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 1 - Solutions Prof. Markus Zahn MIT OpenCourseWare Problem 1.1 A − → → → − − − F = q( E + → v × B ) Lorentz Force Law → − In the steady state F = 0, so → − − → − − → → → − q E = −q − v × B ⇒ E = −→ v ×B � vy îy pos. charge carriers − → v = −vy îy neg. charge carriers → − B = B0ˆiz so → − E = � −vy B0 îx vy B0 îx pos. charge carriers neg. charge carriers B � vH = Φ(x = d) − Φ(x = 0) = − vH = � 0 Ex dx = 0 � d Ex dx d vy B 0 d pos. charges −vy B0 d neg. charges C As seen in part (b), positive and negative charge carriers give opposite polarity voltages, so answer is “yes.” Problem 1.2 By problem � ρ= ρb r b ; ρa ; r<b b < r < a � = 0 for r > a. Also, no σs at r = b, but non zero σs such that E 1 Problem Set 1 6.641, Spring 2009 E - + - v + - + Force caused by E + + + B + + + + + - - v E Force caused by E (which is caused by charge on electrodes) Force caused by B - Force caused by B Figure 1: Figure for 1.1C. Opposite polarity voltages between holes and electrons (Image by MIT OpenCourseWare.) A By Gauss’ Law: � � · d�a = �0 E � ρdV ; Sr = sphere with radius r SR VR � has only radial compoent: E � = Er îr As shown in class, symmetry ensures E LHS of Gauss’ Law: � � · d�a = �0 E SR 2π � 0 π � 0 � � � � �0 Er îr · r2 sin θdθdφîr � �� � d� a in spherical coord. = 2 4πr Er �0 � �� � surface area of sphere of radius r RHS of Gauss’ Law: For r < b: � � r � 2π π � ρdV = VR 0 0 0 ρr 2 r sin θdθdφdr �� � b � diff. vol. element 2 Problem Set 1 = 6.641, Spring 2009 4 πr4 πr4 ρb ρb = b �4 ��b � vol of sphere For r > b & r < a (b < r < a): � � b 2π � π � ρdV = VR 0 0 0 ρb r 2 r sin θdθdφdr + b r � 2π � b 0 � π ρa r2 sin θdθdφdr 0 4πρb b3 4πρa (r3 − b3 ) =� + 3 4� 4 = πρb b3 + πρa (r3 − b3 ) b < r < a 3 B Equating LHS and RHS ⎧ 4 ⎪ ⎨ πr ρb ; b 4πr2 Er �0 = 3 3 ⎪ ⎩ πρb b3 + 4πρa (a − b ) ; 3 ⎧ 2 r ρb ⎪ ⎨ ; 4�0 b Er = 3 3 3 ⎪ ⎩ b ρb + ρa (r − b ) ; 4�0 r2 3�0 r2 r < b b<r<a r<b b<r<a C Again: n̂ · (�0 E a − �0 E b ) = σs � (r = a+ ) = 0 E ρb b3 ρa (a3 − b3 ) + ← by part (a) 4�0 a2 3�0 a2 � � � (r = a− ) , so: σs = îr · −�0 E � � ρb b3 ρa (a3 − b3 ) σs = − + 4�0 a2 3�0 a2 Er (r = a− ) = D r<b Qb = πb3 ρb b < r < a Qa = 43 π(a3 − b3 )ρa Qσ (r = a) = σs 4πa2 = −4πa2 Qσ = Qb + Qa + Qσ = 0 3 � ρb b3 4�0 a2 + ρa (a3 −b3 ) 3�0 a2 � Problem Set 1 6.641, Spring 2009 Figure 2: A diagram of a wire carrying a non-uniform current density and the return current at r = a (Image by MIT OpenCourseWare). Problem 1.3 A We are told current in +z direction inside cylinder r < b Current going through cylinder: � = Itotal = J� · d�a = b � S 0 2π � 0 � � J 2πb2 J0 r ˆ � 0 iz · rdφdriˆz = b 3 � �� � � �� � � J� �|= |K Total current in sheet length of sheet (ie, circumference of circle of radius a) � ’s units are Thus, K �|= |K d� a 2 2 3 J0 πb 2πa = Amps m , whereas J�’s units are Amps m2 J 0 b2 3a � = − J0 b2 iˆz K 3a 4 Problem Set 1 6.641, Spring 2009 B Figure 3: A diagram of the wire with a circle C centered on the z-axis with minimum surface S (Image by MIT