The earth’s magnetic field lines resemble that of a dipole at the center of the earth. If the magnetic moment of this dipole is close to 8 × 1022Am2, the value of earth’s magnetic field near the equator is close to (radius of the earth = 6.4 × 106 m)

Given M= 8 ×1022 Am2 d= Re =6.4 × 106m earth 's magnetic field , B =μ∘4π2Md3 = 4π×10-74π×2×8×1022(6.4×106)3 ≅0.6 Gauss

The earth’s magnetic field lines resemble that of a dipole at the center of the earth. If the magnetic moment of this dipole is close to 8 × 1022Am2, the value of earth’s magnetic field near the equator is close to (radius of the earth = 6.4 × 106 m)

Given M= 8 ×1022 Am2 d= Re =6.4 × 106m earth 's magnetic field , B =μ∘4π2Md3 = 4π×10-74π×2×8×1022(6.4×106)3 ≅0.6 Gauss

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The earth’s magnetic field lines resemble that of a dipole at the center of the earth. If the magnetic moment of this dipole is close to 8 × 10²²
Am², the value of earth’s magnetic field near the equator is close to (radius of the earth = 6.4 × 10⁶ m)

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0.6 Gauss

1.2 Gauss

1.8 Gauss

0.32 Gauss

answer is A.

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Detailed Solution

$G i v e n M = 8 \times 10^{22} A m^{2} d = R_{e} = 6.4 \times 10^{6} m e a r t h' s m a g n e t i c f i e l d, B = \frac{μ_{\circ}}{4 π} \frac{2 M}{d^{3}} = \frac{4 π \times 10^{- 7}}{4 π} \times \frac{2 \times 8 \times 10^{22}}{{(6.4 \times 10^{6})}^{3}} ≅ 0.6 G a u s s$