Q42. A cube of side 1 mm is placed at the centre of a circular current loop of radius 10 cm carrying a current of 2 A. Assuming the magnetic field is effectively uniform over the small cube, the magnetic energy stored inside the cube is α×10⁻¹⁴ J. Taking μ₀ = 4π×10⁻⁷ T·m·A⁻¹ and π = 3.14, what is α?
Options

Question

Q42. A cube of side 1 mm is placed at the centre of a circular current loop of radius 10 cm carrying a current of 2 A. Assuming the magnetic field is effectively uniform over the small cube, the magnetic energy stored inside the cube is α×10⁻¹⁴ J. Taking μ₀ = 4π×10⁻⁷ T·m·A⁻¹ and π = 3.14, what is α?

Options

G1 Physics · Accepted Answer

The magnetic field at the centre of a circular loop is B = μ₀I ÷ (2R). Here, I = 2 A, R = 0.10 m. Therefore, B = (4π×10⁻⁷)(2) ÷ (0.20) = 4π×10⁻⁶ T. Magnetic energy density is u = B² ÷ (2μ₀). The cube volume is V = (1 mm)³ = (10⁻³)³ = 10⁻⁹ m³. Hence the magnetic energy is U = [B² ÷ (2μ₀)]V = 2π×10⁻¹⁴ J ≈ 6.28×10⁻¹⁴ J. Thus α = 6.28. Correct option: A.