Electromagnetism Answers to Problem Set 10 Spring 2006

Transcription

1 Electromgnetism 76 Answers to Problem Set 1 Spring 6 1. Jckson Prob. 5.15: Shielded Bifilr Circuit: Two wires crrying oppositely directed currents re surrounded by cylindricl shell of inner rdius, outer rdius b, nd reltive permebility µ r. ( Determine the mgnetic potentil for two wires; the first is locted t x = d/ nd crries current I in the -z direction nd the second is locted t x = d/ nd crries current I in the z direction. Φ m = I ln ( + d / d cos φ + d / + d cos φ I cos φ, where φ is mesured counter-clockwise from the x xis. (b Find the potentil in the three regions. We my ssume tht only terms in the expnsion of the potentil in cylindricl coordintes proportionl to cos φ contribute: ] Φ m = cos φ < = A + κ B + C ] cos φ < < b = D cos φ b <, where κ = I/( nd where (A, B, C, D re unknown expnsion coefficients to be determined by boundry coefficients on the two surfces = nd = b. These conditions; Φ m continuous nd norml component of B continuous, led to the equtions: A + κ = B + C ( B + C A + κ = µ r B + C b = D b µ r ( B + C b = D b Solving, we find ( b κ ( µ r 1 ] Φ = (b (µ r + 1 (µ r 1 + κ cos φ κ(µ r 1 Φ b = b (µ r + 1 (µ r 1 + b ] κ(µ r b (µ r + 1 (µ r 1 cos φ b ] κµ r 1 Φ c = b (µ r + 1 (µ r 1 cos φ 1

2 Substituting into the erlier expression gives explicit results for Φ m in ech region. In prticulr, outside the shield we find dipole potentil with coefficient proportionl to tht of the two wires (κ; the coefficient of proportionlity is F = b µ r b (µ r + 1 (µ r 1. In problem 5.1, uniform externl field mintined its form but ws reduced in strength inside cylindricl shield. Here n internl dipole field mintins it s form but is reduced in strength outside cylindricl shield. (c For µ r 1 nd b = + t with t b, we find (µ r =, b = 1.5 cm, t =.3 mm F b tµ r =.17. Jckson: Prob. 5.: For conducting plne with circulr hole nd tngentil field H on one side: ( Determine H (1 on the side with H for >. We hve for z = Φ (1 (, φ = H For >, we find H = Φ(1 = H j 1 (kj 1 (kdk sin φ ( = H ( sin 1 1 sin φ. >. H φ = Φ(1 φ = H sin 1 1 ( ] cos φ sin 1 ( sin φ H x = H cos φ H φ sin φ = H = H 3 xy 3 sin φ

3 H y = H sin φ + H φ cos φ = H = H sin 1 ( 3 y + H sin 1 3 ( ] (b Sketch the surfce currents bove nd below the plne. Above the plne both H nd H (1 contribute to the current: (K x, K y = ( H H (1 y, H (1 x ] cos φ while below, only H (1 contributes: (K x, K y = ( H (1 y, H (1 x 3. Jckson Prob. 5.5: A rectngulr loop crrying current I 1 intercts with wire crrying current I. The center of the loop is distnce d from the wire nd two sides of the loop of length re prllel to the wire; the sides of length b mke ngle α with the plne of the wire nd the line from the wire to the center of the loop The direction of the current in the side nerest the wire is in the sme direction s I. Set up coordinte system with the loop in the xy plne nd center of the loop t the origin; 3

4 the sides run prllel to y nd re locted t x = ±b/; the sides b re prllel to the x xis. The wire locted t z = d sin α, x = d cos α nd I flows long +y. In this coordinte system, the vector potentil of the wire hs only y component nd A = I ln(x d cos α + (z d sin α ] ŷ ( The interction energy is W 1 = I 1 dl 1 A = µ I 1 I ln = µ I 1 I ln ( b/ d cos α + ( d sin α ] (b/ d cos α + ( d sin α d + b ] + db cos α d + b db cos α where only the two sides prllel to y contribute. (b Clculte the force on the loop We hve in the xy plne B x (x, = A y d sin α z = z= (x d cos α + d sin α B z (x, = A y (x d cos α x = z= (x d cos α + d sin α The force on the two sides of the rectngle of length b precisely cncel. The x nd z components of the force on the two sides of length re F x = I 1 B z (b/, B z ( b/, ] = µ I 1 I b ( d cos(α b (b 8d cos(αb + 16d 8µ I 1 I b d sin(α F z = I B x (b/ B z ( b/, = (b 8d cos(αb + 16d (c Repet for the cse where the rectngle of sides, b is replced by circle of rdius. In this cse, we write W 1 = I 1 cos φ A y ( cos φ, dφ, where we hve used the fct tht x = cos φ nd dl y = cos φ dφ long the circle. We expnd the A y in series in powers of 1/d nd crry out the integrl term by term to find ( cos(α W 1 = I 1 I µ + cos(3α3 d 8d 3 + cos(5α5 16d cos(7α7 18d cos(9α9 56d 9

