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(C) The magnetic field inside a long solenoid is given by $B = \frac{\mu_0 NI}{L}$.Magnetic flux $\phi$ through the cross-sectional area $A$ is $\phi

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(C) The magnetic field inside a long solenoid is given by $B = \frac{\mu_0 NI}{L}$.Magnetic flux $\phi$ through the cross-sectional area $A$ is $\phi = BA = \frac{\mu_0 NIA}{L}$.The magnetic moment $M$ of the solenoid is defined as $M = NIA$.Substituting $M$ into the flux equation: $\phi = \frac{\mu_0 M}{L}$.Rearranging to solve for $M$: $M = \frac{\phi L}{\mu_0}$.Given $\phi = 1.57 	imes 10^{-6} 	ext{ Wb}$, $L = 0.8 	ext{ m}$, and $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$.$M = \frac{1.57 	imes 10^{-6} 	imes 0.8}{4 	imes 3.14 	imes 10^{-7}}$.$M = \frac{1.57 	imes 0.8 	imes 10^{-6}}{12.56 	imes 10^{-7}} = \frac{1.256 	imes 10^{-6}}{1.256 	imes 10^{-6}} = 1 	ext{ Am}^2$.

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