Exercises with Magnetic Monopoles - Kurt Nalty

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Taking our difference, our first integral becomes µq e q [ m −⃗a z 1 + π ] 8π 4 Our second integral evaluates much like first, <strong>with</strong> an overall change in sign. µq e q m −⃗a z 8π ⃗a z µq e q m 8π = ⃗a z µq e q m 8π ∫ R/2 0 ∫ R/2 0 [ ( ) 2r rR R 2 tan−1 − 2r2 ((R/2) 2 − r 2 ) (R/2) 2 − r 2 (r 2 + (R/2) 2 ) 2 R [ ( ) 2r rR R 2 tan−1 − 2r2 (r 2 − (R/2) 2 ) r 2 − (R/2) 2 (r 2 + (R/2) 2 ) 2 R [ − 2Rr ( ) R 2 + 4r + r2 4Rr 2 R 2 tan−1 4r 2 − R 2 ] dr ] dr − r R + 3 2 tan−1 ( 2r R )] R/2 With this integral, we easily see the lower limit is zero. For the upper limit, we need to be a bit careful. Figure 1 shows a plot of the first tan −1 term of this integral. We have a discontinuity at r = R/2. In our zone, we are negative. We will need to take the first inverse tangent as the negative value here. We have terms in the brackets of Our second integral is ⃗a z µq e q m 8π Our total integral becomes − 1 2 − 1 π 4 2 − 1 2 + 3 π 2 4 = −1 + π 4 [ −1 + π ] µq e q [ m = −⃗a z 1 − π ] 4 8π 4 µq e q m −⃗a z 8π 2 = −⃗a µq e q m z 4π This overall result agrees <strong>with</strong> the previous cylindrical integration, and totally misses the concentration of angular momentum between the charges, as shown previously. 0 16
Figure 1: tan −1 (4rR/(4r 2 − R 2 )) 17
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 and 10: Looking at the radial component in
Page 11 and 12: ∫ ρ=∞ ρ=0 ρ 2 ∫ ρ=∞ (
Page 13 and 14: So, x = r cos θ sin φ y = r sin
Page 15: For the zone R/2 < r < ∞, we have
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

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