problemset2

problemset2 - 6.641 Electromagnetic Fields Forces and...

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± ² ³´ µ · · 6.641 Electromagnetic Fields, Forces, and Motion Spring 2009 Problem Set 2 - Solutions Prof. Markus Zahn MIT OpenCourseWare Problem 2.1 A Surface S Curve C φ y x Figure 1: Surface S and contour C for using Ampere’s law (Image by MIT OpenCourseWare.) Step 1: Find ±eld of z -directed line current, I , at x = y = 0. I = I ˆ i z By symmetry: H = H φ i φ in cylindrical coordinates. By Ampere: ˆ H d l = J d a · · C S current going through S (2 πr ) H φ = I I H φ = ˆ i φ (1) 2 πr ˆ i φ = ¸ y ˆ i x + ¸ x ˆ i y x 2 + y 2 x 2 + y 2 H = I ( y ˆ + x ˆ i y ) 2 π ( x 2 + y 2 ) i x Step 2: Find solution by adding two translated H ±elds: H total H I 2 −→ = H I 1 + = ¹ I 1 º d ) ˆ i x + x ˆ + ¹ I 2 º d ) ˆ i x + x ˆ 2 π x 2 + ( y d ) 2 ( y 2 i y 2 π x 2 + ( y + d ) 2 ( y + 2 i y 2 2 1
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± ² ³ ² ´ µ³ ² ³ ² · ³ Problem Set 2 6.641, Spring 2009 B Step 3: We want feld in y = 0 plane so y 0. i I 1 = I,I 2 = 0 H tot i y −→ = 2 π ( x 2 I + d 4 2 ) d 2 ˆ i x + x ˆ ii I 1 = 2 = I = I d 2 H tot x ˆ i y π ( x 2 + 4 ) iii I 1 = 2 = I H tot = [ d ˆ i x ] 2 π ( x 2 I + d 4 2 ) C F v H ) = q × ( µ 0 so d F = dq v × µ 0 H In our problem the line current is a moving line charge, so dq = λdl d F v µ 0 dl = I H ) dl = λ H × ( µ 0 × ¯ ¸ L F = I H ) dl = ( I H ) L × ( µ 0 × µ 0 0 So Force = I H ) × ( µ 0 Length don’t need to know −−→ since I 1 cannot exert a net ±orce on itsel±. So we need a feld I 2 H I 1 I 2 y + d 2 ˆ i x + · x 2 ˆ i y H I 2 = ´ µ 2 π x 2 + y + d 2 I 1 is at x = 0 ,y = d 2 , so x 0 d 2 above. H = I 2 d ˆ i x = I 2 ˆ i x 2 πd 2 2 πd 2
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± ² ³ ´ ³ ´ ³ ´ ³ ´ µ ² = ² Problem Set 2 6.641, Spring 2009 Force = ( H ) = ( I 1 ˆ ) µ 0 I 2 ˆ i x Length I 1 ) × ( µ 0 i z × 2 πd = µ 0 I 1 I 2 ˆ i y 2 πd ( i ) I 1 = I, I 2 = 0 Force = 0 Length ( ii ) I 1 = I 2 = I Force = µ 0 I 2 ˆ i y Length 2 πd ( iii ) I 1 = I 2 = I Force = + µ 0 I 2 ˆ i y Length 2 πd Problem 2.2 A The idea here is similar to applying the chain rule in a 1D problem d 1 ± ³ d 1 ±´³ df ´ f ( x ) = = dx f ( x ) f ( x ) dx f 2 ( x ) f ( x ) corresponds to | r r | . So, by diff. f ( x ) we get part of the answer to the derivative of f ( 1 x ) . But we can just do it directly too.
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problemset2 - 6.641 Electromagnetic Fields Forces and...

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