Paquete Generales

82. Calculate the magnetic anomaly created by a magnetic dipole buried at depth d, arbitrarily oriented, at an angle to the vertical of a. The negative pole is upwards.

Consider a point P with coordinates (x, z), where x is measured along the horizontal from the projection of the centre of the dipole and z is the vertical from the reference level (the Earth’s surface). The position vector r forms an angle b to the vertical (Fig. 82). The anomalous magnetic potential created by the dipole for this point is

F ¼Cm cosða þ bÞ

r² ¼Cmðsin b cos a þ cos b sin aÞ

r² ð82:1Þ

where (Fig. 82)

cos b¼ zþ d ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þ ðz þ dÞ² q

sin b¼ x

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þ ðz þ dÞ² q

0 5 10 15 20

t (h) –2

–1 0 1

Bθ

B_r

B_l

Br,q,l B0

Fig. 81b

Substituting in Equation (82.1) we obtain

F ¼Cm ðz þ dÞ cos a  x sin a½  x²þ ðz þ dÞ²

h i3=2

To calculate the magnetic anomalyDB:

B ¼ rð
FÞ

The vertical component of the magnetic field anomaly, taking the z-coordinate positive downward, is

Z ¼@ð
FÞ

@z ¼

Cm
x²þ ðz þ dÞ²

cos a 3ðz þ dÞ ðz þ dÞ cos a  x sin a½ 

h i

x²þ ðz þ dÞ²

h i5=2

For points on the Earth’s surface (z ¼ 0)

Z ¼Cm ðx½ ² 2d²Þ cos a þ 3dx sin a x²þ d²

½ ⁵⁼² ð82:2Þ

The component of the magnetic anomaly in an arbitrary horizontal direction x for the Earth’s surface points (z ¼ 0) is given by

X ¼ @ð
FÞ

@x ¼Cm½ð2x² d²Þ sin a  3dx cos a x²þ d²

½ ⁵⁼² ð82:3Þ

z

Z

r d

b

a

P x

X

+ –

Fig. 82

83. Calculate the magnetic anomaly produced at a point with geographical coordinates 38^ N, 30^ W by a horizontal dipole buried at a depth of 10 m with Cm¼ 5  10^5T m³which is in the vertical plane of geographical east, and with the negative pole to the west. Also calculate the total values of the field in the NS, EW, and vertical directions, and total field F, as well as the variations in the magnetic declin-ation and inclindeclin-ation due to the existence of the dipole. Consider the Earth’s magnetic field to be produced by a centred dipole with North Pole at 72^ N, 30^ W, and with B₀¼ 32 000 nT.

We calculate the horizontal and vertical components of the magnetic anomaly from Equations (82.2) and (82.3), taking a ¼ 90º because the dipole is horizontal and x ¼ 0 because the dipole’s centre is beneath the point (Fig. 83a). We call DX and DY the horizontal components in the NS and EW directions, respectively. Because the dipole is on the vertical east–west plane, the north–south component DX ¼ 0,

Z ¼ 0

Y ¼Cm½ð2x² d²Þ sin a  3dx cos a x²þ d²

½ ⁵⁼² ¼ Cm

d³ Substituting

Cm¼ 5  10^5T m³ d¼ 10 m

we obtain

Y ¼ 50 nT j
Bj ¼
Y ¼ 50 nT

X ¼ 0

Z – +

a d

P E

Fig. 83a

To calculate the components of the magnetic anomaly in the direction of the Earth’s magnetic field, F, and their horizontal component, H, we need to determine the magnetic declination and inclination at the point:

H ¼
X cos D^þ
Y sin D^

F ¼
H cos I^þ
Z sin I^ ð83:1Þ

Because the point has the same longitude as the Geomagnetic North Pole (Fig. 83b), f^¼ 90^ ðf_B fÞ ¼ 56:0^

D^¼ 0^ The inclination is given by

tan I^¼ 2 tan f^) I^¼ 71:4^ Substituting these values in Equations (83.1) we obtain

H ¼ 0

F ¼ 0

The total value of the field in the NS, EW, and vertical directions, and total field F are X_T¼ X^þ
X

