Magnetic dipole moment of a square current loop of side 1 cm and carry

Magnetic dipole moment of a square current loop of side 1 cm and carrying a current of 0.10 A is :

3.14 × 10<sup>-5</sup> A m<sup>2</sup>
1.00 × 10<sup>-5</sup> A m<sup>2</sup>
6.28 × 10<sup>-3</sup> A m<sup>2</sup>
4.00 × 10<sup>-3</sup> A m<sup>2</sup>
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UPSC Geoscientist – 2024
The magnitude of the magnetic dipole moment (m) of a planar current loop is given by the product of the current (I) flowing through the loop and the area (A) of the loop: m = IA.
The loop is a square with side length s = 1 cm. Convert the side length to meters: s = 1 cm = 0.01 m = 10⁻² m. The area of the square loop is A = s² = (10⁻² m)² = 10⁻⁴ m². The current is I = 0.10 A.
The magnitude of the magnetic dipole moment is m = I * A = (0.10 A) * (10⁻⁴ m²) = 0.1 * 10⁻⁴ A m² = 10⁻¹ * 10⁻⁴ A m² = 10⁻⁵ A m². This can also be written as 1.00 × 10⁻⁵ A m².
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