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

Answer: (a). Equatorial field of earth's dipole: B = (μ₀)/(4π)M/R³ M = BR³10⁻⁷ = 4 × 10⁻⁵ × (6.4 × 10⁶)³10⁻⁷ (6.4 × 10⁶)³ ≈ 2.6 × 10²⁰ ⇒ M ≈ 1.05 × 10²³ A m² So the answer is (a) 10²³ A m².

The magnetic field of earth at the equator is approximately 4 × 10 -… | JEE Main 2014 (9 Apr, Shift 1)

Q13JEE Main 2014 · 9 APR · SHIFT 1Medium

Earth's Magnetism

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

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Answer: (a)

Equatorial field of earth's dipole: $B = \frac{\mu_0}{4\pi}\frac{M}{R^3}$
$M = \frac{BR^3}{10^{-7}} = \frac{4 \times 10^{-5} \times (6.4 \times 10^6)^3}{10^{-7}}$
$(6.4 \times 10^6)^3 \approx 2.6 \times 10^{20} \Rightarrow M \approx 1.05 \times 10^{23}\ \text{A m}^2$
So the answer is (a) $10^{23}\ \text{A m}^2$ .

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Earth's Magnetism question from JEE Main 2014 (9 Apr, Shift 1)

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