5 The sme series results if we evlute W 1 = I 1 Φ, where Φ is the mgnetic flux through the circle. Note tht the term in prentheses bove cn be written ( = R { } z + z3 8 + z z z9, 56 Where z = d eiα Moreover, 1 1 z z = z + z3 8 + z z z9 56 Thus, we my write { 1 } 1 z W 1 = I 1 I µ R z with z = d eiα This correct nswer is close (but not identicl to the nswer given in the text. Indeed, if we ssumed { 1 } 1 z R = 1 1 (Rz, z Rz then we would recover the result in the text. Find the force. F = î I 1 cos φ B z ( cos φ, dφ ˆk I 1 cos φ B x ( cos φ, dφ Agin, expnding the potentil nd crrying out the integrtions leds to We find F x = µ I 1 I ( cos(α d F z = µ I 1 I ( sin(α d + 3 cos(α 8d + 5 cos(6α6 16d 6 35 cos(8α8 + 18d sin(α 8d + 5 sin(6α6 16d 6 35 sin(8α8 + 18d cos(1α1 56d 1 63 sin(1α1 56d 1 Agin, we cn identify the two series: Consider the function G(z = 1 1 = z 1 z + 3z 8 + 5z z z1 56 5

6 Compring, we find ( F x = µ I 1 I RG d eiα ( F z = µ I 1 I IG d eiα (d Express the energies for lrge d in terms of moments of loops. For the rectngulr loop: W 1 = µ I 1 I d ln + b ] + db cos α d + b db cos α µ I 1 I b cos α d For the circulr loop: ( cos(α W 1 = I 1 I µ d I 1 I µ cos(α d = (I 1 b µ I d cos α = m 1B z + cos(3α3 8d 3 + cos(5α5 16d 5 + = (I 1 µ I d cos α = m 1B z In both cses, the + sign is result of the fct tht the moment nd the norml component of the field re in opposite directions.. Jckson Prob. 5.3: Two identicl circulr loops re locted distnce R prt on common xis, ( Find M 1 using A φ from Prob. 5.1b: A φ (, z = µ I 1 J 1 (kj 1 (ke k z dk W 1 = I dφa φ (, R = µ I 1 I J 1 (kj 1 (ke kr dk Leding to the result M 1 = µ J 1 (k] e kr dk (b Assuming << R, we obtin n symptotic series in R by expnding J ( k in power series Integrting, we find J 1 (k] k M 1 = µ k k k ( 3 R 3 35 R R R 9 + 6

7 (c Find the mutul inductnce for co-plner loops with centers seprted by R. The xil B z field from the loop centered t the origin is B z (z = µ I 1 + z ] 3/ This field cn be derived from sclr potentil Φ m (z = µ ] I 1 z 1 z + = µ I 1 ( z 3 8z z z z 1 + Anlyticlly continuing the potentil leds to. Φ m (r, θ = µ I 1 ( r P 1(cos θ 3 8r P 3(cos θ r 6 P 5(cos θ r 8 P 7(cos θ r 1 P 9(cos θ + We need B z t lrge vlues of r nd θ = /. We find B z (r, / = 1 Φ m r θ θ=/ = µ ( I 1 r r r r r 11 Introduce the vector centered on the second loop. Then we my write bmr = R +, where R is the vecror from the center of the first loop to the center of the second. We my replce r R + + R cos φ where φ is the polr ngle with respect to the center of the second loop nd crry out second expnsion of B z with respect to R. With this in hnd, we clculte the flux Φ the through the second loop. First, integrting B z over the polr ngle φ, we obtin B z dφ = µ { ( I 1 9 R 3 + ( R 5 1 R 7 ( n } 1 R 9 7

8 To evlute the flux through the second loop, we integrte the previous result over Φ = d B z dφ = µ I 1 ( 3 R R R R 9 Given tht Φ = M 1 I 1, we my write M 1 = µ ( 3 R R R R 9 (d clculte the force in ech cse. For the co-plnr loops, the only non-vnishing component of the force on the second loop is F x = I cos φb z ( =, φdφ = µ I 1 I ( 3 R R R R 1 The force is repulsive nd long the line joining the centers of the loops. In the cse of the co-xil loops, only the component F z of the force on the second loop contributes: F z = cos θf r sin θf θ. Now, t the loction of the second loop, components of the force re F r = I B θ (r, θ F θ = I B r (r, θ, where r = + z nd θ = rccos(z/ + z. Therefore, F z = I (cos θb θ (r, θ sin θb r (r, θ Substituting nd expnding the fields in z, one obtins F z = I 1 I µ ( 3 R R R R 1 The force is ttrctive nd long z. 8