OpenCourseWare). Ampere’s Law � � · d�s = H C � J� · d�a + S d dt � � r � · d�a �0 E � �� � field, term is 0 no E Choose C as a circle and S as the minimum surface that circle bounds. Now solve LHS of Ampere’s Law � C � · d�s = H 2π � 0 (Hφ iˆφ ) · (rdφ)iˆφ = 2πrHφ � �� � � �� � � H d� s 5 Problem Set 1 6.641, Spring 2009 We assumed Hz = Hr = 0. This follows from the symmetry of the problem. Hr = 0 because 0. In particular choose S as shown in Figure 4a. � S � · d�a = µ0 H Figure 4: A diagram of the current carrying wire of radius b with the choice for S as well as a diagram of the wire with the choice of contour C (Image by MIT OpenCourseWare). Hz is more difficult to see. It is discussed in Haus & Melcher. The basic idea is to use the contour, C (depicted in Figure 4b), to show that if Hz �= 0 it would have to be nonzero even at ∞, which is not possible without sources at ∞. Now for RHS of Ampere’s Law: r<b � J� · d�a = 2π � S 0 � r 0 � J0 r � ˆ � � � ˆ � iz · r dr dφiz b �� � � �� � � � d� a J� 2J0 r3 π 3b a > r > b = � J� · d�a = 2π � S 0 � b � 0 J0 r � ˆ iz b � � � � 2π � r � � � � � � ˆ · r dr dφiz + 0 · iˆz · r� dr� dφiˆz b �0 �� � 0 2 J0 b2 π 3 Equating LHS & RHS: = 2πrHφ = ⎧ ⎨ 2 3 3b J0 r π ; 2 2 3 J0 b π ; ⎩ 0; r < b a>r>b r>a 6 Problem Set 1 � = H J0 r 2 ˆ 3b iφ J 0 b2 ˆ 3r iφ ⎧ ⎨ ⎩ 6.641, Spring 2009 ; ; 0; r<b a>r>b r>a Problem 1.4 A We can simply add the fields of the two point charges. Start with the field of a point charge q at the origin and let SR be the sphere of radius R centered at the origin. By Gauss: � � → → − − ε0 E · d a = ρdV SR V − In this case ρ = δ(→ r )q, so RHS is � � � � − ρdV = δ(→ r )qdxdydz = q LHS is � − → → ε0 Er · d− a = (ε0 SR Er )(surface area of Sr ) ���� symmetry = 4πr2 ε0 Er Equate LHS and RHS 4πr2 ε0 Er = q → − E = q îr 4πr2 ε0 Convert to cartesian: Any point is given by → − r = x(r, θ, φ)îx + y(r, θ, φ)ˆiy + z(r, θ, φ)îz By spherical coordinates x = r sin θ cos φ y = r sin θ sin φ z = r cos θ − → r = r sin θ cos φîx + r sin θ sin φˆiy + r cos θîz îr � line formed by varying r and fixing φ and θ r̄ = ri¯r Thus, i¯r = sin θ cos φiˆx + sin θ sin φîy + cos θîz =� x x2 + y 2 + z 2 y z iˆx + � iˆy + � iˆz x2 + y 2 + z 2 x2 + y 2 + z 2 7 Problem Set 1 6.641, Spring 2009 so, → − E = 4πε0 (x2 q ˆir + y2 + z2 ) − → → → − − E total = E 1 + E 2 → → → − → − − − E 1 is just E with y → y − d2 . E 2 is just E with y → y + d2 . Problem has y = 0 (i) ⎡ − → − → x z E total = E 1 = ⎣ � îx + � îz − � 2 d2 2 2 2 x + 4 +z x + d4 + z 2 x2 + � � q d = xîx − îy + z îz 3 2 2 4πε0 (x2 + d4 + z 2 ) 2 d 2 d2 4 (ii) → − → → − − E total = E 1 + E 2 =q xîx + z îz � �2 3 2πε0 (x2 + d2 + z 2 ) 2 (iii) − → − → → − E total = E 1 + E 2 −dq îy = � 4πε0 x2 + d2 4 + z2 � 32 B − → − → F = q1 E → − E doesn’t include field of q (i) → − F = 0, by Newton’s third law a body cannot exert a net force on itself. (ii) → − − → → − d F = q1 E = q E 2 (x = 0, y = , z = 0) 2 2¯ 2¯ q iy q iy = = 4πε0 (d2 ) 4πε0 d2 (iii) → − q 2 i¯y F =− 4πε0 d2 8 ⎤ � q îy ⎦ · 2+ 4πε (x 2 0 +z � d2 4 + z2)
