Physics, asked by sonukumardreamer, 11 months ago

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how many earth magnetic moment in joule per time inverse ​


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Answers

Answered by Anonymous
0

Answer:

Earth’s Magnetic Field

Magnetic Potential for a dipole field pointing South

V(r) = m • r / (4 π r

3

) = − m cosθ / (4 π r

2

) = scalar magnetic potential of dipole

field. Field is expanded in spherical harmonics. First term (above) is the dipole term.

m = 8 x 1022 Am2

is dipole moment at center of Earth point south

r = distance from dipole

θ = colatitude

B(r) = − µ0 grad V = vector magnetic field

where µ0 = 4π x 10-7 kg m A-2

s

-2 = magnetic permeability in free space (A=amps)

Br = − µ0 dV/dr = − µ0 m 2 cosθ / (4 π r

3

) radial component of magnetic field (up)

Bθ = − µ0 r

-1 dV/dθ = − µ0 m sinθ / (4 π r

3

) tangential component of field (south)

|B| = sqrt( Br

2 + Bθ

2 ) = µ0 m / (4 π R3

) sqrt(sin2

θ +4cos2

θ) = B0 sqrt(1+3cos2

θ)

size of magnetic field at Earths surface

at r = R = 6371*103 m (surface of earth) define:

B0 = µ0 m / (4 π R3

Earth’s Magnetic Field

Magnetic Potential for a dipole field pointing South

V(r) = m • r / (4 π r

3

) = − m cosθ / (4 π r

2

) = scalar magnetic potential of dipole

field. Field is expanded in spherical harmonics. First term (above) is the dipole term.

m = 8 x 1022 Am2

is dipole moment at center of Earth point south

r = distance from dipole

θ = colatitude

B(r) = − µ0 grad V = vector magnetic field

where µ0 = 4π x 10-7 kg m A-2

s

-2 = magnetic permeability in free space (A=amps)

Br = − µ0 dV/dr = − µ0 m 2 cosθ / (4 π r

3

) radial component of magnetic field (up)

Bθ = − µ0 r

-1 dV/dθ = − µ0 m sinθ / (4 π r

3

) tangential component of field (south)

|B| = sqrt( Br

2 + Bθ

2 ) = µ0 m / (4 π R3

) sqrt(sin2

θ +4cos2

θ) = B0 sqrt(1+3cos2

θ)

size of magnetic field at Earths surface

at r = R = 6371*103 m (surface of earth) define:

B0 = µ0 m / (4 π R3

3 * 10-5 Tesla (T) = 3 * 104 nanoTesla

So the strength of the magnetic field at Earth’s surface varies from 30,000nT at the

equator (θ = 90; cos2

θ=0) to 60,000nT at the poles (θ = 0 or 180; cos2

θ=1).

The inclination angle (I) is defined to be the dip of magnetic field: horizontal=0, down is

positive) so:

tan(I) = -Br / | Bθ | = (B0 2 cosθ ) / (B0 sinθ ) = 2/tanθ = 2 tanλ

negative Br because we measure dip pointing down but positive r is up

θ is colatitude; λ = 90 - θ is latitude

Time variation of magnetic field

Secular variation

Intensity

Direction

Westward drift of non-dipole field

Reversals

Answered by nancypawansharmapami
1
have a nice day ahead
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