YT¼ Y^þ
Y ZT¼ Z^þ
Z F_T¼ F^þ
F

f∗

f P fB

GMNP GNP

Fig. 83b

The vertical Z^and horizontal H^components of the geomagnetic field are given by Z^¼ 2B0sin f^¼ 53 058 nT

H^¼ B0cos f^¼ 17 894 nT

The NS (X^) and horizontal EW (Y^) components of the geomagnetic field and its magnitude F^ are given by

X^ ¼ H^cos D^¼ H^¼ 17 894 nT Y^ ¼ H^sin D^ ¼ 0

F^ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi H^2þ Z^2

p ¼ 55 994 nT

We finally obtain that the total field components are XT¼ 17 894 nT YT ¼ 50 nT Z_T ¼ 53 058 nT F_T ¼ 55 994 nT

The variations in magnetic declination and inclination due to the presence of the buried dipole are

tan D⁰¼YT

X_T) D⁰¼ 0:02^ D^ D⁰¼ 0:02^

tan I⁰¼ZT

HT

) I⁰ 71:4^¼ I^

84. Buried at a point with magnetic latitude 30^N and a depth of 50 m is a horizontal magnetic dipole with Cm ¼ 10⁷nT m³ with the positive pole to the geographical north.

(a) Calculate DF if B₀¼ 30 000 nT and the declination at that point is 15^. Find the ratio DF /F.

(b) How far from the dipole’s centre along the north–south line will the dipole field strength be in the same direction as that of the Earth (take D^¼ 0^).

(a) The component of the magnetic anomaly DF in the direction of the Earth’s magnetic field (the total field anomaly) is given by

F ¼
H cos I^þ
Z sin I^ ð84:1Þ

We first calculate the components in the geographical directions of the magnetic anomaly produced by the buried dipole using Equations (82.2) and (82.3) of Problem 82, substitut-ing a ¼ 90º because the dipole is horizontal, and x ¼ 0 because the dipole’s centre is beneath the point. In this problemDY ¼ 0 because the dipole is on the geographical north–

south vertical plane (Fig. 84). Then

X ¼ Cm d³

Y ¼ 0

Z ¼ 0 Substituting the values

Cm¼ 10⁷nT m³ d¼ 50 m we obtain

X ¼ 80 nT

Substituting D^ ¼ 15º, the component of the magnetic anomaly in the direction of the horizontal component H of the Earth’s magnetic field is

H ¼
X cos D^ ¼ 77 nT

At a point of magnetic latitude f^¼ 30^ the magnetic inclination is tan I^¼ 2 tan f^) I^¼ 49:1^ Substituting in Equation (84.1), the total field anomaly is

F ¼ 50 nT

To calculate the geomagnetic field F^ we first obtain the components H^and Z^: Z^¼ 2B0sin f^¼ 30 000 nT

H^¼ B0cos f^¼ 25 981 nT F^¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

H^2þ Z^2

p ¼ 39 686 nT

P

Z

N

a d

– +

Fig. 84

The ratio of the total field anomaly and the Earth’s total magnetic field is

F

F ¼ 1:26  10^3

(b) If the dipole field strength is in the same direction as that of the Earth then the inclination I⁰due to the dipole is equal to that of the Earth’s field I^, where

tan I⁰¼
Z

H tan I^¼ Z^

H^ Assuming D^¼ 0º, then

H ¼
E

If we substitute in Equations (82.2) and (82.3) of Problem 82, the angle a¼ 90º because the dipole is horizontal, we obtain

Z ¼ 3Cmdx x²þ d² 5=2

X ¼Cmð2x² d²Þ x²þ d² 5=2

We have changed the sign of the vertical component because the negative pole is toward the south.

