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

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$125$

Vedclass · Answer

(C) The magnetic field $B$ inside a solenoid is given by $B = \mu_r \mu_0 n I$,where $n = N/\ell$ is the number of turns per unit length.Given:Area $A = 2\,cm^2 = 2 	imes 10^{-4}\,m^2$Length $\ell = 40\,cm = 0.4\,m$Number of turns $N = 400$Current $I = 0.4\,A$Total magnetic flux $\phi = 4\pi 	imes 10^{-6}\,Wb$Permeability of vacuum $\mu_0 = 4\pi 	imes 10^{-7}\,T\cdot m/A$The formula for magnetic flux is $\phi = B \cdot A = (\mu_r \mu_0 \frac{N}{\ell} I) A$.Substituting the values:$4\pi 	imes 10^{-6} = \mu_r 	imes (4\pi 	imes 10^{-7}) 	imes (\frac{400}{0.4}) 	imes 0.4 	imes (2 	imes 10^{-4})$$4\pi 	imes 10^{-6} = \mu_r 	imes (4\pi 	imes 10^{-7}) 	imes 400 	imes (2 	imes 10^{-4})$$10^{-6} = \mu_r 	imes 10^{-7} 	imes 400 	imes 2 	imes 10^{-4}$$10^{-6} = \mu_r 	imes 10^{-7} 	imes 8 	imes 10^{-2}$$10^{-6} = \mu_r 	imes 8 	imes 10^{-9}$$\mu_r = \frac{10^{-6}}{8 	imes 10^{-9}} = \frac{1000}{8} = 125$.

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