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(A) Given: Number of turns per unit length $ n = 40 	ext{ turns/cm} = 4000 	ext{ turns/m} = 4 	imes 10^3 	ext{ m}^{-1} $.Current $ I = 1 	ext{ A}

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(A) Given: Number of turns per unit length $ n = 40 	ext{ turns/cm} = 4000 	ext{ turns/m} = 4 	imes 10^3 	ext{ m}^{-1} $.Current $ I = 1 	ext{ A} $.The magnetic energy stored per unit volume $ u_m $ is given by the formula $ u_m = \frac{1}{2} \frac{B^2}{\mu_0} = \frac{1}{2} \mu_0 n^2 I^2 $.Substituting the values:$ u_m = \frac{1}{2} 	imes (4\pi 	imes 10^{-7} 	ext{ T m/A}) 	imes (4 	imes 10^3 	ext{ m}^{-1})^2 	imes (1 	ext{ A})^2 $.$ u_m = \frac{1}{2} 	imes 4\pi 	imes 10^{-7} 	imes 16 	imes 10^6 	imes 1 $.$ u_m = 2\pi 	imes 16 	imes 10^{-1} $.$ u_m = 32\pi 	imes 0.1 = 3.2\pi 	ext{ J m}^{-3} $.

$A$ long solenoid with $ 40 $ turns per cm carries a current of $ 1 \,A $. The magnetic energy stored per unit volume is $ J m^{-3} $. (in $\pi$)

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