Exercises with Magnetic Monopoles - Kurt Nalty

More documents

Recommendations

Info

orbit. Some novelties are found, however. The orbital radius is not always an equatorial radius. The electron can orbit at any arbitrary latitude. The next observation is more interesting. A positive charge orbits a positive monopole in a clockwise direction seen from further out, in the lower density B region. If I magically reverse the direction of the electric charge at a point, the new orbit is in a plane at right angles to the previous orbit. A 180 degree change in initial conditions leads to a 90 degree change in the solution. We still orbit in a clockwise direction about the radial flux. I find this fascinating. Here are the formulae for the orbital radius and acceleration of our planar solution. ρ = 1 κ = v2 a a = µq eq m 4π 1 v m ρ = mv 4πr2 µq e q m r 2 We have a little gem hidden in this formula. Suggestively re-arranging our equation, we have ρ = mv 4πr2 µq e q m mvr = µq eq m ρ 4π r This shows that for the case of equatorial orbits, where ρ = r and L = mvr, the classical angular momentum is equal to the field momentum of the electric charge/magnetic monopole pair. Minimum Radius for Equatorial Orbits We have yet another gem hidden in this formula. The maximum value for ρ is ρ = r for an equatorial orbit. Our speed for this system is constant. Rather than looking for ρ, let’s find r as a function of speed and mass. 24
ρ = r = mv 4πr2 µq e q m r = µq eq m 1 4π mv Given the maximum speed is the speed of light, we see we have a minimum separation between electric and magnetic charges for equatorial planar orbits. r min = µq eq m 4π 1 mc Force Ratios We earlier had to dismiss colocated electric and magnetic charges, as the known electron spin greatly exceeds the maximum Maxwell Equation duality spin. Can we go backwards, and estimate the magnetic charge from the known spin? The answer, is yes, of course. L = µq eq m 4π = ¯h/2 q m = ¯h4π 2q e = h µq e = 3.29 · 10 −9 A m So, we have an expression for the magnetic charge, can we figure out ratio of magnetic to electric forces? The answer, is yes, of course. F e = q 2 e 4πɛr 2 F m = µq2 m 4πr 2 F m /F e = µɛq2 m q 2 e = q2 m c 2 q 2 e = 4720 This is in agreement <strong>with</strong> Professor Errede’s notes. 25
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 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: τ = = = = [ µq e q m ⃗a × ⃗r
Page 27 and 28: Without loss of generality, I can p
Page 29: [5] Alexander Russell, “The Magne

Exercises with Magnetic Monopoles - Kurt Nalty

Create successful ePaper yourself

Delete template?

Save as template?