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 / 4πr^{2})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 = -μ_{0}ΔU

this will give us the two components of the field B

so,

vertical component

B_{r} = -Z = (μ_{0}M / 2πr^{3})cosθ_{m}

horizontal component

B_{θ} = -H = (μ_{0}M / 4πr^{3})sinθ_{m}

now, if B_{0} = (μ_{0}M / 4πr^{3})

then,

B_{r} = 2B_{0}cosθ_{m}

and

B_{θ} = B_{0}sinθ_{m}

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 = B_{r} / B_{θ}

now, for a dipole

tani = cotθ_{m}

thus,

**tani = 2tanλ**

**
**