6.641 Electromagnetic Fields, Forces, and Motion
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MIT OpenCourseWare http://ocw.mit.edu 6.641 Electromagnetic Fields, Forces, and Motion Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 6.641 — Electromagnetic Fields, Forces, and Motion Spring 2009 Problem Set 2 - Solutions Prof. Markus Zahn MIT OpenCourseWare Problem 2.1 A y φ x Surface S Curve C Figure 1: Surface S and contour C for using Ampere’s law (Image by MIT OpenCourseWare.) → − Step 1: Find field of z-directed line current, I , at x = y = 0. − → I = I îz → − By symmetry: H = Hφˆiφ in cylindrical coordinates. By Ampere: � � → − − → − → → H ·d l = J · d− a C S � �� � current going through S (2πr)Hφ = I Hφ = I îφ 2πr îφ = � → − H = −y x2 + (1) y2 x îx + � îy 2 x + y 2 I (−yˆix + xîy ) 2π(x2 + y 2 ) → − Step 2: Find solution by adding two translated H fields: − → → − → − H total = H I1 + H I2 � � � � I1 d I2 d ˆ ˆ � −(y − )îx + xiy + � −(y + )îx + xiy � � = 2 2 2π x2 + (y − d2 )2 2π x2 + (y + d2 )2 1 Problem Set 2 6.641, Spring 2009 B Step 3: We want field in y = 0 plane so y → 0. i I1 = I, I2 = 0 − → H tot = I 2π(x2 + � d îx + xˆiy d2 2 ) 4 � ii I1 = I, I2 = I − → H tot = I 2 π(x + � d2 4 ) xîy � iii I1 = I, I2 = −I → − H tot = I + 2π(x2 [dîx ] d2 4 ) C → − → − − F = q→ v × (µ0 H ) so � � �� → − → − → d F = dq − v × µ0 H In our problem the line current is a moving line charge, so dq = λdl � � → − → − → − − d F = λ→ v × µ0 H dl = I¯ × (µ0 H )dl → − F = � 0 L → − − → → → − − I × (µ0 H )dl = ( I × µ0 H )L So Force → → − − = I × (µ0 H ) Length −−→ We don’t need to know HI1 since I1 cannot exert a net force on itself. So we need a field of I2 � � � � d I − y + î + x î 2 x y 2 −−→ � HI2 = � �2 � 2 2π x + y + d2 I1 is at x = 0, y = d2 , so x → 0, y → d 2 above. −I2 → −I2 d − H = îx = îx 2πd2 2πd 2 Problem Set 2 6.641, Spring 2009 Force → → − − = ( I1 ) × (µ0 H ) = (I1ˆiz ) × Length −µ0 I1 I2 îy 2πd (i) I1 = I, � � −µ0 I2 ˆ ix 2πd = I2 = 0 Force =0 Length (ii) I1 = I, I2 = I Force −µ0 I 2 = îy Length 2πd (iii) I2 = −I I1 = I, Force µ0 I 2 =+ îy Length 2πd Problem 2.2 A The idea here is similar to applying the chain rule in a 1D problem d dx � 1 f (x) � � d = df � 1 f (x) �� � � � df −f (x) = 2 dx f (x) − → f (x) corresponds to |→ r −− r � |. So, by diff. f (x) we get part of the answer to the