Derive the relation ..

tan delta= 2tan lamda, where delta is the angle of dip and lamda is the magnetic latitude

The derivation is quite involved and complex so we will attempt to make it short and meaningful...

Now,

the potential due to a dipole is given as

U = (M / 4πr2)cosθ

here M is the magnetic dipole moment and θ is the colatitude angle

now, the magnetic field induction will be given as the gradient of U (three dimensional derivative)

so,

B = -μ0ΔU

this will give us the two components of the field B

so,

vertical component

Br = -Z = (μ0M / 2πr3)cosθm

horizontal component

Bθ = -H =  (μ0M / 4πr3)sinθm

now, if B0 =  (μ0M / 4πr3)

then,

Br = 2B0cosθm

and

Bθ = B0sinθm

now,

we define an angle of inclination which is the angle made between the magnetic field (vertical) and the dip or horizontal component.

so,

tani = Br / Bθ

now, for a dipole

tani = cotθm

thus,

tani = 2tanλ

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