MIT OpenCourseWare http://ocw.mit.edu 6.641 �� Electromagnetic Fields, Forces, and Motion Spring 2005 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Spring 2005 6.641 — Electromagnetic Fields, Forces, and Motion Problem Set 3 - Solutions Prof. Markus Zahn MIT OpenCourseWare Problem 3.1 A (x,y,z) z r + d 2 d 2 +q r r- x -q Figure 1: Addition of potential contributions from 2 point charges. (Image by MIT OpenCourseWare.) We can simply add the potential contributions of each point charge: q q Φ= − 4πε0 r+ 4πε0 r− � d r+ = x2 + y 2 + (z − )2 2 � d r− = x2 + y 2 + (z + )2 2 ⎡ ⎤ q ⎣ 1 1 ⎦ � Φ= −� 4π�0 d 2 d 2 2 2 2 2 x + y + (z − ) x + y + (z + ) 2 2 B p = qd. We must make some approximations. As r → ∞, �r+ , �r− , and �r become nearly parallel. Thus, r+ ≈ r − a = r − r+ ≈ r(1 − d cos θ 2 d cos θ) 2r Similarly, r− ≈ r(1 + d cos θ) 2r 1 Problem Set 3 6.641, Spring 2005 z r+ r r- θ x a= d cosθ 2 Figure 2: Differences in lengths between �r+ , �r− , and �r (Image by MIT OpenCourseWare.) By part (a): Φ = | q 4πε0 d cos θ| � 1 2r � 1 r+ − 1 r− � . If |x| � 1, then 1 1+x so 1 1 1 1 ≈ ≈ d r+ r 1 − 2r cos θ r � � d 1+ cos θ 2r 1 1 1 1 ≈ ≈ d r− r 1 + 2r cos θ r � ⇒ 1− d cos θ 2r � 1 1 1d d − ≈ cos θ = 2 cos θ r+ r− rr r Φ≈ p cos θ qd cos θ = 4πε0 r 2 4πε0 r 2 C ∂Φ 1 ∂Φ 1 ∂Φ ˆ − → E = −�Φ = − îr − îθ − iφ ∂r r ∂θ r sin θ ∂φ p cos θ ∂Φ p sin θ ∂Φ =− ; =− 3 ∂r 2πε0 r ∂θ 4πε0 r 2 ∂Φ =0 ∂φ p cos θ ˆ 1 p sin θ ˆ − → E = ir + iθ 2πε0 r 3 r 4πε0 r 2 � � p − → 2 cos θ îr + sin θ îθ E = 3 4πε0 r 2 ≈1−x Problem Set 3 6.641, Spring 2005 D Er 2 cos θ dr = = = 2 cot θ rdθ Eθ sin θ 1 dr = 2 cot θdθ r � � 1 dr = 2 cot θdθ r ln r = 2 ln(sin θ) + k r = C sin2 θ; when θ = π , r = C = r0 2 Thus, C = r0 , r = sin2 θ r0 Figure 3.2b) Used with permission. Zahn, Markus. In Electromagnetic Field Theory: A Problem Solving Approach, 1987. 3 Problem Set 3 6.641, Spring 2005 Plots of Equipotential and Field Lines θ 0.75 0.5 0.25 -2 r0 = 1 r0 = .5 1 -1 r0 = 2 2 E Field Lines -0.25 -0.5 -0.75 r0 = 0.25 Figure 4: Polar plot of r0 sin2 θ for 0 ≤ θ ≤ π and for r0 = 0.25, 0.5, 1, and 2 meters with 100 volt−1 -m−2 (Image by MIT OpenCourseWare.) = Φ = .0025 Equipotential Lines 2 1 4πε0 p Φ = .01 Φ = 0.04 Φ = .16 -1 -0.5 0.5 1 Φ=0 -1 -2 p cos θ Figure 5: Polar plot of equipotential lines Φ = 4πε 2 for 0 ≤ θ ≤ π, Φ = 0, ±0.0025, ±0.01, ±0.04, ±0.16, 0r −1 4πε0 −2 and ±0.64 volts with p = 100 volt -m (Image by MIT OpenCourseWare.) 