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 $\alpha \times 10^{- 14}\text{ }J$. The value of $\alpha$ is $\_\_\_\_$ . ( $\mu_{0} = 4\pi \times 10^{- 7}Tm/A,\pi = 3.14$ )
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Solution
Sol. [Key Concept: Field at the loop centre $B = \frac{\mu_{0}}{4\pi}\frac{2\pi\ i}{R}$; the stored energy in the small cube is $U = \frac{B^{2}}{2\mu_{0}} \times (cube\ volume)$.]
Magnetic field at center $B = \left( \frac{\mu_{0}}{4\pi} \right)\frac{2\pi i}{R}$
$$= \frac{10^{- 7} \times 2 \times \pi \times 2}{10 \times 10^{- 2}} $$So energy stored
$${U = \frac{B^{2}}{2\mu_{0}} \times V }{= \frac{10^{- 14} \times 4 \times \pi^{2} \times 4 \times 100}{2 \times 4\pi \times 10^{- 7}} \times \left( 1 \times 10^{- 9} \right) }{= 2\pi \times 10^{- 14}\text{ }J }{= 6.28 \times 10^{- 14}\text{ }J }$$
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