Applying the condition, tan I⁰¼ tan I^, we obtain

Z

H ¼
Z

X ¼ Z^ H^ 3Cmdx

x²þ d²5=2

Cm2x² d²Þ

x²þ d²5=2

¼ 3dx

2x² d²¼Z^ H^

2Z^x² 3dH^x Z^d² ¼ 0 Substituting the values

d¼ 50 m Z^¼ 30 000 nT H^¼ 25 981 nT

and solving the equation, we obtain

x1¼ 80 m x2¼ 15 m

We have two solutions: a point 80 m to the north from the surface projection of the dipole’s centre and another 15 m to the south.

85. Located at a point with geocentric geographical coordinates 45^ N, 30^ W, at a depth of 100 m, is a dipole of magnetic moment Cm¼ 1 T m³, tilted 45^ from the horizontal to true north, with the negative pole to the north and downwards. At this point on the surface, the following magnetic field values were observed (in nT):

F¼ 55 101; H¼ 12 413; DF¼ 1268; DH¼ 547.

Determine:

(a) At the indicated point, the main field components X^, Y^, Z^.

(b) At the indicated point, the deviation of the compass with respect to geomagnetic north.

(c) The geocentric geographical coordinates of the North Pole of the Earth’s dipole.

Precision 1 nT.

(a) We calculate first the magnetic anomaly produced by the dipole, applying Equa-tions (82.2) and (82.3) of Problem 82, substituting a ¼ 225º and x ¼ 0. The horizontal component is in the NS direction (DX) (Fig. 85a)

Z ¼2Cm cos a

d³ ¼ 1414 nT

X ¼Cm sin a

d³ ¼ 707 nT

Y ¼ 0

ð85:1Þ

To calculate the declination we use the equation

H ¼
X cos D^) cos D^¼
H

X D^¼ 39:3^

P

N

Z d

a +

45°

–

Fig. 85a

To obtain the Earth’s main field we eliminate the buried dipole contribution from the observed values:

F^ ¼ F 
F ¼ 56 369 nT H^ ¼ H 
H ¼ 11 866 nT

Z^ ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðF^Þ² ðH^Þ² q

¼ 55 106 nT and the X^and Y^components

X^ ¼ H^cos D^¼ 9182 nT Y^ ¼ H^sin D^¼ 7516 nT (b) The observed declination is given by

tan D⁰¼Y

X ¼Y^þ
Y

X^þ
X ) D⁰¼ 37:2^

The deviation of the compass due to the buried dipole with respect to geomagnetic north is D⁰ D^ ¼ 2:1^

(c) We calculate first the geomagnetic latitude of the point from the vertical and horizontal components:

Z^¼ 2B0sin f^ H^¼ B0cos f^

tan f^¼ Z^ 2H^ f^¼ 66:7^

With this value, the declination D^and the geographical coordinates of the point (f, l), we can solve the spherical triangle (Fig. 85b) and obtain the geographical coordinates of the Geomagnetic North Pole:

GMNP

90º –f

q = 90º– f∗

90º –fB

l – lB

180º –l∗

D ∗ GNP

P

Fig. 85b

cosð90  f_BÞ ¼ cosð90  f^Þ cosð90  fÞ þ sinð90  f^Þ sinð90  fÞ cos D^ sin f_B¼ sin f^sin fþ cos f^cos f cos D^

f_B¼ 60:0^

cosð90  f^Þ ¼ cosð90  f_BÞ cosð90  fÞ þ sinð90  f_BÞ sinð90  fÞ cosðl  lBÞ sin f^¼ sin f_Bsin fþ cos f_Bcos f cosðl  l_BÞ

cosðl  lBÞ ¼sin f^ sin f_Bsin f cos f_Bcos f l l_B¼ 30:0^

The correct solution is the negative one because a positive value of the declination implies that the point is to the west of the Geomagnetic North Pole:

D^> 0) l  lB < 0 l_B¼ 0^

86. Located at a point with geographical coordinates 45^N, 30^W, at a depth of 100 m, is a dipole of magnetic moment Cm¼ 1 T m³, inclined 45^to the vertical towards the south, with the positive pole upwards, and in the geographical north–south vertical plane. The Earth’s dipole has its north pole at 60^ N, 0^ E and B₀ ¼ 30 000 nT.

Calculate:

(a) The values of Z, H, F at the given point.

