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(B) The magnetic field $B$ in a solenoid with a core is given by the formula $B = \mu_r \mu_0 H$, where $H$ is the magnetic intensity.Given values are

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$\frac{1}{16\pi}$

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(B) The magnetic field $B$ in a solenoid with a core is given by the formula $B = \mu_r \mu_0 H$, where $H$ is the magnetic intensity.Given values are $B = 1.0 	ext{ T}$, $\mu_r = 400$, and $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$.Rearranging the formula to solve for $H$, we get $H = \frac{B}{\mu_r \mu_0}$.Substituting the values: $H = \frac{1.0}{400 	imes 4\pi 	imes 10^{-7}} = \frac{1}{1600\pi 	imes 10^{-7}} = \frac{1}{16\pi 	imes 10^{-5}}$.This simplifies to $H = \frac{10^5}{16\pi} = \frac{1}{16\pi} 	imes 10^5 	ext{ A/m}$.Comparing this with the given expression $\alpha 	imes 10^5$, we find $\alpha = \frac{1}{16\pi}$.

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