6.641 �� Electromagnetic Fields, Forces, and Motion
advertisement
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 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 = IˆiZ − → By symmetry: H = Hφˆiφ in cylindrical coordinates. By Ampere: � � → − → − − → − J · d→ a H ·d l = S C � �� � current going through S (2πr)Hφ = I I îφ 2πr x −y îx + � îy îφ = � 2 2 2 x +y x + y 2 Hφ = − → H = I 2π(x2 + y2 ) (1) (−y îx + xiˆy ) − → Step 2: Find solution by adding two translated H fields: − → − → − → H total = H I1 + H I2 � � � � d d I1 I2 � −(y − )îx + xîy + � −(y + )îx + xîy � � = 2 2 2π x2 + (y − d2 )2 2π x2 + (y − d2 )2 B Step 3: We want field in y = 0 plane so y → 0. 1 Problem Set 2 6.641, Spring 2005 i I1 = I, I2 = 0 − → H tot = I 2π(x2 + � d îx + xîy 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 d2 4 ) [dˆix ] 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 So L − → − → − → − → I × (µ0 H )dl = ( I × µ0 H )L Force − → − → = I × (µ0 H ) Length − → We don’t need to know H since I1 cannot exert a net force on itself. So we need a field of I2 � � � � d ˆ I − y + i + x î 2 x y 2 − → � H = � �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 Force − → − → = ( I1 ) × (µ0 H ) = (I1ˆiz ) × Length � � −µ0 I2 ˆ ix 2πd 2 Problem Set 2 6.641, Spring 2005 −µ0 I1 I2 ˆ iy 2πd (i) I1 = I, = I2 = 0 Force =0 Length (ii) I1 = I, I2 = I 2 −µ0 I Force = îy Length 2πd (iii) I1 = I, 2 I2 = −I Force µ0 I =+ 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 ∂ ∂ ∂ = î + î + î x y z � � � � → → → → → → → → ∂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 ∂ − → ∂y | r � � � � � ∂ − → − ∂x |→ r −− r | = − � → |→ r −− r |2 y−y → −− r |= − and � → |→ r −− r | 1 � − → → |r −− r ∂ − → ∂z | r � z−z → −− r |= − � . → |→ r −− r | � 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 2005 so: � � � � 1 −(x − x )îx = � 3 + ...similar terms for y and z → → |− r −− r | [(x − x� )2 + (y − y � )2 + (z − z � )2 ] 2 → → Denominators = |− r −− r | 2 . Thus, � � � � → → → → −(− r ) −1 (− r −− r ) 1 r −− = − = → � − � � � � → − − → → |− r −→ r |2 |− r − → r | |→ r −− r | |→ r −− r |3 � 3 −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 � → → r −− r | 4πε0 |− � � 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)rdφdθdr � → → 4πε0 |− 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. (Image by MIT OpenCourseWare.) We only care about z axis in this as well, so, by symmetry, there is no field in x and y directions. � 2π λ0 (adφ) → Φ(− r)= 1 4πε0 (a2 + z 2 ) 2 0 � �� � distance from the charge λ0 adφ to the point z on the z-axis → Φ(− r)= λ0 a 1 2ε0 (a2 + z 2 ) 2 4 Problem Set 2 6.641, Spring 2005 on z-axis. ⎛ ⎞ 0 0 ∂ � ∂ ⎟ → − ⎜ ∂ � → E = −�Φ(− r ) = − ⎝îx � Φ + îy �Φ + ˆiz 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 Eq. (5) with z component only (symmetry) and with ρ(− r )dV → λ0 adφ � Ez (z) = � 2π 0 = = � 2π 0 � λ0 adφ cos θ z , cos θ = 1 2 2 2 4πε0 (z + a ) (a + z 2 ) 2 λ0 az dφ 3 (a2 + z 2 ) 2 4πε0 λ0 az 3 2ε0 (a2 + z 2 ) 2 Limit |z| → ∞ � a2 + z 2 → |z| Φ(z) ≈ λ0 a 2ε0 (a2 + z 2 ) 1 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 2ε0 |z | z<0 2ε0 (a2 + z 2 ) 2 4πε |z|2 0 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 = (r)(2π) (dr)σ0 . Line charge � �� � circumference a r dr Figure 3: A ring of width dr in the disk (Image by MIT OpenCourseWare.) 