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

B=4×10-5T B=(μ0/4π)×(M/r3) =10-7×(M/(6.4×106)3)=4×10-5 M=(4× 10-5× 1018×6.43/10-7) =1.048×103+18+7-5 =1023

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 :

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

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

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

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

$B = 4 \times 1 0^{- 5} T$
$B = \frac{μ _{0}}{4 π} \times \frac{M}{r ^{3}}$
$= 1 0^{- 7} \times \frac{M}{( 6.4 \times 1 0 ^{6} ) ^{3}} = 4 \times 1 0^{- 5}$
$M = \frac{4 \times 1 0 ^{- 5} \times 1 0 ^{18} \times 6. 4 ^{3}}{1 0 ^{- 7}}$
$= 1.048 \times 1 0^{3 + 18 + 7 - 5}$
$= 1 0^{23}$

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