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(C) The magnetic field inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length.Given: $n = 50\,turns/cm =

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$44$

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(C) The magnetic field inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length.Given: $n = 50\,turns/cm = 5000\,turns/m$,$I = I_0 \sin(\omega t)$ with $I_0 = 2.5\,A$ and $\omega = 700\,rad\,s^{-1}$.The area of the square loop is $A = (2.0\,cm)^2 = (0.02\,m)^2 = 4 	imes 10^{-4}\,m^2$.The magnetic flux through the loop is $\Phi = B \cdot A = \mu_0 n I A$.The induced emf is $\varepsilon = -\frac{d\Phi}{dt} = -\mu_0 n A \frac{dI}{dt} = -\mu_0 n A I_0 \omega \cos(\omega t)$.The amplitude of the induced emf is $\varepsilon_0 = \mu_0 n A I_0 \omega$.Substituting the values:$\varepsilon_0 = (4\pi 	imes 10^{-7}) 	imes (5000) 	imes (4 	imes 10^{-4}) 	imes (2.5) 	imes (700)$.$\varepsilon_0 = 4 	imes \frac{22}{7} 	imes 10^{-7} 	imes 5 	imes 10^3 	imes 4 	imes 10^{-4} 	imes 2.5 	imes 700$.$\varepsilon_0 = 4 	imes 22 	imes 10^{-7} 	imes 5 	imes 10^3 	imes 4 	imes 10^{-4} 	imes 2.5 	imes 100$.$\varepsilon_0 = 44 	imes 10^{-4}\,V$.Thus,$x = 44$.

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