Irodov – Problems in General Physics

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3.264. him = V 2F11in/RonR.
3.265. P = v 2 B 2 d 2 RI(R pd/S) 2 ; when R = pd/S, the power
is P = P max = 1 14v 2 B 2 dS1p.
3.266. U = 1/4110/2/n2R2ne = 2pV.
3.267. n = jBleE = 2.5.10^28 m--1; almost 1 : 1.
3.268. u 0 = 1/1B = 3.2.10-3 m 2 /(V•s).
3.269. (a) F = 0; (b) F = (11^0 /4n) 2/pm/r^2 , F1-1. B; (c) F
= (p, 0 /4n) 2Ip,,Ir 2 , F r.
3.270. F = (R^0 l4n) 6nR^2 Ipmxl(R^2 + x^2 )^512.
3.271. F = 312R0PimP2,1n14 = 9 nN.
3.272. 2Bx^3 /R 9 R^2 = 0.5 kA.
3.273. B' =B R 2 sine cos 2 a.


3.274. (a) %, H dS = nR 2 B cos (^9) 1)/N-10;
(b) ic;IB dr = (1— [) B1 sin O.
3.275. (a) Isu,. = (^) (b) 1;ot = xI; in opposite directions.
3.276. See Fig. 24.
3.277. B — R0111R2 I
11 1+14 nr •
3.278. B = 2B 0 [11(1
3.279. B = 3B 9 [t/(2
3.280. H, = NIll = 6 kA/m.
3.281. H^ bBliuond =
=0.10 kA/m.
3.282. When b << R, the per-
meability is 12 2aRBI(p, 0 NI
bB) = 3.7-10 3.
3.283. H = 0.06 kA/m, Rmax z 1.0.10 4.
3.284. From the theorem on circulation of the vector H we
obtain
B tio N I
b 11°3-cd H = 1.51 — 0.987H (kA/m).
Besides, B and H are interrelated as shown in Fig. 3.76. The requir-
ed values of H and B must simultaneously satisfy both relations.
Solving this system of equations by means of plotting, we obtain
H z 0.26 kA/m, B z 1.25 T, and p, = B/R 0 H z 4.10 3.
3.285. F z 112 xSB (^21110) •
3.286. (a) xn, = 1/ V 47i; (b) X = RoFmax V e/a/VB 3.6.10-4.
3.287. A 1 / 2 xVB 2 /R 0 •
3.288. e 1 = By178wIa.
3.289. I = Byll(R 111 ,,), where Rp, = M2/(RI. + B2).
3.290. (a) Acp = 112 (o 2 a 2 mle = 3.0 nV; (b) 1 / 2 o1Ba 2 =
= 20 mV.
c
3.291..c E dr = —^1 / 2 c0Bd 2 = —10 mV.
Fig. 24.
A

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