Lorentz Force Law-Electromegnatic Field-Assignment, Exercises of Electromagnetism and Electromagnetic Fields Theory

Download Lorentz Force Law-Electromegnatic Field-Assignment and more Exercises Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! � � � � C Problem 1.1 A −→ F → v − B ) Lorentz Force Law = q( − E + − ×→→ In the steady state − F = 0, so → q −→ E v − B → v → = −q−→×→⇒ −E = −− × − B→ vy ̂iy pos. charge carriers v =→− −vy ̂iy neg. charge carriers − B iz → = B0ˆ so − E −vyB0îx pos. charge carriers → = vyB0îx neg. charge carriers B � d � 0 vH = Φ(x = d) − Φ(x = 0) = − Exdx = Exdx 0 d vyB0d pos. charges vH = −vyB0d neg. charges As seen in part (b), positive and negative charge carriers give opposite polarity voltages, so answer is “yes.” Problem 1.2 By problem ρbr ; r < b ρ = b ρa; b < r < a Also, no σs at r = b, but non zero σs such that E� = 0 for r > a. 1 docsity.com � � � �� � Problem Set 1 6.641, Spring 2009 - - - - - - - - - - - + + + + + + + + + + + v Force caused by B Force caused by E (which is caused by charge on electrodes) Force caused by E Force caused by B E v E B Figure 1: Figure for 1.1C. Opposite polarity voltages between holes and electrons (Image by MIT Open- CourseWare.) A By Gauss’ Law: �0E� d�a = ρdV ; Sr = sphere with radius r · SR VR As shown in class, symmetry ensures E� has only radial compoent: E� = Er ̂ir LHS of Gauss’ Law: � � 2π � π � � � � �0E� d�a = �0 Er ̂ir r 2 sin θdθdφîr· · SR 0 0 � �� � d�a in spherical coord. = 4πr2 Er�0 surface area of sphere of radius r RHS of Gauss’ Law: For r < b: � � r � 2π � π ρdV = ρr r 2 sin θdθdφdr b � �� �VR 0 0 0 diff. vol. element 2 docsity.com 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 � � � C �H · d�s = S �J · d�a + d dt � r �0 �E �� · d�a � �no E field, term is 0 Choose C as a circle and S as the minimum surface that circle bounds. Now solve LHS of Ampere’s Law � � 2π H� d�s = (Hφîφ) (rdφ)îφ = 2πrHφ· � �� � · � �� �C 0 � d�sH 5 docsity.com � � �� � Problem Set 1 6.641, Spring 2009 We assumed Hz = Hr = 0. This follows from the symmetry of the problem. Hr = 0 because S µ0H � d�a = · 0. In particular choose S as shown in Figure 4a. 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 � � 2π � r � � � � J� · d�a = J0 b r� î z · r�dr�dφî z S 0 0 � �� � � �� � J� d�a 2J0r 3π = 3b a > r > b � � 2π � b � J0r� � � � � 2π � r � � � � J� d�a = î z r �dr�dφî z + 0 î z r �dr�dφî z S · 0 0 b · 0 b · · 0 = 2 J0b 2π 3 Equating LHS & RHS: ⎧ ⎨ 32 b J0r3π ; r < b 2πrHφ = 2 J0b2π ; a > r > b ⎩ 3 0 ; r > a 6 docsity.com � � � � � � � ���� Problem Set 1 6.641, Spring 2009 ⎧ 2 ⎨ J30 rb îφ ; r < b H� = ⎩ J30 rb2 îφ ; a > r > b 0 ; 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 · → = (ε0 )(surface area of Sr → d−a Er ) SR symmetry = 4πr2ε0Er Equate LHS and RHS 4πr2ε0Er = q − E q îr → = 4πr2ε0 Convert to cartesian: Any point is given by −r = x(r, θ, φ)̂ix + y(r, θ, φ)̂ + z(r, θ, φ)̂iz→ iy 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 φˆ + r cos θîz → iy îr � line formed by varying r and fixing φ and θ r̄ = rīr Thus, īr = sin θ cos φî x + sin θ sin φîy + cos θîz = � x î x + � y î y + � z î z x2 + y2 + z2 x2 + y2 + z2 x2 + y2 + z2 7 docsity.com