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

Question

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 $α \times 10^{- 14} J$ . The value of $α$ is ______. $(μ_{0} = 4 π \times 10^{- 7} Tm/A, π = 3.14)$ [2026]

Smart Achievers · Accepted Answer

(1)

$Magnetic field at centre, B = \frac{μ_{0} N I}{2 R} = \frac{μ_{0} I}{2 R} (For N = 1)$

$So energy stored, U = \frac{B^{2}}{2 μ_{0}} \times V$

$U = \frac{1}{2 μ_{0}} \times \frac{μ_{0}^{2} I^{2}}{4 R^{2}} \times {(side)}^{3} = \frac{μ_{0} I^{2}}{8 R^{2}} \times {(10^{- 3})}^{3}$

$U = \frac{4 π \times 10^{- 7} \times 4}{8 \times 10^{- 2}} \times 10^{- 9} = 6.28 \times 10^{- 14} J$

Solution

Q.
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 $α \times 10^{- 14} J$ . The value of $α$ is ______. $(μ_{0} = 4 π \times 10^{- 7} Tm/A, π = 3.14)$ [2026]

1 $6.28$

2 $6.28 \times 10^{- 6}$

3 $628$

4 $6.28 \times 10^{- 4}$

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Solution

Q. 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 J. The value of α is ______. (μ0=4π×10-7 Tm/A, π=3.14) [2026]

1 6.28

2 6.28×10-6

3 628

4 6.28×10-4