Exercises with Magnetic Monopoles - Kurt Nalty

For the zone R/2 < r < ∞, we have the expression
(
)
2
rR
r 3 R 3 tan−1 − 2 (r2 − (R/2) 2 )
r 2 − (R/2) 2 (r 2 + (R/2) 2 ) 2 r 2 R 2
For the zone 0 < r < R/2, we have the expression
(
)
2
rR
r 3 R 3 tan−1 − 2 ((R/2)2 − r 2 )
(R/2) 2 − r 2 (r 2 + (R/2) 2 ) 2 r 2 R 2
We are now ready to finish our integrations.
⃗L = − µq ∫ ∞
[ (
)
eq m
2
2πr 3 rR
R⃗a
16π 2
z
R/2 r 3 R 3 tan−1 − 2 ]
(r2 − (R/2) 2 )
r 2 − (R/2) 2 (r 2 + (R/2) 2 ) 2 rdr
r 2 R 2
− µq ∫ R/2
[ (
)
eq m
2
2πr 3 rR
R⃗a
16π 2 z
0
r 3 R 3 tan−1 − 2 ]
((R/2)2 − r 2 )
(R/2) 2 − r 2 (r 2 + (R/2) 2 ) 2 rdr
r 2 R
∫ 2
µq e q ∞
[ (
)
]
m 2r
rR
⃗L = −⃗a z
8π R/2 R 2 tan−1 − 2r2 (r 2 − (R/2) 2 )
r 2 − (R/2) 2 (r 2 + (R/2) 2 ) 2 dr
R
∫
µq e q R/2
[ (
)
]
m 2r
rR
−⃗a z
8π R 2 tan−1 − 2r2 ((R/2) 2 − r 2 )
(R/2) 2 − r 2 (r 2 + (R/2) 2 ) 2 dr
R
0
Our first integral is
∫
µq e q ∞
[ (
)
m 2r
rR
−⃗a z
8π R/2 R 2 tan−1 r 2 − (R/2) 2
[
µq e q m
= −⃗a z −
2Rr
( )
8π R 2 + 4r + r2 4Rr
2 R 2 tan−1 4r 2 − R 2
]
− 2r2 (r 2 − (R/2) 2 )
(r 2 + (R/2) 2 ) 2 dr
R
− r R + 3 2 tan−1 ( 2r
R
)] ∞
R/2
As we approach infinity, the first term in the bracket disappears, the second
and third terms cancel each other, and we have a residual value from the last
term of 3π/4. For the lower limit, we have a term in brackets of
− 1 2 + 1 π
4 2 − 1 2 + 3 π
2 4 = −1 + π 2
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