derivative of can just do it directly too. � − − |→ r −→ r � | = (x − x� )2 + (y − y � )2 + (z − z � )2 � � � � � � � � 1 ∂ 1 ∂ 1 ∂ 1 = îx + îy + îz � − → → → − − − − ∂x |− ∂y |→ ∂z |→ r −→ r �| |→ r −− r �| r −− r �| r −→ r � | So we can apply the trick above by just considering x, y, and z components separately. � ∂ → ∂ �� − |− r −→ r �| = (x − x� )2 + (y − y � )2 + (z − z � )2 ∂x ∂x x − x� = � (x − x� )2 + (y − y � )2 + (z − z � )2 x − x� = → − → |r −− r � | Similarly, ∂ ∂x ∂ → z − z � ∂ − y − y� − − |→ r −→ r �| = → |− r −→ r �| = → . � and − → − − − ∂y ∂z |r − r | |r −→ r �| � 1 → − |− r −→ r� � = ∂ → − − ∂x |− r −→ r �| − − |→ r −→ r � |2 and so on for y and z. → → |− r −− r � |2 = (x − x� )2 + (y − y � )2 + (z − z � )2 3 1 f (x) . But we Problem Set 2 6.641, Spring 2009 so: � � � − (x − x� )î + (y − y � )iˆ + (z − z � )iˆ x y z 1 � − = 3 − |→ r −→ r �| [(x − x� )2 + (y − y � )2 + (z − z � )2 ] 2 � 3 − − Denominators = |→ r −→ r � | 2 . Thus, � � − − − − −1 1 −(→ r −→ r �) (→ r −→ r �) � − = − � � 3 = − � 2 → → → − → → − → → − − → − |r − r | |r − r | | r − r | | r − r �| −îr� r = → → |− r −− r � |2 B Follows from (A) immediately by substitution. Remember � is derived in terms of unprimed x, y, z. � does not affect x� , y � , z � . C − Φ(→ r)= −� → � V� − ρ(→ r � )dV � − − 4πε0 |→ r −→ r �| C . In this sense, ρ → ∞ at the ring. We can represent ρ( r ) = charge density in mC3 . We have λ in units of m � − this in cylindrical coordinates by ρ(→ r ) = λ0 δ(z)δ(r − a). Then we can evaluate the triple integral � � � λ0 δ(z)δ(r − a)rdrdφdz = → − 4π�0 |− r −→ r �| 2π � 0 λ0 adφ 4π�0 |r − r� | But, we can skip that unnecessary work by simply considering infinitesimal charges (adφ)λ0 around the ring. y x dφ a adφ Figure 2: A ring of line charge with infinitesimal charge elements dq = λ0 adφ. (Image by MIT OpenCourseWare.) We only care about z axis in this problem as well, so, by symmetry, there is no field in the x and y directions. 4 Problem Set 2 → Φ(− r)= − Φ(→ r)= 6.641, Spring 2009 2π � λ0 (adφ) 4πε0 0 1 (a2 + z 2 ) 2 � �� � distance from the charge element λ0 adφ to the point z on the z-axis λ0 a 1 2ε0 (a2 + z 2 ) 2 on the z-axis. 0 0 ⎛ ⎞ ∂ � ∂ ⎟ → − ⎜ ∂ � → E = −�Φ(− r ) = − ⎝îx � Φ + îy �Φ + îz Φ⎠ ∂x ∂y ∂z � � ∂ → − E = −îz ∂z � → − E = îz aλ0 z � λ0 a 1 2ε0 (a2 + z 2 ) 2 3 2ε0 (a2 + z 2 ) 2 − Using the