The magnetic field of earth at the equator is approximately $ 4 \times 10^{-5} \mathrm{~T} $. The radius of earth is $ 6.4 \times 10^{6} \mathrm{~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 \times 10^{-5} \mathrm{~T} $. The radius of earth is $ 6.4 \times 10^{6} \mathrm{~m} $. Then the dipole moment of the earth will be nearly of the order of :, $ 10^{20} \mathrm{Am}^{2} $, $ 10^{16} \mathrm{Am}^{2} $, $ 10^{10} \mathrm{Am}^{2} $, $ 10^{23} \mathrm{Am}^{2} $

MARKS App · Accepted Answer

B = μ 0 4 π · M d 3 = 1 0 - 7 × M 6.4 × 1 0 6 3 ∴ 4 × 1 0 - 5 = 1 0 - 7 × M 6.4 × 1 0 6 3 ∴ M = 6.4 × 1 0 6 3 × 4 × 1 0 - 5 1 0 - 7 = 6.4 3 × 1 0 1 8 × 4 × 1 0 2 = 6.4 3 × 4 × 1 0 2 0 = 1 . 0 4 8 × 1 0 2 3 ≈ 10 23 A m 2

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