4 Problem Set 3 6.641, Spring 2005 Figure 6: The superposition of the previous two plots of perpendicular equipotential and field lines (Image by MIT OpenCourseWare.) Problem 3.2 A We can think of the bird as a perfectly conducting small sphere when it lands on the uninsulated wire, it must become the same potential as the wire. This forces it to acquire a charge. When it flies away, the charge stays with it because air is a poor conductor. B, C For B and C, use the method of images. We can use superposition to get total potential. � � 1 q 1 Φ= − 1 4πε0 [(x − U t)2 + (y − h)2 + z 2 ] 12 [(x − U t)2 + (y + h)2 + z 2 ] 2 where q is the charged bird modeled as a point charge. D By boundary condition found using Gauss’ Law − →a − →b n̂ · (εa E − εb E ) = σs at y = 0 ground plane boundary − →b E =0 5 Problem Set 3 6.641, Spring 2005 y +q bird at (Ut, h, z) Φ=0 on xz-plane x z -q image at (Ut, -h, z) Figure 7: Figure for 3.2 B, C. Method of Images for charged bird taken as a point charge (Image by MIT OpenCourseWare.) a q n b Figure 8: Figure for 3.2 D. Field lines from point charge above a perfectly conducting ground plane (Image by MIT OpenCourseWare.) − →a Because we can consider ground plane to be a perfect conductor, n̂ · E = σs ε0 . σs − → − → (îy ) · ( E (x, y = 0+ , z)) = implies we only care about y components of E ε0 Ey (x, y = 0+ , z) = σs ε0 � ∂ q Ey = − Φ = ∂y 4πε0 (1) (y − h) 3 [(x − U t)2 + (y − h)2 + z 2 ] 2 Evaluate at y = 0 and substitute into (1) above: � � −2h q Ey (x, y = 0, z) = 4πε0 [(x − U t)2 + h2 + z 2 ] 23 So σs = ε0 Ey (x, y = 0, z) σs = −qh 3 2π [(x − U t)2 + h2 + z 2 ] 2 6 − (y + h) 3 [(x − U t)2 + (y + h)2 + z 2 ] 2 � Problem Set 3 6.641, Spring 2005 E Q= � w 0 � 0 l −qh 3 2π [(x − U t)2 + h2 + z 2 ] 2 dxdz For w very small, σs does not change significantly from z = 0 to z = w, so integral in z becomes just multiplication at z = 0. Q= � 0 l −qhw dx 2π [(x − U t)2 + h2 ] Let x� = x − U t ⇒ dx� = dx So: � l−Ut −qhw � Q= 3 dx � 2 2 2 −Ut 2π [((x ) + h ] � � qw l − Ut Ut � Q=− +� 2πh (l − U t)2 + h2 (U t)2 + h2 q (1) (2) l Ut (3) Figure 9: Representative shape q = Q versus U t (Image by MIT OpenCourseWare.) F dQ −qw i= = dt 2πh � −U h2 � U h2 3 + 3 [(l − U t)2 + h2 ] 2 [(U t)2 + h2 ] 2 � � qwR U h2 −U h2 V = −iR = + 3 2πh [(l − U t)2 + h2 ] 32 [(U t)2 + h2 ] 2 v l /U t Figure 10: Voltage V versus time across small electrode resistance R (Image by MIT OpenCourseWare.) 7 Problem Set 3 6.641, Spring 2005 P = origin II y a I III x IV b Figure 11: Magnetic field from rectangular line current (Image by MIT OpenCourseWare.) Problem 3.3 A 1 − → H = 4π 1 − → H = 4π → J(− r � ) × îr� r � dv → → |− r −− r � |2 � a y= � 2 (I îy ) × � b 2 îx � − yiˆy dy b x= � 2 � � (I îx ) × −xîx − a2 îy dx �� � � �� � � 12 + a 2 a 2 2 2 +x +x 2 2 �� � � �� � � 21 + 2 2 b b y=− a x=− 2b 2 + y2 + y2 z=0 z=0 2 2 b y= a x=− 2 2 �� � � �� � � � 2 − → − → normalization for î � |r −r | r r � � � � b a x= y= � 2 (−I îy ) × − b îx − y îy dy � 2 (−I îx ) × −xîx + a îy dx 2 2 �� �2 �� � � �� �2 � 12 + � �� � � 12 2 2 b a b a 2 2 2 2 b + x + x + x + x y=− a x=− 2 2 2 2 2 2 x= 2b z=0 ⎡ y=− a 2 z=0 1 ⎢ ⎣2 4π � x= 2b � ⎤ y= a 2 a I(−îz )dx b I(−îz )dy ⎥ � 32 + 2 � 23 ⎦ �� �2 a 2 2 �� a �2 y=− 2 b 2 2 + x + x 2 2 � a2 ⎤ � 2b ⎡ � � � � I îz ⎢ ax by � ⎥ � ⎥ =− + ⎣ � � �� � 1 � 1 � � � � ⎦ � b �2 � b �2 4π 2 � 2 � a 2 a 2 2 2 � � + x + y 2 2 2 2 b a = x=− 2b −2 −2 ⎡ ⎤ 4ab I îz ⎢ − → ⎥ � �4ab H =− ⎣ �� � � 12 + �� �2 � � � 12 ⎦ � � 4π 2 2 2 � a2 − a2 + 2b b2 2b + a2 � 2 � 2 − → −2I a + b îz H = 1 πab(a2 + b2 ) 2 8 Problem Set 3 6.641, Spring 2005 1 − → −2I(a2 + b2 ) 2 H = îz πab B Figure 12: Line current in circular coil (Image by MIT OpenCourseWare.) − → I = −I îφ 1 − → H = 4π = 1 4π � π 0 � � 2π 0 � −I îφ × (−îr )adφ a2 −îz Iadφ a2 I − → H = − îz 2a C I a P Figure 13: Line current with semi-circular bump (Image by MIT OpenCourseWare.) 1 − → H = 4π � 0 π (−I îφ ) × (−ˆir )adφ a2 9 Problem Set 3 6.641, Spring 2005 I − → H = − îz 4a D Figure 14: Line current with rectangular bump (Image by MIT OpenCourseWare.) As in part (c), contributions from segments I and V are zero (see Fig. 14). Segments II, III, and IV are just like part (a), except integrals in y are from 0 to a and only one integral in x and ( a2 ) → a. − → H = � � �2 � 12 −I a2 + 2b πab ˆiz Problem 3.4 A îr� r � π � 2π � π � 2π (K0 îφ ) × ˆir� r R2 sin θdφdθ → → |− r −− r � |2 0 0 − → → − → r −− r� r � = − = − = −î�r � → → |→ r −− r | |− r �| 1 − → H = 4π − → K0 H = 4π K0 =− 4π 0 � 0 π 0 � → (− r = 0) 2 � (îφ ) × (−îr )R � sin θdφdθ 2 R� � 2π (sin θ)(cos θ cos φîx + cos θ sin φîy − sin θ îz )dφdθ 0 Any term� with an odd power of sin or cos in φ integrates to 0 in φ because integral is over one period. − → − → π 0 H = îz 2πK sin2 θdθ = K40 π îz = H 4π 0 10 Problem Set 3 6.641, Spring 2005 B This requires us to integrate an infinite number of infinitesimal current shells of the type in (a) from r = R1 to R2 . K0 − → H = � � �� � (J0 dr) π ˆ − → J0 π iz dr ⇒ H = (R2 − R1 )îz 4 4 R2 R1 Problem 3.5 Figure 15: Line current above perfectly conducting plane with image current (Image by MIT OpenCourseWare.) A − → µ0 A = 4π � � ∞ −∞ µ0 Idz � îz � − 4π x2 + (y − d)2 + (z − z � )2 � ∞ −∞ � Idz � îz � x2 + (y + d)2 + (z − z � )2 Let ξ ≡ z − z ⇒ dξ = dz . Both integrands are even functions in ξ. � � � µ0 I ∞ 1 1 − → � A = îz dξ −� 2π 0 x2 + (y − d)2 + ξ 2 x2 + (y + d)2 + ξ 2 ��ξ=∞ � �� x2 + (y − d)2 + ξ 2 ξ µ0 I � ln +� = îz 2π x2 + (y − d)2 x2 + (y − d)2 ξ=0 � �� ��ξ=∞ x2 + (y + d)2 + ξ 2 µ0 I ξ � − îz ln +� 2π x2 + (y + d)2 x2 + (y + d)2 ξ=0 �� � x2 + (y + d)2 − → µ0 I ln � îz A = 2π x2 + (y − d)2 11 Problem Set 3 6.641, Spring 2005 B 1 − → − → H = �× A µ0 � � � � I(y − d) Ix Ix I(y + d) − îx − − îy = 2π(x2 + (y + d)2 ) 2π(x2 + (y − d)2 ) 2π(x2 + (y + d)2 ) 2π(x2 + (y − d)2 ) C � −Hx �y=0+ = Kz − → K =− Id îz + d2 ) π(x2 D Force comes from the image current − → − → F = (Ilîz ) × (µ0 H (x = 0, y = d)) = µ0 I 2 l îy 4πd F µ0 I 2 = îy l 4πd 12