Exercises with Magnetic Monopoles - Kurt Nalty

More documents

Recommendations

Info

Before we do the next step, I want to point out that the right hand portion after the minus sign is alway positive definite. Thus, I put some absolute value signs in the expression that follows. Inner = (z2 + R 2 /4) − |z2 − R 2 /4| 2R 2 z 2 2R 2 z 2 = (z2 + R 2 /4 − |z 2 − R 2 /4|) 2R 2 z 2 = (4z2 + R 2 − |4z 2 − R 2 |) 8R 2 z 2 Five Zones for the Integral The inner radial integral really has five zones along the z axis. The first is from positive infinity to just outside our positive pole. The second is at the positive pole. The third spans between the positive and negative poles. The fourth is at the negative pole, while the fifth extends from the negative pole to negative infinity. In zones one, three and five, we have an integral of the form above. Let’s evaluate this inner integral. For zones one and five, we have z 2 > R 2 /4, and the inner integral is Inner = 4z2 + R 2 − (4z 2 − R 2 ) 8R 2 z 2 = 2R2 8R 2 z 2 = 1 4z 2 For zone three, we have z 2 < R 2 /4, and the integral is Inner = 4z2 + R 2 − (R 2 − 4z 2 ) 8R 2 z 2 = 8z2 8R 2 z 2 = 1 R 2 We see we have continuity at the poles <strong>with</strong> this expression, but we should really evaluate at points two and four for thoroughness. For this point, I choose z = R/2 and the integral specializes to 10
∫ ρ=∞ ρ=0 ρ 2 ∫ ρ=∞ ( ρ2 + (z − R/2) 2) 3/2 ( ρ2 + (z + R/2) 2) ρdρ = ρ 2 3/2 ρ=0 (ρ 2 ) 3/2 ( ρ 2 + (R) 2) ρdρ 3/2 = ∫ ρ=∞ ρ=0 dρ (ρ 2 + R 2 ) 3/2 Now ∫ ρ=∞ ρ=0 dρ (ρ 2 + R 2 ) 3/2 = [ ] ρ=∞ ρ R 2√ ρ 2 + R 2 ρ=0 = 1 R 2 In a similar fashion, at the other pole, we get the same value. This is satisfying, as our values are continuous across the poles. We are now able to finish evaluating our integral for total field angular momentum. ⃗L = −⃗a z µq e q m R 8π ∫ z [ ∫ ρ=∞ ρ=0 ] ρ 2 ( ρ2 + (z + R/2) 2) 3/2 ( ρ2 + (z − R/2) 2) 3/2 d(ρ2 ) dz ⃗L = −⃗a z µq e q m R 8π = −⃗a z µq e q m R 8π = −⃗a z µq e q m R 8π [ ∫ ∞ ∫ dz R/2 R/2 4z + 2 −R/2 [ [ − 1 ] ∞ [ z ] R/2 + + 4z R/2 R 2 −R/2 [( 0 + 2 ) ( ) R + + 4R R 2 ∫ dz −R/2 R + 2 −∞ [ − 1 4z ] dz 4z 2 ] −R/2 −∞ ( 2 4R + 0 )] ] ⃗L = −⃗a z µq e q m 4π This results agrees <strong>with</strong> Goldhaber and Jackson. We have the same magnitude, irregardless of distance, and the momentum points from the positive 11
Page 1 and 2: Exercises with Magnetic Monopoles K
Page 3 and 4: Field Angular Momentum due to Charg
Page 5 and 6: ⃗∇(ρ e ρ m ) −1 . Each form
Page 7 and 8: We now write the field angular mome
Page 9: Looking at the radial component in
Page 13 and 14: So, x = r cos θ sin φ y = r sin
Page 15 and 16: For the zone R/2 < r < ∞, we have
Page 17 and 18: Figure 1: tan −1 (4rR/(4r 2 − R
Page 19 and 20: if the ratio of magnetic to electri
Page 21 and 22: Electron Motion in a Radial Magneti
Page 23 and 24: τ = = = = [ µq e q m ⃗a × ⃗r
Page 25 and 26: ρ = r = mv 4πr2 µq e q m r = µq
Page 27 and 28: Without loss of generality, I can p
Page 29: [5] Alexander Russell, “The Magne

Exercises with Magnetic Monopoles - Kurt Nalty

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?