5 Problem Set 2 6.641, Spring 2005 density = λ0 = total charge length = σ0 dr So: λ0 = σ0 dr σ0 rdr Φ= 1 2ε0 (r 2 + z 2 ) 2 � a � a rdr σ0 rdr σ0 Φtotal = = 2 + z 2 ) 21 2 + z 2 ) 12 2ε 2ε (r (r 0 0 0 0 ��r=a � �� σ σ0 �� 2 � 0 r + z2 � = a2 + z 2 − |z | = 2ε0 2ε0 r=0 � � 1 σ0 1 − → iz E = −�Φtotal = z √ −√ 2ε0 a2 + z 2 z2 E As z → ∞, a2 (a + z ) → z + ; 2z 2 2 Φtotal → 1 2 2 − 21 2 (a + z ) 1 → z � � a2 1− 2 2z πa2 σ0 4�0 πz 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 � � � σ0 σ0 z 1 ε0 Ez → − 0 = 2−σ 0 2ε0 |z| 2ε 0 z>0 z<0 just like sheet charge. Problem 2.3 A By the divergence theorem: i � V − → � · (� × A )dV = � S − → → (� × A ) · d− a where S encloses V . By Stokes’ Theorem: 6 Problem Set 2 6.641, Spring 2005 S Figure 4: Closed surface S (Image by MIT OpenCourseWare.) S � C � Figure 5: Open surface S (Image by MIT OpenCourseWare.) ii � S � − → → (� × A ) · d− a = � C → − → − A ·d l Suppose S is as in figure 4 � and S is as in figure 5 � � i.e. S is the same as S, except for the curve C, which makes S slightly unclosed. Now consider limit as C → 0 (Figure 6) 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 7 Problem Set 2 6.641, Spring 2005 � � � � � � ∂Ay ˆ ∂Ax ∂Az ˆ ∂Ay ∂Ax ˆ ∂Az − ix + − iy + − iz ∂y ∂z ∂z ∂x ∂x ∂y � � � � � � ∂Ay ∂ ∂Ax ∂Az ∂ ∂Ay ∂Ax ∂ ∂Az + + = − − − ∂x ∂y ∂z ∂y ∂z ∂x ∂z ∂x ∂y − → � · (� × A ) = � · = �� ∂ 2 Az ∂ 2 Ay ∂ 2 Ax ∂ 2 Az ∂ 2 Ay ∂ 2 Ax − + − + − ∂x∂y ∂x∂z ∂y∂z ∂y∂x ∂z∂x ∂z∂y = 0 because of interchangeability of partial derivatives In cylindrical coordinates A¯ = Ar¯ir + Aφ¯iφ + Az¯iz � � � � � � ∂Ar 1 ∂(rAφ ) ∂Ar 1 ∂Az ∂Aφ ∂Az + īφ + īz � × Ā = īr − − − 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 = r ∂ 2 Aφ 1 ∂Aφ 1 ∂ 2 Ar 1 ∂ 2 Az ∂ 2 Aφ 1 ∂Aφ 1 ∂ 2 Ar 1 ∂ 2 Az − − + − + + − r ∂r∂φ r ∂r∂z r ∂z r ∂φ∂z r ∂r∂φ ∂r∂z r ∂z r ∂φ∂z = 0 Problem 2.4 A Q= � ∞ λ(z)dz = � a −a −∞ B Φ(r̄) = � V� � λ0 z λ0 z 2 ��+a =0 dz = a 2a −a � ρ(r̄ )dv 4πε0 |r̄ − r̄ � | z a 0 −a Figure 7: Axis showing extent of line charge −a < z < a for Problem 2.4B (Image by MIT OpenCourseWare.) Φ(r = 0, z) = � a −a � � λ0 z dz 4πε0 |z − z � |a 8 Problem Set 2 Φ(r = 0, z) = ¯ = −�Φ E 6.641, Spring 2005 λ0 4πε0 a � a −a � � � � � � z z+a λ0 dz = z ln − 2a for z > a z − z� z−a 4πε0 a � � � � λ0 ∂ z+a z ln − 2a îz 4πε0 a ∂z z − a � � � � z+a 2az λ0 ¯ iz =− ln − (z + a)(z − a) 4π ε0 a z−a � � �� λ0 2az z+a − ln = īz 4πε0 a (z + a)(z − a) z − a = C 1 1 z → ∞ : ln(1 + x) = x − x2 + x3 . . . x � 1 2 3 � � λ0 −a 1 a 2 1 a 3 a 1 a 2 1 a 3 Φ(z → ∞) = z( − ( ) + ( ) − ( − ( ) − ( ) )) − 2a z 2 z 3 z z 2 z 3 z 4πε0 a 2 2 a λ0 2 a3 λ0 z 3 = 3 4πε0 a 3 z 4πε0 z 2 � � λ0 2az a 2 a 1 a 2 1 a 3 a 1 a 2 1 a 3 Ez (z → ∞) = (1 + ( ) )) − ( − ( ) + ( ) + + ( ) + ( ) ) z z 2 z 3 z z 2 z 3 z 4πε0 a z 2 = = 4 2 a λ0 λ0 4 a 3 ( ) = 3 4πε0 a 3 z 4πε0 z 3 It’s a solution for dipole along z axis D 2 2 a λ0 3 Check: Zahn, pp. 139-140 � � p̄ = r̄dq ⇒ pz = p= all q a −a zλ(z)dz = � a −a λ0 z 2 λ0 z 3 �� a 2 dz = = λ0 a 2 a 3a −a 3 9
