Classical Calculation for Mutual Inductance of Two ... - Kurt Nalty

The positions and separations are
r left = L cos φa x + L sin φa y − l 2 a z
r right = R cos θa x + R sin θa y + l 2 a z
∆r = (L cos φ − R cos θ) a x + (L sin φ − R sin θ) a y − la z
r 2 = L 2 + R 2 + l 2 − 2LR (cos φ cos θ + sin φ sin θ)
= L 2 + R 2 + l 2 − 2LR cos (φ − θ)
r = √ L 2 + R 2 + l 2 − 2LR cos (φ − θ)
We now can write our expression for the mutual inductance.
∮ ∮
dt · ds
M = µ
= µ ∮ ∮
dt · ds
s t 4πr 4π s t r
= µ ∮ ∮
4π
= µ
4π
= µ
4π
∮
φ
θ
( ∮
φ θ
∮
φ
( ∮
θ
LR (cos (φ − θ)) dφdθ
√
L2 + R 2 + l 2 − 2LR cos (φ − θ)
)
LR (cos (φ − θ)) dθ
√ dφ
L2 + R 2 + l 2 − 2LR cos (φ − θ)
)
LR (cos (θ − φ)) dθ
√ dφ
L2 + R 2 + l 2 − 2LR cos (θ − φ)
where the last step uses the even nature of the cosine to absorb a minus sign.
During the inner integration over θ, φ is kept constant. We now do a change
of variable,
γ = θ − φ
dγ = dθ inside the θ integral
We can simplify the mutual inductance expression.
M = µ ∮ ( ∮
4π φ γ
= µ ∮
4π
= µ ∮
2
γ
φ
( ∮
dφ
γ
LR cos γdγ
√
L2 + R 2 + l 2 − 2LR cos γ
)
dφ
)
LR cos γdγ
√
L2 + R 2 + l 2 − 2LR cos γ
LR cos γdγ
√
L2 + R 2 + l 2 − 2LR cos γ
This integral is known in terms of the complete elliptic functions K and E.
The reference integral is
∮
cos θdθ
√ = 4√ )
]
a + b
[(1 − β2
K(β) − E(β)
a − b cos θ b
2
√
2b
where β =
a + b
3