Download irodov_problems_in_general_physics_2011 and more Study Guides, Projects, Research Physics in PDF only on Docsity!
/i'Dpom = 0.7 ms, where p is the resistivity, pois the
density of copper.
3.319. Li — 1/21:: In = 0.26 p,H/m.
3.320. L p,N2a In (1 +1). 3.321. L1 = tioldb = 25 nH/m. 3.322. Li Pi'. In 1. 3.323. (a) I = na2B/L; (b) A = 1/2n2a4B2/L. 3.324. I = I (1 + = 2 A. 3.325. / = jia8B — 50 A. [to ln — 2 )
3.326. I= + (1— 1) e-tliR/L].
3.327. I =1- (1— e—t-R/29.
3.328.1.," R (.1,1L+ 2 L2)
3.329. Lie= 1' 421 In (1 -1-÷).
3.330. Li^
p, 0
2N
inb
3.331. (a) L 2
(^1) /211, 03-ta2/b; (b) 021 = 1121-Lona2//b. 3.332. pm= 2aRqlp,oN. 3.333. L12^---% 1 12 p,ona4113. 3.334. /2 = aLR12 (1^ e-tR/L2).
3.335. / Q 2R2(1+ Rol R) —
J.
3.336. W =^ 0.5 J.
3.337. W = BlIa2a2b = 2.0 J, where H = 1 /2 NI/nb. 3.338. (a) Wgap /Wmpb/nd = 3.0; (b) L^ aNdi2 —^ 0.15 H. 3.339. Wi= Rdt,26)2a2/831. 3.340. E = (^) solo = 3.108V/m. 3.341. wm/ive = go110002a^ 4/ 2^ 1.1.10-15. 3.343. (a) (^) Ltotat = 2L; (^) (b) Ltotat = L/2. 3.344. L12=1/- L1L2.
3.346. W12= P2b2 /1/2cos 0. 3.347. (^) (a) ja = --i; (b) Id = qieogP• 3.348. The displacement current should be taken into account in addition to the conduction current. 3.349. Em = InzleocoS = 7^ V/cm. 3.350. H = Hmcos (cot ±a,), where and a is determined from the formula tan a = com/a.
R' (Li+ (^) L2)
117 n -7 2; V 62 + (8°"))
319
3.351. (^) fd-={
1/2:Eir for r < R,
1 12BR2Ir^ for r >^ R.
Here h = p,on/mo)2sin cot.
3.352. (a) jd = 42itqrv3 ; (b) jd
3.353. (^) xn,=-. (^) 0, id max — 4,,qva 3 •
3.354. H— q 4nr3 • Evri
3.355. (a) If B (t), then V X E —OBI& 0. The spatial derivatives of the field E, however, may not be equal to zero (V x E 0) only in the presence of an electric field. (b) If B (t), then V X E = 0. But in the uniform field V x E = 0. (c) It is assumed that E = of (t), where a is a vector which is independent of the coordinates, f (t) is an arbitrary function of time.
Then —awat = V X E = 0, that is the field B does not vary with
time. Generally speaking, this contradicts the equation V x H =
= apiat for in this case its left-hand side does not depend on time
whereas its right-hand side does. The only exception is the case when f (t) is a linear function. In this case the uniform field E can be time-dependent. 3.356. Let us find the divergence of the two sides of the equation
V X H = j °plat. Since the divergence of a rotor is always equal
to zero, we get 0 = V•j + 4F(V •D). It remains to take into
account that V•D = p. 3.357. Let us consider the divergence of the two sides of the first equation. Since the divergence of a rotor is always equal to
zero, V • (mat) = 0 or --a-Fa(V•B) = 0. Hence, V•B = const which
does not contradict the second equation. 3.358. V X E 3.359. E' = 3.360. a = eovB = 0.40 pC/m2. 3.361. p = —2eo)B= —0.08 nC/m3, a = eoczo)B= 2 pC/m2.
3.362. B=
[r] [yr]r3 •
3.364. E' = br/r2, where r is the distance from the z' axis.
3.365. B' = (^) c2r
a [rv] ' where r is the distance from the z' axis.
3.367. (a) E' = E V 132 c°82 aVO 2 =9 kV/m; tan a'
whence a 51'; (b) B' —
PE sin a -14 IA. c171-
qv 4nr
tan a
V1 —,P
320
3.380. (a) p=qrB; T =m0c2 (1/-1+ (qrBlmoc)2—1);^ (c) w = C^2
(1) (^) mv
leB 3.384. r 2p I sin (cp/2) I , where p —m eBy sin a, cos a
3.385. rmax ---- aevo/b,^ where^ b=^ )1°— 2n m I.
3.386. v —^ V^ qlm^ V rB (b I a) ' =r2B2 In^ (b la) 3.387. (a) yn = 913, ; (b) tan a = 2nEn •
2n2mEn 2 voB
3.388. z = 1 tan V - 7B2q—T;-^1 y;^ for z <1 this equation reduces to y = (^) (2mElq12B2) z2. 3.389. F = mEl I qB =^ 20 [IN. 3.390. Al —^
2 nmE (^) tan cp 6 cm. eB a (a+2b) B 3.391. (^) 2EAx •
3.392. (a) x = a (cot —^ sin cot); y =^ a (1 —^ cos cot), where^ a = = mElqB2, co qB/m. The trajectory is a cycloid (Fig. 26). The
y
2u
,z• Fig. 26.
motion of the particle is the motion of a point located at the rim of a circle of radius a^ rolling without slipping along the^ x^ axis so that its centre travels with the velocity v =^ E/B; (b)^ s = 8mE/gB2; (c) (v x) = EIB.
3.393. V^ = 2^ nt 21-'4n^1 -) 2ln -a- b
3.394. B< b'
2b a2 VIL 11 V.
3.395. y = t^ sin^ cot, x= 2(÷0,^ (sin cot— cot cos cot),^ where
a = qEmlm.^ The trajectory has the form of unwinding spiral. 3.396. V > 2n2v2mrAr/e = 0.10 MV.
r [1+ (moclqrB)2]• 3.381. T = imoc2, 5 keV and 9 MeV respectively. 3.382. Al = lac 2mv I eB2 cos a= 2.0 cm. 8n2v 3.383. t2 (B2 —B1)2 •
3.397. (a) T =
(e
7E2Vm
2 r B)2 =--- 12 MeV; (b)
T v MHz. r 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 vnis 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/eBc2 = 9. 3.400. w = wo/l/ I +^ at,^ where coo =^ qB1m, a = qBAWInm2c2. 3.401. v = 112rqB1m, p = r/2. 3.402. N = W led) = 5.106revolutions,^ s =^ 2ItrN = 8.103km. 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,
- 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/-213o/3a.
3.405. dEldr = B (r0) — 112(B) = 0. 3.406. OW = 2ar2eB/At = 0.10 keV. 3.407. (a) W= Oil (reBlmoc)2—1) moc2; (b) s= WAtIreB. 4.1. (a) See Fig. 27; (b) (vx/a(o)2 (xla)2= 1^ and wx= —w2x.
Fig. 27.
4.2. (a) The amplitude is equal to a/2, and the period is T = at/w, see Fig. 28a; (b) vx = 4w2x (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•
