Irodov – Problems in General Physics

3.368. (a) E' = 13B — 1.4 nV/m;

cv'1—R2

(b) B' =B

1-132 co

t32

s2 a 0.9 T, 51°.

3.370. B' B 1 — (E I cB) ^.e^2 ., 0.15 mT.
3.371. Suppose the charge q moves in the positive direction of
the x axis of the reference frame K. Let us pass into the frame K'
at whose origin of coordinates this charge is at rest (the x and x'
axes of the two frames coincide and the y and y' axes are parallel).
In the frame K' the field of the charge has the simplest form: E'

= 4n

1

so q 3

r', with the following components in the plane x, y

q , 1 q

4neo r's x 4neo is Y •

Now let us make the reverse transition to the initial frame K. At the

moment when the charge q passes through the origin of coordinates

of the frame K, the x and y projections of the vector r are related

to the x and y' projections of the vector r' as

x = r cos 0 = x^1 1/ 1 — (v/c)^2 , y = r sin 0 = y',

Besides, in accordance with the formulas that are reciprocal to

Eqs. (3.6i),

Ex= Ex, Ey= E;11/- 1-- (vIc) 2.

Solving simultaneously all these equations, we obtain

1 qr 1 — f3 2

E. Exi E yi — 452s 0 r3 (1— (^02) sine 0) 2 / 2
Note that in this case (v = const) the vector E is collinear with
the vector r.
3
3.372. v = V 9 I 2 alelm= 16 km/s.
2
3.373. tan a =a 4 V 2eV 3
3.374. (a) x = 2Eola; (b) w
3.375. t 1l T (T+2m 0 C2) 3.0 ns.
ceE
3.376. w —
eE
(^) m, (1+ Thnoc 2 ) 3 '
3.377. (a) tan 0 =
eEt
V1 — (vo/c) 2 , where e and mo are the charge
movo
and the mass of a proton; (b) vx = vo/V1 + (1— vVc^2 )(eEtImoc^2 )^2.
3.378. a = arcsin (dB V -,v ) = 30°.
3.379. (a) v= reBlm= 100 km/s, T = 2=1 eB = 6.5 Rs; (b) v
ci I/ 1 + (moc/reB1 2 = 0.51 c, T — , 271m9 — 4.1 ns.^
eB 1— (v/cV^2 )
21-9451
(^321)