midterm207_sol

midterm207_sol - Cornell University Department of Physics...

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Unformatted text preview: Cornell University Department of Physics Classical Electrodynamics P561 Second Exam - solutions by Johannes Heinonen November 29, 2007 1. Multipole moments a) The charge density can be written as ρ = ρ √ 4 πY 00 ( θ, φ ) + ρ 1 radicalbigg 4 π 3 Y 10 ( θ, φ ) . So we get the sperical multipole moments q 00 = integraldisplay d Ω drr 2 Y ∗ 00 ρ = √ 4 π R 3 3 ρ = radicalbigg 1 4 π q q 1 m = integraldisplay d Ω drr 3 Y ∗ 1 m ρ = radicalbigg 4 π 3 R 4 4 ρ 1 δ m = radicalbigg 3 4 π ( p z δ m + p x ± ip y √ 2 δ m, ± 1 ) q 2 m = integraldisplay d Ω drr 4 Y ∗ 2 m ρ = 0 ∼ Q lm , where we used some abusive notation for the x,y-components of the dipole equation. Thus the multimpole moments in cartesian coordinates read q = 4 π 3 ρ R 3 vector p = π 3 ρ 1 R 4 ˆ z Q lm = 0 The electric field far away from the setup can be calculated from Jackson (4.11) to be E r = 4 π 1 1 √ 4 π q 1 r 2 1 √ 4 π + 8 π 3 radicalbigg 3 4 π p z 1 r 3 radicalbigg 3 4 π cos θ = q r 2 + 2 p z cos θ r 3 = 4 πρ 3 R 3 r 2 + 2 πρ 1 3 R 4 cos θ r 3 E θ = − 8 π 3 radicalbigg 3 4 π p z 1 r 3 ∂ ∂θ radicalbigg 3 4 π cos θ = 2 p z sin θ r 3 = πρ 1 3 R 4 sin θ r 3 E ϕ = 0 or in cartesian coordinates vector E = q r 2 vector r r + 3( vector p · vector r ) vector r − r 2 vector p r 5 = 4 π 3 ρ R 3 r 2 ˆ r + π 3 ρ 1 R 4 r 4 (3 z ˆ r − r ˆ z ) . (Note: We don’t have to care about possible delta functions at vectorx = 0, since our expansion in multipoles assumes that we are far away from the source anyway.) b) Using vectorω = ω ˆ z and vectorv = vectorω × vector r we get the current density vector J = ρvectorω × vector r and therefore the magnetic dipole moment vectorm = 1 2 c integraldisplay dvector rvector r × vector J = 1 2 integraldisplay dvector r θ ( R − r ) ρvector r × ( vector r × vectorω ) = 1 2 c integraldisplay dvector r θ ( R − r ) ρ bracketleftbig r 2 vectorω − ( vector r · vectorω ) vector r bracketrightbig = 1 2 c integraldisplay d Ω dr r 2 θ ( R − r ) parenleftBigg ρ √ 4 πY ∗ 00 + ρ 1 radicalbigg 4 π 3 Y ∗ 10 parenrightBigg ωr 2 √ 4 πY 00 1 − ωz x y z . Now we can use that xz and yz is only proportinal to Y 2 , ± 1 , thus those integrals van- ish (or equivalently, the integral over sin ϕ and cos ϕ vanishes) and furthermore we know that z 2 = r 2 2 3 ( √ 4 π 2 Y 00 + radicalBig 4 π 5 Y 20 ). So we end up with vectorm = 1 2 c integraldisplay d Ω dr r 2 θ ( R − r ) parenleftBigg ρ √ 4 πY ∗ 00 + ρ 1 radicalbigg 4 π 3 Y ∗ 10 parenrightBigg 2 3 ωr 2 √ 4 πY 00 1 = 1 3 4 π c ρ R 5 5 ω ˆ z vectorm = 4 π 15 · ρ R 5 c vectorω The magnetic field created by this magnetic dipole is vector B = 3( vectorm · vector r ) vector r − r 2 vectorm r 5 = 4 π 15 ρ R 5 r 4 (3 z ˆ r − r ˆ z ) ....
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This note was uploaded on 10/25/2009 for the course ECE 4300 taught by Professor Lipson/pollock during the Fall '08 term at Cornell.

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midterm207_sol - Cornell University Department of Physics...

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