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(A) The magnetic field at the centre of a circular loop is given by $B = \frac{\mu_0 I}{2R}$.Substituting the given values: $B = \frac{4\pi 	imes 10^

Vedclass · Accepted Answer

$6.28$

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(A) The magnetic field at the centre of a circular loop is given by $B = \frac{\mu_0 I}{2R}$.Substituting the given values: $B = \frac{4\pi 	imes 10^{-7} 	imes 2}{2 	imes 0.1} = 4\pi 	imes 10^{-6} 	ext{ T}$.The magnetic energy density $u$ is given by $u = \frac{B^2}{2\mu_0}$.$u = \frac{(4\pi 	imes 10^{-6})^2}{2 	imes 4\pi 	imes 10^{-7}} = \frac{16\pi^2 	imes 10^{-12}}{8\pi 	imes 10^{-7}} = 2\pi 	imes 10^{-5} 	ext{ J/m}^3$.The volume of the cube $V$ is $(1 	ext{ mm})^3 = (10^{-3} 	ext{ m})^3 = 10^{-9} 	ext{ m}^3$.The magnetic energy stored $U$ is $U = u 	imes V$.$U = (2\pi 	imes 10^{-5}) 	imes 10^{-9} = 2\pi 	imes 10^{-14} 	ext{ J}$.Using $\pi = 3.14$, $U = 2 	imes 3.14 	imes 10^{-14} = 6.28 	imes 10^{-14} 	ext{ J}$.Comparing this with $\alpha 	imes 10^{-14} 	ext{ J}$, we get $\alpha = 6.28$.

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