Exercises with Magnetic Monopoles - Kurt Nalty

Without loss of generality, I can place the positive magnetic charge at
(0, 0, R/2) and the negative magnetic charge at (0, 0, −R/2). The resulting
magnetic dipole field is
⃗B = µq m
4π
[
⃗r − (R/2)⃗a z
(
x2 + y 2 + (z − R/2) 2) 3/2 − ⃗r + (R/2)⃗a z
(
x2 + y 2 + (z + R/2) 2) 3/2
]
For the special case of the midplane between the two charges, we have
⃗B z=0 = µq m
4π
The force on an electric charge at ⃗r is
⃗F = q e (⃗v × B)
[
⃗
= µq eq m
4π
−R⃗a z
(x 2 + y 2 + R 2 /4) 3/2
⃗v × (⃗r − (R/2)⃗a z )
(
x2 + y 2 + (z − R/2) 2) 3/2 − ⃗v × (⃗r + (R/2)⃗a z )
(
x2 + y 2 + (z + R/2) 2) 3/2
]
The acceleration on an electric charge at ⃗r is
⃗a = ⃗ F
m
= µq eq m
4πm
[
]
⃗v × (⃗r − (R/2)⃗a z )
(
x2 + y 2 + (z − R/2) 2) − ⃗v × (⃗r + (R/2)⃗a z )
3/2 (
x2 + y 2 + (z + R/2) 2) 3/2
⃗κ =
The curvature is
⃗a × ⃗v
v 3
= µq eq m
4πm
[
]
[⃗v × (⃗r − (R/2)⃗a z )] × ⃗v
v ( 3 x 2 + y 2 + (z − R/2) 2) − [⃗v × (⃗r + (R/2)⃗a z)] × ⃗v
3/2
v ( 3 x 2 + y 2 + (z + R/2) 2) 3/2
For the special case of motion confined to the midplane, we have
[
]
⃗κ z=0 = µq eq m 1 −R⃗a z
4π mv (x 2 + y 2 + R 2 /4) 3/2
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