61) The vertical field intensity anomaly ΔZ due
to a vertically polarized vertical dyke is given by
$\triangle Z =
2Mt(\frac{Z_{1}}{Z_1^2+x^{2}}-\frac{Z_{2}}{Z_2^2+x^{2}})$
Where M is the magnitude of intensity of magnetization. All relevant parameters are provided in the figure below. The dyke has 1% magnetite (magnetic susceptibility of magnetite = 0.5 SI units) distributed homogenously. Then the magnitude of Peak vertical field intensity over the dyke is ______ nT.
(Thanks to Chandrasekhar, ANU)
Solution:.
The vertical field intensity anomaly ΔZ due to a vertically
polarized vertical dyke is given by
$\triangle Z =
2Mt(\frac{Z_{1}}{Z_1^2+x^{2}}-\frac{Z_{2}}{Z_2^2+x^{2}})$ ---------(1)
Where M is the intensity of the
magnetization.
M is defined as Product of the Susceptibility and the field
intensity
M=K*H= 0.5*60,000 nT
=
30,000nT.
But here the dyke has only 1 % of magnetite = 1% of 30,000 nT
= $30,000\times\frac{1}{100}$
=300
nT.
The magnitude peak will occur at X=0, substitute the Value X=0 in
equation(1)
$\triangle Z =
2Mt(\frac{Z_{1}}{Z_1^2+x^{2}}-\frac{Z_{2}}{Z_2^2+x^{2}})$
$\triangle Z =
2Mt(\frac{Z_{1}}{Z_1^2+0^{2}}-\frac{Z_{2}}{Z_2^2+0^{2}})$
$\triangle Z = 2Mt(\frac{1}{Z_1}-\frac{1}{Z_2})$
Substitute the values of M, t,Z1,
and Z2 which we get from the given figure in the above equation
we get
$\triangle Z = 2(300)(10)(\frac{1}{50}-\frac{1}{150})$
$\triangle Z = 6000(\frac{150-50}{7500})$
$\triangle Z = 6000(\frac{100}{7500})$
$\triangle Z = 80nT$
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