Q27. What is the approximate earth's dipole moment, if the earth's magnetic field at the Equator is

Q: What is the correct answer for CUET UG Physics Q27?

The correct answer is c. The magnetic field (B) at the Earth's equator due to its magnetic dipole moment (m) is given by the formula B = (μ₀ / 4π) * (m / r³), where μ₀ is the permeability of free space (μ₀ / 4π = 10⁻⁷ T·m/A) and r is the radius of the Earth. Given B = 0.5 G = 0.5 × 10⁻⁴ T. The diameter of the Earth is d = 15.0 × 10⁶ m, so the radius r = d/2 = 7.5 × 10⁶ m. Rearranging the formula to solve for the magnetic

Q: Q27. What is the approximate earth's dipole moment, if the earth's magnetic field at the Equator is approximately 0.5 G and diameter is 15.0 × 10⁶m? (a) 5.9 x 108 Am² (b) 6.5 x 1030 Am² (c) 2.1 x…

The correct answer is c. The magnetic field (B) at the Earth's equator due to its magnetic dipole moment (m) is given by the formula B = (μ₀ / 4π) * (m / r³), where μ₀ is the permeability of free space (μ₀ / 4π = 10⁻⁷ T·m/A) and r is the radius of the Earth. Given B = 0.5 G = 0.5 × 10⁻⁴ T. The diameter of the Earth is d = 15.0 × 10⁶ m, so the radius r = d/2 = 7.5 × 10⁶ m. Rearranging the formula to solve for the magnetic

Q: What are the options for this CUET UG Physics question?

A. 5.9 x 10⁸ Am² · B. 6.5 x 10³⁰ Am² · C. 2.1 x 10²³ Am² · D. 1.05 x 10²³ Am²

Question

Q27. What is the approximate earth's dipole moment, if the earth's magnetic field at the Equator is approximately 0.5 G and diameter is 15.0 × 10⁶m?
(a) 5.9 x 108 Am²
(b) 6.5 x 1030 Am²
(c) 2.1 x 1023 Am²
(d) 1.05 x 1023 Am²

FirstInTest · Accepted Answer

The magnetic field (B) at the Earth's equator due to its magnetic dipole moment (m) is given by the formula B = (μ₀ / 4π) * (m / r³), where μ₀ is the permeability of free space (μ₀ / 4π = 10⁻⁷ T·m/A) and r is the radius of the Earth. Given B = 0.5 G = 0.5 × 10⁻⁴ T. The diameter of the Earth is d = 15.0 × 10⁶ m, so the radius r = d/2 = 7.5 × 10⁶ m. Rearranging the formula to solve for the magnetic dipole moment m: m = B * (4πr³) / μ₀ = B * r³ / (μ₀ / 4π). Substituting the values: m = (0.5 × 10⁻⁴ T) * (7.5 × 10⁶ m)³ / (10⁻⁷ T·m/A). m = (0.5 × 10⁻⁴) * (421.875 × 10¹⁸) / 10⁻⁷ = 210.9375 × 10⁻⁴⁺¹⁸⁺⁷ = 210.9375 × 10²¹. Approximately, m ≈ 2.1 × 10²³ Am². Therefore, option (c) is the correct answer. Options (a), (b), and (d) are incorrect values, with different orders of magnitude.

Answer

5.9 x 10⁸ Am²

Answer

6.5 x 10³⁰ Am²

Answer

2.1 x 10²³ Am²

Answer

1.05 x 10²³ Am²

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