(b) Where does the compass point to at that same point?

(a) We calculate first the geomagnetic latitude corresponding to the point by sin f^¼ sin f_Bsin fþ cos f_Bcos f cosðl  lBÞ

f^¼ 66:7^

From this value we obtain the geomagnetic components Z^, H^and the total main field F^: Z^¼ 2B0sin f^¼ 55 107 nT

H^¼ B0cos f^¼ 11 866 nT F^¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðH^Þ²þ ðZ^Þ² q

¼ 56 370 nT The geomagnetic declination D^ is given by

sin D^ ¼ cos f_Bsinðl  lBÞ cos f^ D^¼ 39:2^

and the geomagnetic inclination I^ by

tan I^¼ 2 tan f^) I^¼ 77:8^

The magnetic anomaly created by the dipole buried at depth d is given by Equations (85.1) of Problem 85. Substituting Cm¼ 1 T m³, d¼ 100 m, and a ¼ 45º, we obtain

Z ¼2Cm cos a

d³ ¼ 1414 nT

X ¼Cm sin a

d³ ¼ 707 nT

Y ¼ 0 The field anomalies DH andDF are given by

H ¼
X cos D^¼ 548 nT

F ¼
H cos I^þ
Z sin I^¼ 1266 nT Finally, the observed values are

Z¼ Z^þ
Z ¼ 53 693 nT F¼ F^þ
F ¼ 55 104 nT H¼ H^þ
H ¼ 12 414 nT

(b) To calculate in what direction the compass points we need the value of the observed declination D’ including the effects of the geomagnetic field and the buried dipole:

tan D⁰¼Y

X ¼Y^þ
Y X^þ
X Y^¼ H^sin D¼ 7500 nT X^ ¼ H^cos D¼ 9195 nT D⁰¼ 37:1^

87. Located at a point on the Earth with geographical coordinates 45^ N, 30^ E, at a depth of 100 m, is a dipole of magnetic moment Cm ¼ 10⁷nT m³, tilted 45^ to the vertical towards the south, with the positive pole downwards, and contained in the plane of true north. The Earth’s field is produced by a centred dipole tilted 30^from the axis of rotation in the plane of the 0^ meridian, with B₀¼ 30 000 nT. Calculate the total values of F, Z, and H observed at the point of the surface above the centre of the buried dipole.

We first calculate the geographical coordinates of the Geomagnetic North Pole and the geomagnetic latitude

f_B¼ 90^ 30^¼ 60^ l_B¼ 0^

sin f^¼ sin f_Bsin fþ cos f_Bcos f cosðl  lBÞ f^¼ 66:7^

The geomagnetic field components Z^, H^ and the total main field F^ are given by Z^¼ 2B0sin f^¼ 55 107 nT

H^¼ B0cos f^¼ 11 866 nT F^¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðH^Þ²þ ðZ^Þ² q

¼ 56 370 nT We calculate the geomagnetic declination D^ by

sin D^¼ cos f_Bsinðl  l_BÞ cos f^ D^¼ 39:2^ The inclination I^is given by

tan I^¼ 2 tan f^) I^¼ 77:8^

We obtain the magnetic anomaly produced by the buried dipole applying Equations (82.2) and (82.3) of Problem 82, substituting a¼ 45º and x ¼ 0. The horizontal component is in the NS direction (DX):

Z ¼2Cm cos a

d³ ¼ 14 nT

X ¼Cm sin a

d³ ¼ 7 nT

Y ¼ 0 The field anomalies DH and DF are given by

H ¼
X cos D^¼ 5 nT

F ¼
H cos I^þ
Z sin I^ ¼ 13 nT Finally the observed values are

Z¼ Z^þ
Z ¼ 55 121 nT F¼ F^þ
F ¼ 56 383 nT H¼ H^þ
H ¼ 11 861 nT

88. Buried at a point with geographical latitude 20^ N and the same longitude as the geomagnetic pole, at a depth of 200 m, is a sphere of 50 m radius of material with magnetic susceptibility 0.01. The Earth’s field is produced by a centred dipole tilted 10^ from the axis of rotation and magnetic moment M¼ 10³⁰g cm³(Earth’s radius:

6000 km). Calculate:

(a) The anomaly produced by induced magnetization in the sphere at a point on the Earth’s surface above the centre of the sphere. Give the vertical and horizontal components in units of nT.

