Irodov – Problems in General Physics

(Joyce) #1

3.236. B = Ron//1/1 (2R//) 2.
3.237. (a) B 1 / 2 p,^0 n1 (1 — x/V x 2 + R 2 ), where x> 0 outside
the solenoid and x < 0 inside the solenoid; see Fig. 23; (b) xo =


R (1— 21)12V 1(1— 5R.

3.238. B = .{(1-411h)11 (1— (h12311)^2 = 0.3 mT, r<R,
(11 0 /431) 21/r, r > R.
3.239. r1 x N/Jt = 8.10 2.

= 1.0 pWb/m.

3.240. (^) (110/47t) = 70 B/6a
3.241. (121 = (1 20 0 /2 = kn/S/2, •^8
where (Do is the flux of the vec-^ 0.
for B through the cross-section of^ 0#
the solenoid far from its ends.
3.242. (I) = (11 0 /4n) 2INh In 1=^0.^2
= 8 [Mb.
3.243. p m = 23-cR 3 BI11 0 =
= 30 mA • m 2.
3.244. pm = d 2 =
=0.5 A • m^2.
3.245. a B
LOIN (b/a) 7 N;
() 2 (b— a)
(b) Pm = 113 nIN (a 2 ab b 2 ) = 15 mA • m^2.
3.246. (a) B = 1 / 2 p. 0 60)R; (b) pm =^1 /^4 260)R^4.
3.247. B = 2 / 3 p. 0 60)R = 29 pT.
3.248. pm = 1 / 5 qR 2 co; pml M = q/2m.
3.249. B = 0.
3.250. FmlFe = 11 080722 = (v/c) 2 = 1.00.10-6.
3.251. (a) F (^1) = 11 0 .1 2 /4R = 0.20 mN/m; (b) F1 = 120/21211 =
= 0.13 mN/m.
3.252. B = ad^2 aml4RI = 8 kT, where am is the strength of
copper.
3.253. B = (2pgS I I) tan 0 = 10 mT, where p is the density
of copper.
3.254. B = 0.4 T.
3.255. (a) F = 4.^0 .11^0 1n (41^2 — 1) = 0.40 p.N;^ (b)^ A =
= (11oall^0 /n) In )(21^ 1)/(2ri — 1)] = 0.10 J.
3.256. R Yμ 0 /s 0 (ln i)/n = 0.36 kS2.
3.257. F 1 = p, 01 .2/3-1, 2 R.
3.258. F1= 4
2111 2
In (1 + b/a).
3.259. F 1 = /3 1 /21..t 0.
3.260. In all three cases F 1 = (B: — B:)/211^0. The force is direct-
ed to the right. The current in the conducting plane is directed
beyond the drawing.
3.261. Ap = IBla = 0.5 kPa.
3.262. p = p. 0 /^2 /8n^2 R^2.
3.263. p = 1/2p.0n212•
0
Fig. 23.
,Z'/?

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