A solenoid has a core made of material with relative permeability \(400\). The magnetic field produced in the interior of solenoid is \(1.0\) T. The magnetic intensity in SI units is \(\alpha \times 10^5\). The value of \(\alpha\) is ______.
(Free space permeability \(\mu_0=4\pi \times 10^{-7}\) SI units.)

The magnetic field \(B\) inside the solenoid is related to the magnetic intensity \(H\) by the formula:

\(B = \mu H = \mu_0 \mu_r H\)

Rearranging for \(H\), we get:

\(H = \dfrac{B}{\mu_0 \mu_r}\)

Substituting the given values (\(B = 1.0\) T, \(\mu_r = 400\), and \(\mu_0 = 4\pi \times 10^{-7}\) T m/A):

\(H = \dfrac{1.0}{4\pi \times 10^{-7} \times 400}\)

\(H = \dfrac{1.0}{1600\pi \times 10^{-7}}\)

\(H = \dfrac{1}{16\pi \times 10^{-5}}\)

\(H = \dfrac{1}{16\pi} \times 10^5\) A/m

Comparing this with the given expression \(H = \alpha \times 10^5\), we find:

\(\alpha = \dfrac{1}{16\pi}\)

Answer: \(\dfrac{1}{16\pi}\)