(b) The total anomaly for a point on the Earth’s surface 100 m south of the above point.

(a) We first calculate the geomagnetic co-latitude (y) and latitude (f^) of the point, knowing that it is in the same meridian as the Geomagnetic North Pole (Fig. 88a):

y¼ 90^ f^ ¼ 90^ 10^ 20^¼ 60^ f^¼ 30^

The geomagnetic field is given by B₀¼M

a³ ¼ 4630 nT Z^¼ 2B0sin f^¼ 4630 nT H^¼ B0cos f^¼ 4009 nT

F^¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðH^Þ²þ ðZ^Þ² q

¼ 6124 nT The inclination is given by

tan I^¼ 2 tan f^) I^¼ 49^

The magnetic anomaly created by a sphere is the same as the anomaly created by a magnetic dipole oriented in the same direction as the geomagnetic field, that is, tilted 90^– I^to the vertical and with the negative pole upwards (Fig. 88b). So we use Equations (82.2) and (82.3) taking a¼ 41º and x ¼ 0, but we change the sign of Equation (82.3) because the negative pole is toward the north. The horizontal component is DX ¼ DH because the dipole is in the magnetic north-vertical plane andDY ¼0:

Z ¼2Cm cos a d³

H ¼Cm sin a d³

GMNP GNP

10°

20°

q = 90° – f∗

P

Fig. 88a

To calculate Cm we use the magnetic susceptibility w and the volume V of the sphere:

Cm¼ wF^V ¼ 3:2  10⁷nT m³

Substituting this value in the equation of the components of the magnetic anomaly we obtain

Z ¼2Cm cos a

d³ ¼2 3:2  10⁷ cos 41^ 8 10⁶ ¼ 6 nT

X ¼Cm sin a

d³ ¼3:2 10⁷ sin 41^ 8 10⁶ ¼ 3 nT

(b) The anomaly created by the sphere at a point at a distance x¼100 m to the south of the above point is given by

Z ¼Cmðx² 2d²Þ cos a þ 3dx sin a

x²þ d² 5=2 ¼ 0:8 nT

H ¼Cmð2x² d²Þ sin a  3dx cos a

x²þ d² 5=2 ¼ 3:3 nT The total magnetic field anomaly is therefore

F ¼
H cos I^þ
Z sin I^ ¼ 3:4 nT P

Z

F ∗

GN MN

41°

49°

Fig. 88b

89. Buried at a point on the Earth at magnetic latitude 45^N, at a depth d, is a vertical dipole of magnetic moment M (Cm) with negative pole upwards. If M/d³¼ 10B0(B₀is the geomagnetic constant of the main field) calculate how far along the magnetic meridian the direction of the buried dipole’s field will coincide with that of the Earth (the terrestrial dipole field).

First we calculate the geomagnetic field components and the inclination by Z^¼ 2B0sin f^¼ ffiffiffi

p2 B₀nT H^¼ B0cos f^¼

ffiffiffi2 p

2 B0nT tan I^¼Z^

H^¼ 2 ) I^¼ 63:4^

The components of the magnetic field created by the dipole are given by Equations (82.2) and (82.3) of Problem 82, putting a¼ 0º:

Z ¼Cmx² 2d²

x²þ d²5=2

H ¼ Cm3dx

x²þ d²5=2

If the buried dipole’s field coincides with that of the Earth the magnetic inclinations due to both have to be equal and so

Z

H¼ tan I^¼ 2

x²þ 6xd  2d²¼ 0 ) x ¼ d 3  ffiffiffiffiffi p28



Of the two solutions, x¼ 2.3d and x ¼ 8.3d, only the positive corresponds to the equal direction of the two fields.

In document Plantilla para el Montaje Dinamico de los Productos de la Coleccion Multisaber. (página 159-166)