A solid cube of side 1 mm is placed at the centre of a circular loop of radius 10 cm carrying a current of 2 A.
The magnetic energy stored inside the cube is:
a × 10−14 J
Find the value of a.
Given:
μ0 = 4π × 10−7 Tm/A
π = 3.14
Magnetic field at centre of circular loop:
B = μ₀I/2R
= (4π×10−7)(2) / (2×0.1)
= 4π×10−6 T
Magnetic energy density:
u = B²/(2μ₀)
Substituting values:
u ≈ 62.8 × 10−6 J/m³
Volume of cube:
= (10−3)³
= 10−9 m³
Energy stored:
U = uV
= 6.28 × 10−14 J
Hence:
a = 6.28
Final Answer
6.28