Irodov – Problems in General Physics

3.397. (a) T =

(e

7E2Vm

2 r B)2 =---

12 MeV; (b)

T
v MHz.
r 2
3.398. (a) t= meV = 17 ps; 17 (b) s^ 4a9v2mr2
3eV 0.74 km.
N
Instruction. Here s ti E vn - 1/n, where vn is the velocity of

the particle after the nth passage across the accelerating gap.

Since N is large,^ lin z
J

V n dn.

3.399. n = 2avW/eBc^2 = 9.

3.400. w = wo /l/ I + at, where coo = qB1m, a = qBAWInm^2 c^2.

3.401. v = 112 rqB1m, p = r/2.

3.402. N = W led) = 5.10 6 revolutions, s = 2ItrN = 8.10 3 km.

3.403. On the one hand,

dp (^) eE= e
dt 2ar dt '
where p is the momentum of the electron, r is the radius of the orbit,
41) is the magnetic flux acting inside the orbit.
On the other hand, dp/dt can be found after differentiating the
relation p = erB for r = const. It follows from the comparison of
the expressions obtained that dBoldt = 1 /2 d (B) / dt. In particular,
this condition will be satisfied if Bo =^1 /^2 (B).
3.404. 7. 0 =1/- 213 o/3a.
3.405. dEldr = B (r 0 ) — 112 (B) = 0.
3.406. OW = 2ar^2 eB/At = 0.10 keV.
3.407. (a) W= Oil (reBlmoc) 2 —1) moc 2 ; (b) s= WAtIreB.
4.1. (a) See Fig. 27; (b) (vx/a(o) 2 (xla)^2 = 1 and wx = —w 2 x.
Fig. 27.
4.2. (a) The amplitude is equal to a/2, and the period is T
= at/w, see Fig. 28a; (b) vx = 4w^2 x (a — x), see Fig. 28b.
4.3. x = a cos (wt a) — 29 cm, vx = — 81 cm/s, where
a = Vx: + (Vx0/6))2? a = arctan (— vx0/0)xo)•
21•