The magnetic field of earth at the equator is approximately 4 × 10-5 T. The radius of earth is 6.4 × 106 m. Then the dipole moment of the earth will be nearly of the order of:

Question

The magnetic field of earth at the equator is approximately 4 × 10-5 T. The radius of earth is 6.4 × 106 m. Then the dipole moment of the earth will be nearly of the order of: (A) 1023 A m2 (B) 1020 A m2 (C) 1016 A m2 (D) 1010 A m2

Tardigrade Admin · Accepted Answer

Given, B = 4 ×10-5 T RE = 6.4 ×106m Dipole moment of the earth M = ? B = (μ0/4π) (M/d3) 4 ×10-5 = (4 π×10-7 × M/4 π×(6.4 ×106)3) ∴ M ≅ 1023 Am2

Q. The magnetic field of earth at the equator is approximately $4 \times 1 0^{- 5} T$ . The radius of earth is $6.4 \times 1 0^{6} m$ . Then the dipole moment of the earth will be nearly of the order of:

$1 0^{23} A m^{2}$

$1 0^{20} A m^{2}$

$1 0^{16} A m^{2}$

$1 0^{10} A m^{2}$

Given, $B = 4 \times 1 0^{- 5} T$
$R_{E} = 6.4 \times 1 0^{6} m$
Dipole moment of the earth M = ?
$B = \frac{μ _{0}}{4 π} \frac{M}{d ^{3}}$
$4 \times 1 0^{- 5} = \frac{4 π \times 1 0 ^{- 7} \times M}{4 π \times ( 6.4 \times 1 0 ^{6} ) ^{3}}$
$∴ M ≅ 1 0^{23} A m^{2}$

Q. The magnetic field of earth at the equator is approximately 4×10−5T. The radius of earth is 6.4×106m. Then the dipole moment of the earth will be nearly of the order of:

1023Am2

1020Am2

1016Am2

1010Am2