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Introduction to Electrodynamics - ch05

# Introduction to Electrodynamics - ch05 - Chapter 5...

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Chapter 5 Magnetostatics Problem 5.1 Sincev x B points upward, and that is also the direction of the force, q must be I positive. I To find R, in termsof a and d, use the pythagorean theorem: a2 + d2 (R - d)2 + a2 = R2 =? R2 - 2Rd + d2 + a2 = R2 =? R = . 2d p = qBR = I qB (a2 +~) 2d { r"'",,~ RV The cyclotron formula then giyes Problem 5.2 The general solution is (Eq. 5.6): y(t) = CI cos(u;t) + C2 sin(u;t) + ~t + C3; z(t) = C2 cos(u;t) - CI sin(u;t) + C4. (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 = O.So I y(t) = Et/ B; z(t) = 0.1 Does this make sense? The magnetic force is q(v x B) == -q(E/B)Bz == -qE, which exactly cancels the electric force; since there is no net 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

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90 CHAPTER 5. MAGNETOSTATICS 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 + r.vt), Zo = O. z 4~A y -/3 (c) Problem 5.3 (a) From Eq. 5.2, F = q[E + (v x B)] = 0 => E = vB => I v = ~ .1 q v m (b) FromEq. 5.3, mv = qBR =>;;, = FiR = ~ Problem 5.4 Suppose f flows counterclockwise (if not, change the sign of the answer). The force on the left side (toward the left) cancels the force on the right side (toward the right); the force on the top is laB = lak(a/2) = lka2/2, (pointing upward), and the force on the bottom is laB = -lka2j2 (also upward). So the net force is F = I Ika2 z.1 Problem 5.5 (a) I K = 1 , I because the length-perpendicular-to-flow is the circumference. 21ra (b) J = ~ => 1 = / J da = a / !s ds de/> = 21ra / ds = 21raa => a = 2 1 ; J = 1 2 1 .1 s s 1ra 1ras Problem 5.6 (a) v = r.vr, so I K = f7r.vr.1 (b) v = r.vrsin8~ => I J = pr.vrsin8~, I where p == Qj(4j3)1rR3. Problem 5.7 : = ~ Iv pr dr = / (a;:) r dr = - / (V . J)r dr (by the continuity equation). Now product rule #5 says V . (xJ) = x(V . J) + J . (Vx). But Vx = x, so V. (xJ) = x(V .J) + Jx. Thus Iv(V' J)xdr = Iv V . (xJ) dr - Iv Jx dr. The first term is Is xJ . da (by the divergencetheorem), and since J is entirely insideV, it is zeroon the surface S. Therefore Iv(V' J)xdr = - Iv Jx dr, or, combiningthis with the y and z components, Iv(V' J)rdr = - Iv J dr. Or, referring back to the first line, ii = / J dr. Qed Problem 5.8 . . I~oll (a) Use Eq. 5.35, wIth z = R,82 = -81 = 45°, and four sIdes: B = ~. (b) z = R, 82= -81 = ;, and n sides: B = ~~~ sin(1r/n).
91 . n/-LoI ( 7r ) I/-LOII . (c) For small B, sm B ~ B. So as n -t 00, B -t 2;R, ;: = 2R (same as Eq. 5.38, wIth z = 0). Problem 5.9 (a) The straight segments produce no field at P. The two quarter-circles give B = I ¥ (~- i) I(out).

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