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# problemset1 - 6.641 Electromagnetic Fields Forces and...

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± C 6.641 Electromagnetic Fields, Forces, and Motion Spring 2009 Problem Set 1 - Solutions Prof. Markus Zahn MIT OpenCourseWare 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 v y ˆ i y pos. charge carriers v = v y ˆ i y neg. charge carriers B i z = B 0 ˆ so E v y B 0 ˆ i x pos. charge carriers = v y B 0 ˆ i x neg. charge carriers B ² d ² 0 v H = Φ( x = d ) Φ( x = 0) = E x dx = E x dx 0 d v y B 0 d pos. charges v H = v y B 0 d 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 ρ b r ; r < b ρ = b ρ a ; b < r < a Also, no σ s at r = b , but non zero σ s such that E = 0 for r > a . 1

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± ² ³´ µ 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: 0 E ± d±a = ρdV ; S r = sphere with radius r · S R V R As shown in class, symmetry ensures E ± has only radial compoent: E ± = E r ˆ i r LHS of Gauss’ Law: ± 2 π ± π · ¶ · 0 E ± = 0 E r ˆ i r r 2 sin θdθdφ ˆ i r · · S R 0 0 ² ³´ µ da in spherical coord. = 4 πr 2 E r 0 surface area of sphere of radius r RHS of Gauss’ Law: For r < b : ± ± r ± 2 π ± π ρdV = ρr r 2 sin θdθdφdr b ² ³´ µ V R 0 0 0 diﬀ. vol. element 2
±² ³ ´ µ · ¸ ¹ C Problem Set 1 6.641, Spring 2009 4 πr 4 πr 4 ρ b = ρ b = 4 b b vol

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problemset1 - 6.641 Electromagnetic Fields Forces and...

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