文档介绍：Chapter 5
ostatics
Problem
Sincev x B points upward, and that is also the direction of the force, q must be I positive. I To findR, in
terms of a and d, use the pythagorean theorem:
a2 + d2
(R - d)2 + a2 R2 R2 2Rd + d2 + a2 R2
= =? - = =?R = 2d .
The cyclotron formula then giyes r"'",,~
p = qBR = IqB (a2 +~)
2d RV{
Problem
The general solution is (Eq. ):
z(t) C2cos(u;t) - CI sin(u;t) + C4.
y(t) = CI cos(u;t) + C2 sin(u;t) + ~t + C3; =
(a) y(O) = z(O) = OJ y(O) = E/ Bj i(O) = O. Use these to determine CI, C2, C3, and C4.
y(O)= 0 =? CI + C3 = OJ y(O) = u;C2 + E/B = E/B =? C2 = OJ z(O) = 0 =? C2 + C4 = 0 =? C4 = 0;
i(O)= 0 =?CI = 0, and hence also C3 = I y(t) = Et/ B; z(t) = Does this make sense? The ic
forceis q(v x B) ==-q(E/B)Bz ==-qE, which exactly cancels the electric force; since there is force,
theparticle moves in a straight line at constant speed. ..(
(b) Assumingit starts from the origin, so C3 = -CI, C4 = -C2, we have i(O) = 0 =? CI = 0 =? C3 = 0;
y(O)= 2~ =? C2u; + ~ = 2~ =? C2 = - 2~B = -C4; y(t) = - 2~B sin(u;t) + ~ t;
E E E. E
z(t) = - 2u;B cos(u;t) + 2u;B' or y(t) = 2u;B [2u;t- sm(u;t)]; z(t) = 2u;B [1 - cos(u;t)]. Let (3 == E/2u;B.
Then y(t) = (3[2u;t - sin(u;t)]; z(t) = (3[1 - cos(u;t)]; (y - 2(3u;t) = -(3 sin(u;t), (z - (3) = -(3 cos(u;t) =?
(y- 2(3VJt)2+ (z - (3)2= (32. This is a circle of radius (3 whose center moves to the right at constant speed:
Yo=2(3VJt; Zo = (3.
. . E E E E E
(c) z(O) = y(O) = B =? -ClUJ = B =? CI = -C3 = - u;B j C2u; + B = B =? C2 = C4 = O.
89
90 CHAPTER 5. OSTATICS
E E E E. E E
y(t) = - wB cos(wt) + Bt + wB; z(t) = wB sm(wt). y(t) = wB [1+ wt - cos(wt)] j z(t) = wB sin(wt).
Let /3 ==EjwB; then [y - /3(1 + wi)] = -(3cos(wt), z = {3sin(wt)j [y - {3(1+ wtW + z2 = {32. This is a circle
of radius {3 whose center is at Yo = (3(1 + ), Zo = O. z
y
-/3
4~A (c)
Problem
(a) From Eq. , F = q[E + (v x B)] = 0 =>E = vB =>I v = ~ .1
q v m
(