JEE Main 2026 Apr 6 Shift 1, Physics Q17: Magnetic Flux

JEE Main2026Apr 6, Shift 1Physics

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 $α \times 1 0^{- 14}$ J. The value of $α$ is ________.

( $μ_{0} = 4 π \times 1 0^{- 7}$ T m/A, $π = 3.14$ )

A
$6.28$
B
$6.28 \times 1 0^{- 6}$
C
$628$
D
$6.28 \times 1 0^{- 4}$

Solution

Magnetic field at the centre of a circular loop: $B = \frac{μ _{0} I}{2 R} = \frac{4 π \times 1 0 ^{- 7} \cdot 2}{2 \cdot 0.1} = 4 π \times 1 0^{- 6}$ T.

Energy density: $u = \frac{B ^{2}}{2 μ _{0}} = \frac{( 4 π \times 1 0 ^{- 6} ) ^{2}}{2 \cdot 4 π \times 1 0 ^{- 7}} = \frac{16 π ^{2} \times 1 0 ^{- 12}}{8 π \times 1 0 ^{- 7}} = 2 π \times 1 0^{- 5}$ J/m $^{3}$ .

Cube volume: $V = (1 0^{- 3})^{3} = 1 0^{- 9}$ m $^{3}$ .

Energy stored: $U = u V = 2 π \times 1 0^{- 14}$ J $= 6.28 \times 1 0^{- 14}$ J.

$∴ α = 6.28$ .

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JEE Main 2026 Apr 6 Shift 1, Physics Q17: Magnetic Flux

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