Introduction to - 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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 terms of 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, the particle 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. 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...
View Full Document

This note was uploaded on 02/17/2010 for the course PHYSICS mae142 taught by Professor Kim during the Spring '10 term at American College of Computer & Information Sciences.

Page1 / 23

Introduction to - 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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online