.

  2. Find their common axial displacement then and compare their radii and pitches.

Question 9 - Solution

Two helical particles can regain their common transverse phase after different turn counts

Diagram shown in the original SVG worksheet.

(a) Method and calculation

T2=32T1,n1T1=n2T2⇒2n1=3n2.T_2=\tfrac32T_1,\quad n_1T_1=n_2T_2\Rightarrow2n_1=3n_2.n1=3,n2=2,treturn=3T1.n_1=3,\quad n_2=2,\quad t_{\mathrm{return}}=3T_1.

(b) Interpretation and further calculation

Δz=utreturn=3uT1.\Delta z=ut_{\mathrm{return}}=3uT_1.r2/r1=m2/m1=3/2,p2/p1=T2/T1=3/2.r_2/r_1=m_2/m_1=3/2,\quad p_2/p_1=T_2/T_1=3/2.

Physical check

Their transverse radii differ, but integer-turn returns remove that spatial difference at the specified instant.