A small 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 α×10−14J. The value of α is ____. (μ0=4π×10−7Tm/A,π=3.14)
(A)6.28✓ Correct
(B)6.28×10−6
(C)628
(D)6.28×10−4
Answer: (A)
Solution
The field at the centre of the loop is B=2Rμ0i=2×0.14π×10−7×2=4π×10−6 T, uniform over the tiny cube. The stored energy is the energy density times the cube volume: U=2μ0B2V=2×4π×10−7(4π×10−6)2×(10−3)3=2π×10−14 J. Hence α=6.28.
JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option