equation from the Problem 2.2 Statement with z component only (symmetry) and with ρ(→ r � )dV � → λ0 adφ � 2π λ0 adφ cos θ z Ez (z) = , cos θ = 1 4πε0 (z 2 + a2 ) (a2 + z 2 ) 2 0 � 2π = 0 = λ0 az (a2 + 3 z2) 2 dφ 4πε0 λ0 az 3 2ε0 (a2 + z 2 ) 2 Limit |z | → ∞ � a2 + z 2 → |z | Φ(z) ≈ λ0 a 2ε0 (a2 + 1 z2) 2 ≈ 2πλ0 a Q ≈ 4πε0 |z| 4πε0 |z| Q = 2πλ0 a (total charge on loop). Φ(z) looks like potential from point charge in far field. � Q z>0 λ0 az λ0 az 2 0 |z| Ez = = 4πε−Q 3 ≈ 3 2 2 2ε | z | 2 z<0 2ε0 (a + z ) 0 4πε |z|2 0 5 Problem Set 2 6.641, Spring 2009 a r dr Figure 3: A line charge ring of width dr in the disk (Image by MIT OpenCourseWare.) D From (C), Φ = λ0 r 1 2ε0 (r 2 +z 2 ) 2 for a ring of radius r. But now we have σ0 , not λ0 . How do we express λ0 in terms of σ0 ? Take a ring of width dr in the disk (see figure). Total charge in the ring = (r)(2π) (dr)σ0 . � �� � circumference Line charge density = λ0 = So: λ0 = σ0 dr total charge length = σ0 dr σ0 rdr dΦ = 1 2ε0 (r2 + z 2 ) 2 � a � a σ0 rdr σ0 rdr Φtotal = 1 = 1 2 2 2 2ε 2 0 0 (r + z 2 ) 2 0 2ε0 (r + z ) ��r=a � σ0 �� 2 σ0 �� 2 � = = r + z2 � a + z 2 − |z | 2ε0 2ε0 r=0 � � σ0 1 1 − → √ √ E = −�Φtotal = z − iz 2ε0 a2 + z 2 z2 E As z → ∞, 1 (a2 + z 2 ) 2 → |z| + Φtotal → a2 ; 2|z | 1 (a2 + z 2 )− 2 → 1 |z| � 1− πa2 σ0 4�0 π |z | 2 ¯ → πa σ0 iz E 4πε0 z 2 just like a point charge of σ0 πa2 . F As a → ∞, z in the Φtotal → √ a2 + z 2 can be neglected, so σ0 [a − |z|] 2ε0 6 a2 2z 2 � Problem Set 2 6.641, Spring 2009 � � � σ0 σ0 z 1 0 − 0 = 2ε Ez → −σ0 2ε0 |z| 2ε z>0 z<0 0 just like a sheet charge. Problem 2.3 A By the divergence theorem: i � V − → � · (� × A )dV = � S → − − (� × A ) · d→ a where S encloses V . ii By Stokes’ Theorem: � � → → − − − → − → (� × A ) · d a = A ·d l S� C Suppose S is as in figure 4 S Figure 4: Closed surface S (Image by MIT OpenCourseWare.) and S � is as in figure 5 S ! C Figure 5: Open surface S � bounded by contour C (Image by MIT OpenCourseWare.) i.e., S � is the same as S, except for the contour curve C, which makes S � slightly unclosed. Now consider limit as C → 0 (Figure 6) 7 Problem Set 2 6.641, Spring 2009 Figure 6: Limit as C → 0 (Image by MIT OpenCourseWare.) � → � − → − → − → In limit C → 0, S � → S. If C is 0, then C A · d l = 0. By equation (ii), S (� × A ) · d− a = 0. By � − → equation (i), V � · (� × A )dV = 0. Since V can be any volume, argument of integral must be identically 0. → − � · (� × A ) = 0 B − → A = Axˆix + Ayˆiy + Azˆiz �� � � � � � � ∂Az ∂Ay ∂Ax ∂Az ˆ ∂Ay ∂Ax → − − � · (� × A ) = � · − îx + − iy + îz ∂x ∂y ∂y ∂z ∂z ∂x � � � � � � ∂ ∂Az ∂Ay ∂ ∂Ax ∂Az ∂ ∂Ay ∂Ax − − − = + + ∂x ∂y ∂z ∂y ∂z ∂x ∂z ∂x ∂y ∂ 2 Az ∂ 2 Ay ∂ 2 Ax ∂ 2 Az ∂ 2 Ay ∂ 2 Ax − + − + − ∂x∂y ∂y∂z ∂y∂x ∂z∂x ∂z∂y ∂x∂z = 0 because of interchangeability of partial derivatives = In cylindrical coordinates Ā = Ar īr + Aφ īφ + Az īz � � � � � � 1 ∂Az ∂Aφ ∂Ar ∂Az 1 ∂(rAφ ) ∂Ar − − − � × Ā = ¯ir + ¯iφ + ¯iz r ∂φ ∂z ∂z ∂r r ∂r ∂φ � � � � � � ¯ = 1 ∂ ∂Az − r ∂Aφ + 1 ∂ ∂Ar − ∂Az + ∂ 1 ∂(rAφ ) − 1 ∂Ar � · (� × A) r ∂r ∂φ ∂z r ∂φ ∂z ∂r ∂z r ∂r r ∂φ 1 ∂ 2 Az r ∂ 2 Aφ 1 ∂Aφ 1 ∂ 2 Ar 1 ∂ 2 Az ∂ 2 Aφ 1 ∂Aφ 1 ∂ 2 Ar − − − + − + + r ∂r∂φ r ∂r∂z r ∂φ∂z ∂r∂z r ∂z r ∂φ∂z r ∂z r ∂r∂φ = 0 = Problem 2.4 (Zahn, Problem 23, Chapter 1) A Cartesian Cylindrical Spherical hx = 1 hr = 1 hr = 1 hφ = r hθ = r hy = 1 hz = 1 hφ = r sin θ hz = 1 8 Problem Set 2 6.641, Spring 2009 B df = ∂f ∂f ∂f du + dv + dw ∂u ∂v ∂w = �f · d� = �f · [hu duīu + hv dv īv + hw dwīw ] 1 ∂f 1 ∂f 1 ∂f ; (�f )v = ; (�f )w = hu ∂u hv ∂v hw ∂w 1 ∂f 1 ∂f 1 ∂f �f = īu + īv + īw hu ∂u hv ∂v hw ∂w (�f )u = c) dSu = hv hw dvdw; dSv = hu hw dudw; dSw = hu hv dudv dV = hu hv hw dudvdw D Divergence � � Φ= A¯ · dS = S 1,u � + 2,v+Δv 3,w+Δw � Au hv hw dvdw 1� ,u−Δu � Av hu hw dudw − � + = Au hv hw dvdw − Av hu hw dudw 2� ,v Aw hu hv dudv − � Aw hu hv dudv 3� ,w � � Au hv hw |u − Au hv hw |u−Δu Av hu hw |v+Δv − Av hu hw |v Aw hu hv |w+Δw − Aw hu hv |w + + ΔuΔvΔw Δu Δv Δw � � ¯ · dS A A¯ · dS S S � · Ā = lim = Δu→0,Δv→0,Δw→0 ΔV hu hv hw ΔuΔvΔw � � 1 ∂(hv hw Au ) ∂(hu hw Av ) ∂(hu hv Aw ) = + + hu hv hw ∂u ∂v ∂w ¯ u (Image by MIT OpenCourseWare.) Figure 7: Surface for determining (� × A) Curl � ¯u= (� × A) Ā· d� Δv→0,Δw→0 hv hw ΔvΔw lim L 9 Problem Set 2 � L 6.641, Spring 2009 A¯ · d� = [Av hv Δv|w − Av hv Δv|w+Δw ] + [Aw hw Δw|v+Δv − Aw hw Δw|v ] (� × Ā)u = = 1 hv hw � 1 Δv→0,Δw→0 hv hw � lim ∂(hw Aw ) ∂(hv Av ) − ∂v ∂w Av hv |w − Av hv |w+Δw Aw hw |v+Δv − Aw hw |v + Δw Δv � � Similarly � ∂(hu Au ) ∂(hw Aw ) − ∂w ∂u � � 1 ∂(hv Av ) ∂(hu Au ) ¯ − (� × A)w = hu hv ∂u ∂v ¯v= (� × A) 1 hu hw � E �2 f = � · (�f ) = 1 hu hv hw � ∂ ∂u � hv hw ∂f hu ∂u � + ∂ ∂v 10 � hu hw ∂f hv ∂v � + ∂ ∂w � hu hv ∂f hw ∂w ��
