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JEE Main 2026 · 6 Apr · Shift 1, Physics question 42

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×10−14\alpha\times10^{-14} J. The value of α\alpha is ________.

(μo=4π×10−7\mu_o=4\pi\times10^{-7} Tm/A, π=3.14\pi=3.14)

Choose one option
Answer and worked solution

Official answer: A

Field at the centre.
B=μ0I2r=4π×10−7×20.2=1.256×10−5 T.B=\frac{\mu_0I}{2r}=\frac{4\pi\times10^{-7}\times2}{0.2}=1.256\times10^{-5}\text{ T}.

Energy density.
u=B22μ0=6.28×10−5 J/m3.u=\frac{B^2}{2\mu_0}=6.28\times10^{-5}\text{ J/m}^3.

Energy in the cube (10−910^{-9} m³): 6.28×10−146.28\times10^{-14} J, so α=6.28\alpha=6.28.

(A)

Topics:Energy in an inductorBiot-Savart law
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